Category: CBSE Class 12

Here we are providing 1 Mark Questions for Accountancy Class 12 Chapter 4 Reconstitution of Partnership Firm: Retirement/Death of a Partner are the best resource for students which helps in class 12 board exams.

Here we are providing Class 12 Maths Important Extra Questions and Answers Chapter 6 Applications of Derivatives. Class 12 Maths Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Maths Chapter 6 Important Extra Questions Applications of Derivatives

Applications of Derivatives Important Extra Questions Very Short Answer Type

Question 1.
For the curve y = 5x- 2x³, if increases at the rate of 2 units/sec., find the rate of change of the slope of the curve when x = 3. (C.B.S.E. 2017)
Solution:
The given curve is y = 5x – 2x³
∴ \(\frac{d y}{d x}\) = 5 – 6x²
i.e. m = 5 – 6x²,
where ‘m’ is the slope.
∴ \(\frac{d m}{d t}\) = —12x\(\frac{d x}{d t}\) =-12x(2) = -24x
∴ \(\left.\frac{d m}{d t}\right]_{x=3}\)= -24(3) = -72.
Hence, the rate of the change of the slope = -72.

Question 2.
Without using the derivative, show that the function f(x) = 7x – 3 is a strictly increasing function in R. (N.C.E.R.T.)
Solution:
Let x₁ and x₂∈ R.
Now x₁ > x₂
⇒ 7x₁ > 7x₂
⇒ 7x₁ – 3 > 7x₂ – 3
⇒ f(x₁) > f(x₂).
Hence, ‘f ’ is strictly increasing function in R.

Question 3.
Show that function:
f(x) = 4x³ – 18×2 – 27x – 7 is always increasing in R. (C.B.S.E. 2017)
Solution:
We have :f(x) = 4x³ – 18×2 – 27x – 7
∴ f(x) = 12x² – 36x + 27 = 12(x² – 3x) + 27
= 12(x² – 3x + 9/4) + 27 – 27
= 12(x – 3/2)²∀ x∈ R.
Hence, f(x) is always increasing in R.

Question 4.
Find the slope of the tangent to the curve:
x = at²,y = 2at at t = 2.
Solution:
The given curve is x – at², y = 2at.
∴ \(\frac{d x}{d t}\) = 2at
\(\frac{d x}{d t}\)= 2a
∴ \(\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{2 a}{2 a t}=\)
Hence, slope of the tangent at t = 2 is: \(\left.\frac{d y}{d x}\right]_{t=2}=\frac{1}{2}\)

Question 5.
Find the maximum and minimum values, if any, of the following functions without using derivatives:
(i) f(x) = (2x-1)² + 3
(ii) f(x)= 16x² – 16x + 28
(iii) f(x) = -|x+ 1| + 3
(iv) f(x) = sin 2x + 5
(v) f(x) = sin (sin x).
Solution:
(i) We have :
f(x) = (2x – 1)² + 3.
Here D_f = R.
Now f(x) ≥ 3.
[∵ (2x – 1)² ≥ 0 for all x ∈ R]
However, maximum value does not exist.
[∵ f(x) can be made as large as we please]

(ii) We have :
f(x) = 16x² – 16x + 28.
Here D_f = R.
Now f(x) = 16 (x² – x + \(\frac{1}{4}\) + 24
= (16(x – \(\frac{1}{2}\) )² + 24
⇒ f(x) ≥ 24.
[∵ 16(x – \(\frac{1}{2}\) )² ≥ 0 for all x ∈ R
Hence, the minimum value is 24.
However, maximum value does not exist.
[ ∵ f(x) can be made as large as we please]

(iii) We have :
f(x) = – 1x + 11 + 3
⇒ f(x) ≤ 3.
[ ∵ -|x + 1| ≤ 0]
Hence, the maximum value = 3.
However, the minimum value does not exist.
[∵ f(x) can be made as small as we please]

(iv) We have :
f(x) = sin2x + 5.
Since – 1 ≤ sin 2x ≤ 1 for all x ∈ R,
– 1+5 ≤ sin2x + 5 ≤ 1+5 for all x∈ R
⇒ 4 ≤ sin2x + 5 ≤ 6 for all x ∈ R
⇒ 4 ≤ f(x) ≤ 6 for all x ∈ R.
Hence, the maximum value = 6 and minimum value = 4.

(v) We have :
f(x) = sin (sin x).
We know that – 1 ≤ sin x ≤ 1 for all x ∈ R
⇒ sin(-1) ≤ sin(sinx) ≤ sin 1 for all x ∈ R
⇒ – sin 1 ≤ f(x) ≤ sin 1.
Hence, maximum value = sin 1 and minimum value = -sin 1.

Question 6.
A particle moves along the curve x² = 2y. At what point, ordinate increases at die same rate as abscissa increases? (C.B.S.E. Sample Paper 2019-20)
Sol. The given curve is x² = 2y …(1)
Diff.w.r.t.t, 2x\(\frac{d x}{d t}\) = 2 \(\frac{d y}{d t}\)
⇒ 2x\(\frac{d x}{d t}\) = 2 \(\frac{d x}{d t}\)
∵ \(\frac{d y}{d t}=\frac{d x}{d t}\) given
From(1), 1 = 2y ⇒ y = \(\frac{1}{2}\)
Hence, the reqd. point is (1, \(\frac{1}{2}\) )

Applications of Derivatives Important Extra Questions Long Answer Type 1

Question 1.
A ladder 13 m long is leaning against a vertical wall. The bottom of the ladder is dragged away from the wall along the ground at the rate of 2 cm/sec. How fast is the height on the wall decreasing when the foot of the ladder is 5 m away from the wall? (C.B.S.E. Outside Delhi 2019)
Solution:
Here, \(\frac{d x}{d t}\) = 2 cm/sec.

Hence, the height is decreasing at the rate of 5/6 cm/sec.

Question 2.
Find the angle of intersection of the curves x² + y² = 4 and (x – 2)² + y²= 4, at the point in the first quadrant (C.B.S.E. 2018 C)
Solution:
The given curves are :
x² + y² = 4 ………….(1)
(x – 2)² + y² = 4 ………….. (2)

From (2),
y = 4 – (x – 2)²
Putting in (1),
x² + 4-(x-2)² = 4
⇒ x² – (x – 2)² = 0
⇒ (x + x-2)(x-x + 2) = 0
⇒ (2x – 2)(2) = 0
⇒ x = 1.
Putting in (1),
1 + y² = 4
⇒ y = √3
∴ Point of intersection = (1, √3 )

Question 3.
Find the intervals in which the function: f(x) = – 2x³ – 9x² – 12x + 1 is (i) Strictly increasing
(ii) Strictly decreasing. (C.B.S.E. 2018 C)
Solution:
Given function is:
f(x) = – 2x³ – 9x² – 12x + 1.
Diff. w.r.t. x,
f'(x) = -6x² – 18x – 12
= -6(x + 1) (x + 2).

Now, f'(x) – 0
⇒ x = -2, x = -1
⇒ Intervals are (-∞ – 2), (-2, -1) and (-1, ∞).
Getting f’ (x) > 0 in (-2, -1)
and f'(x) < 0 in (-∞, -2) u (-1, ∞)
⇒ f(x) is strictly increasing in (-2, -1) and strictly decreasing in (-∞, 2) u (-1, ∞).

Question 4.
A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 metres. Find the dimensions of the window to admit maximum light through the whole opening. (C.B.S.E. 2018 C)
Solution:
Let ‘x’ and ‘y’ be the length and breadth of the rectangle ABCD.
Radius of the semi-circle = \(\frac { x }{ 2 }\) .
Circumference of the semi-circle = \(\frac{\pi x}{2}\)

For Max ./Min. of A (x), A’ (x) = 0
\(\frac{20-(2+\pi)(2 x)}{4}+\frac{2 \pi x}{8}=0\)
20-(2 + π)(2x) + πx = 0
20 + x(π – 4 – 2π) = 0
20 – x(4 + π) = 0
x = \(\frac{20}{4+\pi}\)

And radius of semi-circle = \(\frac{10}{4+\pi}\)

Question 5.
The length ‘x’ of a rectangle is decreasing at the rate of 3 cm/m and width ‘y’ is increasing at the rate of 2 cm/m. When x = 10 cm and y = 6 cm, find the rate of change of:
(a) the perimeter and
(b) the area of the rectangle. (N. C.E.R.T.)
Solution:
We have: \(\frac{d x}{d t}\) = -3 cm/m …(1)
and \(\frac{d y}{d t}\) = 2 cm/m …(2)

a) Perimeter, p = 2x + 2y.
∴ \(\frac{d p}{d t}=2 \frac{d x}{d t}+2 \frac{d y}{d t}\)
= 2(-3) + 2(2)
[Using (1) and (2)]
= – 6 + 4 = -2.
Hence \(\left.\frac{d p}{d t}\right]_{x=10 \atop y=6}\) = -2cm/m.

(b) Area, A = xy.
∴ \(\frac{d \mathrm{~A}}{d t}=x \frac{d y}{d t}+y \frac{d x}{d t}\) = x(2) + y(-3)
[Using (1) and (2)]
Hence \(\left.\frac{d p}{d t}\right]_{x=10 \atop y=6}\) = 10(2) + 6(-3)
= 20-18
= 2cm²/m.

Question 6.
A stone is dropped into a quiet lake and waves move in circles at a speed of 4 cm per second. At the instant when the radius of the circular wave is 10 cm, how fast is the enclosed area increasing ? (N.C.E.R.T.)
Solution:
Let ‘r’ be the radius of the circular wave.
Then A = π r², where A is the enclosed area at time t.
Differentiating w.r.t. t, we have:
\(\frac{d \mathrm{~A}}{d t}=2 \pi r \frac{d r}{d t}=2 \pi r(4)\) [∵ \(\frac{d r}{d t}\) = 4cm/s]
= 8πr.
When r = 10cm, \(\frac{d A}{d t}\)= 8K (10) = 80π cm²/s.
Hence, the enclosed area is increasing at the rate of 80π cm²/s.. when r = 10 cm.

Question 7.
Find the intervals in which the function f(x) is (i) strictly increasing (ii) strictly decreasing:
f(x) = x³ – 12x² + 36x + 17. (C.B.S.E. 2009 C)
Solution:
We have :
f(x) = x³ – 12x² + 36x + 17 – 12x² + 36x + 17.
∴ f'{x) = 3x² – 24x + 36.

(i) For f(x) to be strictly increasing function of x: f'(x) > 0
⇒ 3x² – 24x + 36 > 0
⇒ x² – 8x+ 12 > 0
⇒ (x – 2) (x – 6) > 0
⇒ x < 2 or x > 6.
Hence, f(x) is increasing in the interval (-∞ 2) ∪ (6, ∞).

(ii) For f (x) to be strictly decreasing function of x:
f'(x) <0 ⇒  3x² – 24x + 36 < 0 ⇒ x² – 8x + 12 < 0 ⇒ (x – 2) (x – 6) < 0 ⇒ 2 < x < 6.
Hence, f(x) is decreasing in the interval (2,6).

Question 8.
Find the intervals in which the function :
f(x) = 3x⁴ – 4x³ – 12x² + 5 is:
(a) strictly increasing
(b) strictly decreasing. (C.B.S.E. 2014)
Solution:
We have: f(x) = 3x⁴ – 4x³ – 12x² + 5.
∴ f'(x) = 12x³ – 12x² – 24x
= 12x(x² – x – 2)
= 12x(x – 2)(x + 1).
When x < – 1.
f”(x) = 12 (-ve) (-ve) (-ve) = -ve.
Thus f(x) is strictly decreasing.
When -1 < x < 0.
f'(x) = 12(-ve)(-ve)(+ve) = +ve.
Thus f(x) is strictly increasing.

When 0 < x < 2. f'(x) = 12(+ve)(-ve)(+ve)=-ve. Thus f(x) is strictly decreasing. When x > 2.
f'(x) – 12(+ve)(+ve)(+ve) = +ve.
Thus f(x) is strictly increasing.
Combining,/(x) is
0a) strictly increasing in (-1,0) u (2, ∞) and
(b) strictly decreasing in (-∞, -1) u (0,2).

Question 9.
Find the point at which the tangent to the curve
y = \(\sqrt{4 x-3}\) 1 has its slope \(\frac{2}{3}\) (N.C.E.R.T.)
Solution:
The given curve is :

By the question,
slope = \(\frac{2}{3}\)
\(\frac{2}{\sqrt{4 x-3}}=\frac{2}{3}\)
⇒ \(\sqrt{4 x-3}\) = 3.
Squaring, 4x – 3 = 9
⇒ 4x = 12
⇒ x = 3.
Putting in (1),
y = \(\sqrt{4(3)-3}=\sqrt{9}-1\)
= 3-1 = 2.
Hence, the reqd. point is (3,2).

Question 10.
Find the equations of all lines having slope 2 and being tangents to the curve:
\(y+\frac{2}{x-3}=0\) (N.C.E.R.T)
Solution:
The given curve is y + \(y+\frac{2}{x-3}=0\) = 0
y = \(-\frac{2}{x-3}\)
∴ \(\frac{d y}{d x}=\frac{2}{(x-3)^{2}}\)
By the question, \(\frac{2}{(x-3)^{2}}\) = 2
⇒ (x-3)² = 1
x – 3 = ± 1
⇒ x = 2, 4.

When x = 2, then from (1),
y = \(\frac{-2}{2-3}=\frac{-2}{-1}\)
= 2.
When x = 4, then from (1),
y = \(\frac{-2}{4-3}=\frac{-2}{1}\) = 2

Thus there are two tangents to the given curve with slope 2 and passing through (2, 2) and (4,-2).
∴ The equation of the tangent through (2,2) is: y-2 = 2(x-2)
⇒ y-2x + 2 = 0
and the equation of the tangent through (4,-2) is:
y-(-2) = 2(x-4)
⇒ y-2x+ 10 = 0.

Question 11.
Find the equation of the tangent to the curve y = \(\sqrt{3 x-2}\) which is parallel to the line 4x – 2y + 5 = 0. Also, write the equation of normal to the curve at the point of contact.
(C.B.S.E. 2019)
Solution:
(i) The given curve is y = \(\sqrt{3 x-2}\) …(1)
∴ Slope of the tangent, \(\frac{d y}{d x}=\frac{3}{2 \sqrt{3 x-2}}\)
Slope of the given line 4x – 2y + 5 = 0 is \(-\frac{4}{-2}\) = 2
Since the tangent is parellel to (2)
∴ \(\frac{3}{2 \sqrt{3 x-2}}\) = 2 (∵ m₁ = m₂)
3 = 4\(\sqrt{3 x-2}\)
9 = 16(3x – 2)
= 48x = 41
x = \(\frac{41}{48}\)

⇒ 6(4y-3) = 48x-41
⇒ 48x-24y = 23.

(ii) Slope of normal x Slope of tangent = – 1
⇒ Slope of normal = – 1/2
∴ The equation of the normal is :
\(y-\frac{3}{4}=-\frac{1}{2}\left(x-\frac{41}{48}\right)\)

Question 12.
Find the equations of the tangent and the normal to the curve 16x² + 9y² = 145 at the point (x₁, x₂ where x₁ = 2 and y₁ > 0. (C.B.S.E. 2018)
Solution:
The given curve is :
16x² + 9y² = 145 …(1)
Since (x_1s, y₁,) lies on (1),
.-. 16x²+9y] = 145
⇒ 16(2)² + 9y₁² = 145
⇒ 9y₁² = 145-64
⇒ 9y₁² = 81
⇒ y₁² =9
y₁ = 3 [∵ y₁ > 0]
Thus, the point is (2, 3).
Diff (1) w.r.t x ,

(i) The equation of tangent is :
y – 3 = \(\frac{-32}{27}\) (x-2)
⇒ 27y – 81 = -32x + 64
⇒ 32x + 27y = 145

(ii) The equation of normal is :
y – 3 = \(\frac{27}{32}\)(x – 2)
⇒ 32y – 96 = 27x – 54
⇒ 27x – 32y + 42 =0.

Question 13.
Find the equations ofthe tangent and normal to the curve given by:
x = a sin³ θ, y = a cos³θ at a point, where θ = \(\frac{\pi}{4}\) (1C.B.S.E. 2014)
Solution:
The given curve is:
x = a sin³ θ, y = a cos³θ

∴ Slope of the tangent at θ = \(\frac{\pi}{4}\)

and slope of the normal = 1.
When θ = \(\frac{\pi}{4}\), then from (1),

∴ The equations of the tangent and normal to (1) at θ = \(\frac{\pi}{4}\) are :

Question 14.
Find the equation of the tangent to the curve
ay² = x³ at the point (am², am³) (C.B.S.E. 2019 C)
Solution:
The given curve is ay² = x³
y = a^-1/23/2

∴ The equation of the tangent is:
y – y’ = \(\frac{d y}{d x}\) (x – x’)
y – am³ = \(\frac{3 m}{2}\)(x – am²)
2y – 2 am³ = 3 mx – 3 am³
3mx – 2y = am³.

Question 15.
Find the equation of the tangent to the curve x² + 3y = 3, which is parallel to the line y – 4x + 5 = 0. (C.B.S.E. 2009 C)
Solution:
The given curve is x² + 3y = 3
y = \(\frac{1}{3}\) (3 – x²
∴ \(\frac{d y}{d x}=\frac{1}{3}(0-2 x)=-\frac{2}{3} x\)
which is the slope of the tangent.
But the tangent is parallel to the line
y – 4x + 5 = 0, whose slope is 4.
Thus \(-\frac{2}{3}\)x = 4 [∵ m₁ = m₂]
⇒ x = -6.
From(l), y = \(\frac{1}{3}\) (3 – 36) = -11.
Thus the point of contact is (- 6, – 11).
The equation of the tangent is :
y + 11 = 4(x + 6)
⇒ y + 11 = 4x + 24
⇒ 4x – y+ 13 = 0.

Question 16.
Find the equations of the normal to the curve y = 4x³ – 3x + 5, which are perpendicular to the line 9x – y + 5 = 0.
(C.B.S.E. Sample Paper 2018-19)
Solution:
The given curve is :
y = 4x² – 3x + 5.
Let the required normal be at (x₁ y₁,).
Slope of the tangent = \(\frac{d y}{d x}\) = 12x² – 3.
m₁ = slope of the normal

and m₂ = Slope of the line = 9.
Since, the normal is perpendicular to the lines
m₁ m₂ = -1
\(\frac{-1}{12 x_{1}^{2}-3} \times 9\) = -1
12x₁² = 9
x₁² = 1
x₁ = ±1.

When x₁ = 1, then y₁ = 4-3 + 5 = 6.
When x₁ = -1, then y₁ = -4 + 3 + 5 = 4.
Thus the points are (1, 6) and (-1,4).
∴ The equations of the normal are :
y – 6 = \(-\frac { 1 }{ 9 }\) (x-1) i.e.,x + 9y = 55
and y – 4= \(-\frac { 1 }{ 9 }\) (x+ 1) i.e.,x + 9y = 35.

Question 17.
Using differentials, find the approximate value of \(\sqrt[3]{0.026}\) , upto three places of decimals.
Solution:
Let y = f(x) = x^1/3
Take x = 0.027,
x + Δx = 0.026
Δx = 0.026 – 0.027 = -0.001.
x = 0.027,
y = \(\sqrt[3]{0 \cdot 027}\) =0.3.
Let dx = Δx = -0.001.

Question 18.
If the radius of a sphere is measured as 9 m with an error of 0.03 m, then find the approximate error in calculating its surface area. (N.C.E.R.T.; A.I.C.B.S.E. 2011)
Solution:
Let ‘r’ be the radius of the sphere and Ar be the error in measuring the radius.
Then r = 9m and Δr = 0.03 m.
Now S, the surface area of the sphere is given by:
S = 4πr²
∴ \(\left(\frac{d S}{d r}\right)\) = 8πr
Now dS = \(\left(\frac{d S}{d r}\right)\)Δr = (8πr)Δr
= (8π)(9))(0.03) = 2.16πm²
Hence, the approximate error in calculating the surface area is 2.16πm²

Question 19.
Find the absolute maximum and the absolute minimum value of the function given by:
f(x) = sin²x – cos x, x ∈ [0, π]. (A.I.C.B.S.E. 2015)
Solution:
We have :
f(x) = sin²x – cos x.
∴ f'(x) – 2 sin x cos x + sin x
= sinx(2cosx+ 1).
Now f'(x) 0
⇒ sin x (2 cos x + 1) = 0
⇒ sin x = 0 or cosx = \(-\frac{1}{2}\)
x = 0, π or x = \(\frac{2 \pi}{3}\)
Now f(0) = 0 – 1 = -1

Hence, absolute maximum value is \(\frac{5}{4}\) and absolute minimum values is – 1.

Question 20.
Find all the points of local maxima and local minima of the function ‘f’ given by: f(x) = 2x³ – 6x² + 6x + 5. (N.C.E.R.T.)
We have: f(x) = 2x³ – 6x² + 6x + 5.
f'(x) = 6x² – 12x + 6 = 6(x – 1)².
f”(x) = 12 (x – 1).
Now f'(x) = 0 gives x = 1.
Also, f”(1) = 0.
Thus x = 1 is neither a point of maxima nor of minima
Now f” (x) = 12.
And f'”(x)]_x=1 = 1 = 12 ≠ 0.
Hence, x = 1 is a point of inflexion.

Applications of Derivatives Important Extra Questions Long Answer Type 2

Question 1.
Water is leaking from a conical funnel at the rate of 5 cm Vs. If the radius of the base of funnel is 5 cm and height 10 cm, find the rate at which the water level is dropping when it is 2.5 cm from the top.
Solution:
Here 5 cm is the radius and 10 cm is the height of the conical funnel.
Let ‘r’ be the radius of the base and ‘h’ the height at any stage.

∴ V, volume of water in conical funnel
= \(\frac{1}{3}\) πr²h …………. (1)
Now Δs OBC and OED are similar

∴ From (1), V, volume of water in conical funnel

When the water level is 2.5 cm from the top, then
h = 10 – 2.5 = 7.5 cm.

Hence, the rate at which the water level is 16 .
dropping is \(\frac{16}{45 \pi}\) cm/s.

Question 2.
A man is moving away from a tower 41.6 m high at the rate of 2 m/s. Find the rate at which the angle of elevation of the top of tower is changing when he is at a distance of 30 m from the foot of the tower. Assume that the eye level of the man is 1.6 m from the ground.
Solution:
Let AB (= 41.6 m) be the tower.
Let the man be at a distance of ‘x’ metres from the tower AB at any time t. If ‘0’ be the angle of elevation at time t, then :

⇒ tan θ = \(\frac { 40 }{ x }\)
⇒ x = 40 cot θ …………. (1)
\(\frac{d x}{d t}\) = -40 cosec² θ. \(\frac{d \theta}{d t}\)
⇒ 2 = -40 cosec².θ\(\frac{d \theta}{d t}\) [∵ \(\frac{d x}{d t}\) = 2(given) ]

When x = 30, then from (1),
30 = 40 cot θ
⇒ cot θ = \(\frac{3}{4}\)
so that cosec² θ = 1+ cot\frac{3}{4} = 1 + \(\frac{9}{16}=\frac{25}{16}\)
Putting in (2),

Hence, the angle of elevation of the top of the
tower is reducing at the rate of \(\frac{4}{125}\) radians/s.

Question 3.
Find the intervals in which:
f(x) = sin 3x – cos 3r, 0 < x < π, is strictly increasing or strictly decreasing. (C.B.S.E. 2016)
Solution:
We have : f'(x) = sin 3x – cos 3x.
f'(x) = 3 cos 3x + 3 sin 3x.
Now f(‘x) = 0
⇒ 3 cos 3x + 3 sin 3x = 0
⇒ cos 3x = – sin 3x
⇒ tan 3x = – 1

⇒ f is strictly decreasing.
In \(\left(0, \frac{\pi}{4}\right), f^{\prime}(x)\) < 0 f is strictly increasing. ⇒ f is strictly decreasing. In \(\left(\frac{7 \pi}{12}, \frac{11 \pi}{12}\right)\)f'(x) >0
⇒ f is strictly increasing.
In \(\left(\frac{11 \pi}{12}, \pi\right)\), f'(x) < 0
⇒ f is strictly decreasing.
Hence, ‘f’ is strictly increasing in

Question 4.
Show that the equation of normal at any point ‘t’ on the curve:
x = 3 cos t – cos³t and y = 3 sin t – sin³t is : 4(y cos³ t-x sin³t) = 3 sin 4t. (C.B.S.E. 2016)
Solution:
The given curve is :
x = 3 cos t – cos³t and y = 3 sin t – sin³t.
\(\) = -3 sin t – 3 cos²t (- sin t)
= -3 sin t (1- cos²t)
= -3 sin t sin² t
= -3 sin³t
and \(\frac{d y}{d t}\) = 3 cos t – 3 sin²t cos t
= 3 cos t (1- sin²t)
= 3 cos t cos² t = 3 cos² t.
= \(\frac{3 \cos ^{3} t}{-3 \sin ^{3} t}=-\frac{\cos ^{3} t}{\sin ^{3} t}\)
∴ Slope of normal = \(\frac{\sin ^{3} t}{\cos ^{3} t}\)
∴ The equation of the normal at ‘t’ is: y – (3 sin t – sin³ t)
= \(\frac{\sin ^{3} t}{\cos ^{3} t}\)(x – 3 cos t + cos³ t)
⇒ y cos³ t – 3 sin t cos³ t + sin³ t cos³ t
⇒ x sin³ t – 3 cos t sin³ t + sin³ t cos³ t
⇒ y cos³ t – x sin³t
= 3 sin t cos t (cos³t – sin³t)
⇒ y cos³ t – x sin³ t = \(\frac{3 \sin 2 t \cos 2 t}{2}\)
⇒ y cos³ t – x sin³ t = \(\frac{3}{4}\) sin 4t
⇒ 4(y cos³ t – x sin³ t) = 3 sin 4t, which is true.

Question 5.
A cuboidal shaped godown with square base is to be constructed. Three times as much cost per square metre is incurred for constructing the roof as compared to the walls. Find the dimensions of the godown if it is to enclose a given volume and minimize the cost of constructing the roof and the walls.
(C.B.S.E. Sample Paper 2018-19 C)
Solution:
Let the length and breadth of the base = x and the height of the godown = y.
If C be the cost of construction and V, the volume.
∴ C = k[3x² + 4xy] …(1),
where k > 0 is constant of proportionality
⇒ x2y = V (constant) …(2)
⇒ y = \(\frac{\mathrm{V}}{x^{2}}\) …………. (3)
Putting the value of y from (3) in (1), we get:

Question 6.
A given quantity of metal is to be cast into a solid half circular cylinder (i.e., with rectangular base and semi-circular ends). Show that in order that the total surface area may be minimum, the ratio of the length of the cylinder to the diameter of its circular ends is π: (π + 2). (C.B.S.E. Sample Paper 2019-20)
Solution:
Let ‘r’ be the radius and ‘h’, the height of the cylinder.
∴ V = Volume of half cylinder = \(\frac { 1 }{ 2 }\) πr²h …….(1)

Total surface area, S = S₁ (Surface area of semi¬circular ends) + S₂ (Curved surface of half cylinder) + S₃ (Surface area of rectangular base having height h and width 2r)
( \(\frac { 1 }{ 2 }\)πr² \(\frac { 1 }{ 2 }\)πr²) + \(\frac { 1 }{ 2 }\)(2πrh) + 2rh
= πr² + πrh + 2rh
= πr² + (π + 2)r\(\frac{2 \mathrm{~V}}{\pi r^{2}}\)
[Using (1)]
∴ S = πr² + \(\frac{2 \mathrm{~V}}{\pi r}\) (π+2) ……. (2)
∴ \(\frac{d \mathrm{~S}}{d r}=2 \pi r-\frac{2 \mathrm{~V}}{\pi r^{2}}(\pi+2)\)

= 2π + 4π = 6π
which is +ve
∴ S is minimum when π²r³ = V(π + 2)
π²r³ = \(\frac { 1 }{ 2 }\)πr²(π + 2)
\(\frac{h}{2 r}=\frac{\pi}{\pi+2}\)
Hence, h:2r = π: {π + 2), which is true.

Question 7.
A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m3. If the building of tank costs ₹70 per sq. metre for the base and ₹45 per sq. metre for the sides, what is the cost of least expensive tank? (C.B.S.E. 2019)
Solution:
Let ‘x’ and ‘y’ be the length and breadth respetively of the tank.
And depth of the tank = 2 m.
∴ Volume of the tank = 2xy.

By the question,
2xy = 8
⇒ xy = 4 …(1)
Nowarea of the base = xy
and area of the sides = 2 (x + y) (2) = 4 (x + y)
∴ Cost of construction
= ₹(70xy+ 45 (4(x + y)) …(2)
= ₹(70xy+ 1800 + y))
∴ C, the cost of construction
= 70 (4) +180( x + \(\frac{4}{x}\)) [Using (1)]
= 280 + 180 ( x + \(\frac{4}{x}\))
∴ \(\frac{d \mathrm{C}}{d x}=180\left(1-\frac{4}{x^{2}}\right)=180\left(\frac{x^{2}-4}{x^{2}}\right)\)

For max ./min.,
\(\frac{d \mathrm{C}}{d x}\) = 0
x² – 4 = 0
x = ±2.
x = 2. [∵ Length can’t be -ve]

From(1), y = \(\frac{4}{2},\) = 2.
Thus the tank is a cube of side 2 m.
∴ Least cost of construction
= ₹[280 + 180(2 + \(\frac{4}{2},\) )] [From (3)]
= ₹ (280 + 720) = ₹1000. [From (3)]

Question 8.
An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quanity of water. Show that the cost of the material will be least when the depth of the tank is half of its width. (C.B.S.E. 2018)
Solution:
Let ‘x’ be the side of the square base and V the depth of the tank.
Now, V = x²y …(1)
Now S, surface area = x² + 4xy
∴ C, cost is proportional to surface area = k(x² + 4xy)

Hence, the cost of material least when depth is half of its width.

Question 9.
Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius ‘r’ is \(\frac { 4r }{ 2 }\). Also find the maximum volume of the cone.
(C.B.S.E. 2019 (Delhi))
Solution:
(i) Let ‘x’ be the distance of the base BC from the centre O of the sphere.

∴ Height of the cone = AM = r + x
andradius of the base = BM = \(\sqrt{r^{2}-x^{2}}\).
∴ V, the volume of the cone
= \(\frac{1}{3}\)π(r² – x2)(r+x).
∴ \(\frac{d \mathrm{~V}}{d x}\)= \(\frac{1}{3}\)π(r² – x²) + \(\frac{1}{3}\)π( – 2x)(r + x)
= \(\frac{\pi}{3}\)(r + x)(r – x – 2x)
= \(\frac{\pi}{3}\)(r + x)(r – 3x)
and \(\frac{d^{2} V}{d x^{2}}\) = \(\frac{\pi}{3}\)(r + x)( – 3) + \(\frac{\pi}{3}\)(1) (r – 3x)
= \(\frac{\pi}{3}\)( – 3r – 3x + r – 3x)
= \(\frac{\pi}{3}\)( – 2r – 6x).
For V to be max, \(\frac{d \mathrm{~V}}{d x}\) = 0 and \(\frac{d^{2} \mathrm{~V}}{d x^{2}}\) < 0
Now \(\frac{d \mathrm{~V}}{d x}\) = 0 gives :
\(\frac{\pi}{3}\) (r + x )(r – 3x) = 0
⇒ x = -r, \(\frac{r}{3}\)
but x ≠ -r
x = \(\frac{r}{3}\)
Also \(\left.\frac{d^{2} \mathrm{~V}}{d x^{2}}\right]_{x=\frac{r}{3}}=\frac{\pi}{3}(-2 r-2 r)=-\frac{4}{3} \pi r\)
Which is -ve
Hence, V is max. when x = \(\frac{r}{3}\) i.e., when altitude
r +x = r + \(\frac{r}{3}\) = \(\frac{4r}{3}\)

(ii) Maximum volume of the cone

Question 10.
If the lengths ofthree sides ofa trapezium other than base are equal to 10 cm, then find the area of the trapezium when it is maximum. (N.C.E.R.T.; A.I.C.B.S.E. 2010)
Solution:

Let ABCD be the trapezium such that:
AD = BC = CD = 10 cm.
Draw DL and CM perpendiculars on AB.
Let AL = x cm.
Now ΔALD ≅ Δ BMC.
∴ MB = AL = x cm.
Also LM = 10 cm.
And DL = CM = \(\sqrt{100-x^{2}}\)
Let ‘A’ be the area of the trapezium.
Then A (x) = 1/2 (10 + (10 + 2x)) \(\sqrt{100-x^{2}}\)
[Area of Trap. = \(\frac{1}{2}\) (sum of II sides) {height))
= \(\frac{1}{2}\)(2x + 20) \(\sqrt{100-x^{2}}\)
= (x + 10) \(\sqrt{100-x^{2}}\) ………(1)
A'(x) = (x + 10)

Now A'(0) = 0
⇒ -2x² – 10A+ 100 = 0
⇒ x² + 5x – 50 = 0
= (x+ 10) (x-5) = 0
=» x = -10,5.
But x ≠ – 10.
[ ∵ x, being the distance, can’t be – ve]
Thus A = 5 cm.
And

Thus A (A) is maximum when x = 5.
Hence, A (5) = (5 +10) \(\sqrt{100-25}\)
[Using (1)]
= 15√75 = 75√3 cm².

Question 11.
A metal box with a square base and vertical sides is to contain 1024 cm³. The material for the top and bottom costs₹ 5 per cm² and the material for the sides costs ₹ 2.50 per cm² Find the least cost of the box. (C.B.S.E 2017)
Solution:
Let ‘x’ cm be the side of square base and ‘y’ cm the height.
∴ volume of the box = x x x x y = x²y.
By the question,
x²y = 1024 …(1)
Now the cost, C = 5(x²) + \(\frac { 5 }{ 2 }\) (4xy)
= 5x²+ 10xy
= 5x² + 10x (\(\frac{1024}{x^{2}}\)) …………(2) [Using (1)]

Now for least cost, \(\frac{d \mathrm{C}}{d x}\) = 0
⇒ 10x – \(\frac{10240}{x^{2}}\) = 0
⇒x³ = (1024)^1/3
⇒ x = (1024)^1/3
And \(\frac{d^{2} \mathrm{C}}{d x^{2}}\) > 0 for x = (1024)^1/3
Hence, least cost =5 (1024)^2/3 + \(\frac{10240}{(1024)^{1 / 3}}\)
= 5(1024)^2/3 + 10(1024)^2/3
= ₹ 15(1024)^2/3

Question 12.
Show that the height of the cylinder, open at the top, of given surface area and greatest volume is equal to the radius of its base. (C.B.S.E. 2010, 2019 C)
Solution:
Let ‘r’ and ‘h’ be the radius and height respectively of the cylinder.
∴ S, the surface area = πr² + 2πrh
h = \(\frac{\mathrm{S}-\pi r^{2}}{2 \pi r}\) …………(1)
And V, the volume = πr²h
i.e. V = πr² \(\left(\frac{\mathrm{S}-\pi r^{2}}{2 \pi r}\right)\) [Using(1)]
V = \(\frac{r}{2}\) (S – πr²)
V = \(\frac{1}{2}\)(Sr – πr³)
∴ \(\frac{d \mathrm{~V}}{d r}\) = \(\frac{r}{2}\) (S – 3πr²) …….. (2)
and \(\frac{d^{2} \mathrm{~V}}{d r^{2}}=-\frac{3 \pi}{2}\) (2r) = -3 πr ………(3)
For greatest volume , \(\frac{d \mathrm{~V}}{d r}\) = 0 and \(\frac{d^{2} \mathrm{~V}}{d r^{2}}\) < 0
Now \(\frac{d \mathrm{~V}}{d r}\) = 0
= \(\frac{r}{2}\) (S – 3πr²) = 0
S = 3πr²

Hence, height = radius of the base.

Question 13.
An Apache helicopter of enemy is flying along the curve given by: y = x² + 7.
A soldier, placed at (3,7), wants to shoot down the helicopter when it is nearest to him. Find the nearest distance.
(NCERT; C.B.S.E. 2019 C)
Solution:
Let P (x, x² + 7) be the position of the helicopter at any instant.
Also A is (3,7).

Let f(x) = (x – 3)² + x⁴.
f'(x) = 2(x-3) + 4x³ = 4x³ + 2x – 6
= 2(2x³ + x – 3) = 2(x – 1)(2x² + 2x + 3).
Now f'(x) = 0
x – 1 = 0, or 2x² + 2x+3 = 0
⇒ x = 1.
[∵ Roots of 2x² + 2x + 3 = 0 are not real] Thus, there is only one point x = 1.
And f(1) = (1 -3)²+ 14 = 4 + 1 = 5.
∴ Distance between soldier and helicopter is √5
Now √5 is either max. value or min. value.
Since \(\sqrt{f(0)}=\sqrt{(0-3)^{2}+(0)^{4}}=3>\sqrt{5}\)
⇒ √5 is min. value of \(\sqrt{f(x)}\).
Hence, √5 is the reqd. minimum distance between the soldier and the helicopter.

Question 14.
Show that the triangle of maximum area that can be inscribed in a given circle is an equilateral triangle. (C.B.S.E. Sample Paper 2019-20)
Solution:
Let O be the centre and V, the radius of the circle in which AABC is inscribed.

For maximum area, the vertex A should be at a maximum distance from the base BC
⇒ A must lie on the diameter, which is perp. to BC
⇒ ΔABC is isosceles.
Let ∠BAC = θ. Then ∠BOC = 2θ
⇒ DOC = θ.
Now BC = 2DC = 2OC sin θ = 2rsine…(1)
and AD = AO + OD = AO + OC cos θ
= r + r cos θ = r (1 + cos θ) …(2)
If ‘A’ be the area of the triangle, then:
A = \(\frac { 1 }{ 2 }\) (BC) (AD) = \(\frac { 1 }{ 2 }\) (2r sin θ) r (1 + cos θ)
[Using (1) & (2)]
⇒ A = r² (sin θ + sin θ cos θ).
= r²(cos θ + cos² θ – sin² θ)
= r² (cos²θ + cos θ) = r² (2cos² θ -1 + cos θ)
= r² (2cos ²θ + cos θ – 1) …………. (3)
and \(\frac{d^{2} \mathrm{~A}}{d \theta^{2}}\) = r²(4cox θ sin θ – sin θ) ……(4)
For ‘A’ to be maximum \(\frac{d \mathrm{~A}}{d \theta}\) = 0 and \(\frac{d^{2} \mathrm{~A}}{d \theta^{2}}\) < 0
Now, \(\frac{d \mathrm{~A}}{d \theta}\) = 0
⇒ r² (2 cos² θ + cos θ – 1) = 0
⇒ 2 cos² θ + cos θ – 1 = 0
⇒ (2 cos θ -1) (cos θ + 1) = 0

Thus, A is maximum when θ = \(\frac{\pi}{3}\)
But ΔABC is isosceles.
Hence, for maximum area, triangle is equilateral.

Question 15.
Find the point on the circle x² + y² = 80, which is nearest to the point (1,2). (C.B.S.E. 2019)
Solution:
Let P (√80 cos θ, √80 sin θ)
i.e. (4√5 cosθ, 4√5 sin θ) be a point on the circle x² + y² = 80.
And, A(1, 2) is the given point.
|AP| = \(\sqrt{(4 \sqrt{5} \cos \theta-1)^{2}+(4 \sqrt{5} \sin \theta-2)^{2}}\) ………….. (1)
AP is least⇒ AP² is least
D = AP² = (4√5cos θ – 1)
(4√5 sinθ – 2)² …………. (2)
\(\frac{d D}{d \theta}\)= 2(4√5 cos θ – 1)(- 4√5sin θ)
+ 2 (4θ5sin θ – 2)(4θ5cos θ)
= – 160 sin θ cos θ + 8√5 sin θ + 160 sin θ cos θ – 16√5 cos θ ………… (3)
= 8√5 (sin θ-2 cos θ) ………… (3)
and \(\frac{d^{2} \mathrm{D}}{d \theta^{2}}\) = 8√5 (cos θ + 2 sin θ) …(4)
Now \(\frac{d \mathrm{D}}{d \theta}\) = 0
⇒ 8√5(sin θ – 2 cos θ) = 0
⇒ sin θ – 2 cos θ = 0 tan θ = 2

Thus, D is least when tan θ = 2
Hence, from (1), least distance

Here we are providing Class 12 Maths Important Extra Questions and Answers Chapter 13 Probability. Class 12 Maths Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Maths Chapter 13 Important Extra Questions Probability

Probability Important Extra Questions Very Short Answer Type

Question 1.
If A and B are two independent event, prove that A’ and B are also independent.
(C.B.S.E. Sample Paper 2018-19)
Solution:
Since A and B are independent events, [Given]
.-. P (A ∩ B) = P (A) . P(B) …(1)
Now P(A’∩B) = P (B) – P (A ∩ B)
= P(B) – P (A) P( ∩ B) [Using (1)]
= (1 – P(A)) P(B) = P(A’) P(B).
Hence, A’ and B are independent events.

Question 1.
One card is drawn from a pack of 52 cards so that each card is equally likely to be se-lected. Prove that the following cases are in-dependent :
(a) A : “The card drawn is a spade”
B : “The card drawn is an ace.”
(N.C.E.R.T.)
(b) A : “The card drawn is black”
B : “The card drawn is a king.”
(.N.C.E.R.T.)
Solution:
(a) P(A) =\(\frac{13}{52}=\frac{1}{4}\),P(B) = \(\frac{4}{52}=\frac{1}{13}\)
P(A∩B) = \(\frac{1}{52}=\frac{1}{4} \cdot \frac{1}{13}\) = p(A).p(B)
Hence, the events A and B are independent

(b) P(A) = \(\frac{26}{52}=\frac{1}{2}\), P(B) = \(\frac{4}{52}=\frac{1}{13}\)
P(A∩B) = \(\frac{2}{52}=\frac{1}{26}=\frac{1}{2} \cdot \frac{1}{13}\) = P(A).P(B)
Hence, the events A and B are independent

Question 3.
A pair of coins is tossed once. Find the probability of showing at least one head.
Solution:
S, Sample space = {HH, HT, TH, TT}
where H ≡ Head and T ≡ Tail.
∴ P (at least one head) = \(\frac { 3 }{ 4 }\) .

Question 4.
P(A) = 0.6, P(B) = 0. 5 and P(A/B) = 0.3, then find P(A∪ B)
(C.B.S.E. Sample Paper 2018-19)
Solution:
We have: P(A/B) = 0.3
\(\frac{P(A \cap B)}{P(B)}\) = 0.3
\(\frac{P(A \cap B)}{0.5}\) = 0.3
P (A ∩ B) = 0.5 x 0.3 = 0.15.
Now, P(A∪B) = P(A) + P(B) – P(A ∩ B)
= 0.6 + 0.5 – 0. 15
Hence, P (A ∪ B) = 1.1 – 0.15 = 0.95.

Question 5.
One bag contains 3 red and 5 black balls. Another bag contains 6 red and 4 black balls. A ball is transferred from first bag to the second bag and then a ball is drawn from the second bag. Find the probability that the ball drawn is red. (C.B.S.E. Sample Paper 2018-19)
Solution:
P(Red transferred and red drawn or black trans¬ferred red drawn)
\(\begin{array}{l}
=\frac{3}{8} \times \frac{7}{11}+\frac{5}{8} \times \frac{6}{11} \\
=\frac{21}{88}+\frac{30}{88}=\frac{51}{88}
\end{array}\)

Question 6.
Evaluate P(A ∪ B), if 2P(A) = P(B) = \(\frac { 5 }{ 13 }\) and P(A/B) = \(\frac { 2 }{ 5 }\) (C.B.S.E. 2018 C)
Solution:
P(A/B) = \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}\) P(A∩B) = \(\frac{2}{11}\)
P(A∪B) = P(A) + P(B) – (A ∩ B)
= \(\frac{5}{26}+\frac{5}{13}-\frac{2}{13}=\frac{11}{26}\)

Probability Important Extra Questions Short Answer Type

Question 1.
Given that A and B are two independent events such that P(A) = 0.3 and P(B) = 0.5. Find P(A/B). (C.B.S.E. 2019 C)
Solution:
We have:
P(A)= 0.3 and P(B) = 0.5.
Now P(A ∩ B) = P(A). P(B)
[∵A and B are independent events]
= (0.3) (0.5) = 0.15.
Hence, P(A/B) = \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{0.15}{0.5}\) =0.3.

Question 2.
A bag contains 3 white and 2 red balls, another bag contains 4 white and 3 red balls. One ball is drawn at random from each bag.
Find the probability that the balls drawn are one white and one red.
(C.B.S.E. 2019 C)
Solution:
Reqd. probability
= P(White, Red) + P (Red, White)
\(\frac{3}{5} \times \frac{3}{7}+\frac{2}{5} \times \frac{4}{7}=\frac{9}{35}+\frac{8}{35}=\frac{17}{35}\)

Question 3.
The probabilities of A, B and C solving a problem independently are \(\frac{1}{2}, \frac{1}{3}\) and \(\frac{1}{4}\) respectively. If all the three try to solve the problem independently, find the probability that the problem is solved. (C.B.S.E. 2019 C)
Solution:
Given: P(A) = \(\frac{1}{2}\), P(B) = \(\frac{1}{3}\) and P(C) = \(\frac{1}{4}\)

Probability that the problem is solved
= Probability that the problem is solved by at least one person
= 1 – \(\mathrm{P}(\overline{\mathrm{A}}) \mathrm{P}(\overline{\mathrm{B}}) \mathrm{P}(\overline{\mathrm{C}})\)
= 1 – \(\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}\)= 1 – \(\frac{1}{4}=\frac{3}{4}\)

Question 4.
A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event “number is even” and B be the event “number is marked red”. Find whether the events A and B are independent or not. {Delhi 2019)
Solution:
Here, A : number is even i.e.,
A = {2,4,6}
and B : number is red i.e.,
B = {1,2,3}
∴ P(A) = \(\frac{3}{6}=\frac{1}{2}\) and P(B) = \(\frac{3}{6}=\frac{1}{2}\)
And,
P(A ∩ B) = P(Number is even and red) = \(\frac { 1 }{ 6 }\) .
Thus, P(A ∩ B) ≠ P(A). P(B)
[∵ \(\frac{1}{6} \neq \frac{1}{2} \times \frac{1}{2}\) ]
Hence, the events A and B are not indepedent.

Question 5.
A die is thrown 6 times. If “getting an odd number” is a success, what is the probability of (i) 5 successes (ii) at most 5 successes? (Delhi 2019)
Solution:
Probability of getting an odd number is one 3 1
trial = \(\frac{3}{6}=\frac{1}{2}\) = p ( say)
Probability of getting an even number in one 3
trial = \(\frac{3}{6}=\frac{1}{2}\) = g(say) o l
Also, n = 6.

(i) P(5 successes) = P(5) = ⁶C₅ q¹ p⁵
(ii) P(at most 5 successes)
= P(0) + P(1) + … + P(5) = 1 – P(6)
= 1 – ⁶C₆ q⁰ p⁶
= 1 – \(\frac{1}{64}=\frac{63}{64}\)

Question 6.
The random variable ‘X’ has a probability distribution P(X) of the following form, where ‘k’ is some number :
\(\mathbf{P}(\mathbf{X}=\boldsymbol{x})=\left\{\begin{array}{l}
\boldsymbol{k}, \text { if } \boldsymbol{x}=\mathbf{0} \\
2 k, \text { if } x=1 \\
3 k, \text { if } x=2 \\
0, \text { otherwise. }
\end{array}\right.\)
Determine the value of ‘P. (Outside Delhi 2019)
Solution:
We have : P(X = 0) + P(X = 1) + P(X = 2) = 1
⇒ k + 2k + 3k = 1
⇒ 6k = 1.
Hence, k = \(\frac { 1 }{ 6 }\).

Question 7.
Out of 8 outstanding students of a school, in which there are 3 boys and 5 girls, a team of 4 students is to be selected for a quiz competiton. Find the probability that 2 boj and 2 girls are selected. (Outside Delhi 2019)
Solution:
Read, probability = \(\frac{{ }^{3} \mathrm{C}_{2} \times{ }^{5} \mathrm{C}_{2}}{{ }^{8} \mathrm{C}_{4}}\)
= \(\frac{3 \times 10}{70}=\frac{3}{7}\)

Question 8.
12 cards numbered 1 to 12 (one number on one card), are placed in a box and mixed up thoroughly. Then a card is drawn at ran¬dom from the box. If it is known that the number on the drawn card is greater than 5, find the probability that the card bears an odd number. {Outside Delhi 2019)
Solution:
Let the events be as :
A : Card bears an odd number.
B : Number on the card is greater than 5.
A∩B = {7, 9, 11}.
Hence, P(A/B) = \(\frac{P(A \cap B)}{P(B)}\)
= \(\frac{3 / 12}{7 / 12}=\frac{3}{7}\)

Question 9.
The probability of solving a specific problem independently by A and B are \(\frac { 1 }{ 3 }\) and \(\frac { 1 }{ 5 }\) respectively. If both try to solve the problem independently, find the probability that the problem is solved.
(iOutside Delhi 2019)
Solution:
P (Problem is solved)
= 1 – P (Problem is not solved)
= 1 – \(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\)
= 1 – \(\frac{2}{3} \times \frac{4}{5}\)
= 1 – \(\frac{8}{15}=\frac{7}{15}\)

Probability Important Extra Questions Long Answer Type 1

Question 1.
A black and a red die are rolled together. Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4. (C.B.S.E. 2018)
Solution:
Let the events be as:
E : Sum of numbers is 8
F : Number of red die less than 4.
E : {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
F = {(1, 1), (2, 1), … (6, 1), (1, 2), (2, 2), … (6, 2), (1, 3), (2, 3), … (6, 2) (6, 3)}
and E ∩ F = {(5, 3), (6, 2)}
P(E) = \(\frac{5}{36}\), P(F) = \(\frac{18}{36}\)
and P(E ∩ F) = \(\frac{2}{36}\).
Hence, P(E/F) = \(\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\frac{2 / 36}{18 / 36}\)
= \(\frac{2}{18}=\frac{1}{9}\)

Question 2.
Two numbers are selected at random (with-out replacement) from the first five positive integers. Let X denote the larger of the two numbers obtained. Find the mean and vari-ance of X.
Solution:
The first five positive integers are 1, 2, 3, 4 and 5.
We select two positive numbers in 5 x 4 = 20 way.
Out of three, two numbers are selected at ran-dom.
Let ‘X’ denote the larger of the two numbers.
X can be 2, 3, 4 or 5.
∴ P (X = 2) = P (Larger number is 2)
{(1, 2), (2,1)} = \(\frac{2}{20}\)
Similarly, P(X = 3) = \(\frac{4}{20}\) ,
P(X = 4) = \(\frac{6}{20}\)
and P(X = 5) = \(\frac{8}{20}\)
Hence, the probability distribution is:

Question 3.
The probabilities of two students A and B coming to the school in time are \(\frac{3}{7}\) and \(\frac{5}{7}\) respectively. Assuming that the events, ‘A coming in time’ and ‘B coming in time’ are independent, find the probability of only one of them coming to the school in time. (A.I.C.B.S.E. 2013)
Solution:
We have : P(A) = Probability of student A coming to school in time = \(\frac{3}{7}\)
P(B) = Probability of student B coming to school in time = \(\frac{5}{7}\)
∴ \(\mathrm{P}(\overline{\mathrm{A}})=1-\frac{3}{7}=\frac{4}{7}\)
and \(\mathrm{P}(\overline{\mathrm{B}})=1-\frac{5}{7}=\frac{2}{7}\)
∴ Probability that only one of the students coming to school in time
= P(A ∩ \(\overline{\mathrm{B}}\)) + P( \(\overline{\mathbf{A}}\) ∩B)
= P(A)P(\(\overline{\mathrm{B}}\)) + P(\(\overline{\mathrm{B}}\))PB)
[∵ A and B are independent => A and \(\overline{\mathrm{B}}\) and \(\overline{\mathrm{A}}\) and B are also independent]
= \(\left(\frac{3}{7}\right)\left(\frac{2}{7}\right)+\left(\frac{4}{7}\right)\left(\frac{5}{7}\right)=\frac{26}{49}\)

Question 4.
A speaks truth in 80% cases and B speaks truth in 90% cases. In what percentage of cases are they likely to agree with each other in stating the same fact? (C.B.S.E. Sample Paper 2019-20)
Solution:
P(A) = \(\frac{80}{100}=\frac{4}{5}\)
and P( B) = \(\frac{90}{100}=\frac{9}{10}\)
P(\(\overline{\mathrm{A}}\)) = 1 – P(A) = 1 – \(\frac{4}{5}=\frac{1}{5}\)
P(\(\overline{\mathrm{B}}\)) = 1 – P(B) = 1 – \(\frac{9}{10}=\frac{1}{10}\)
∴ P(Agree) = P(Both speak the truth or both tell a lie)

Hence, the reqd. percentage = 74%.

Question 5.
A problem in Mathematics is given to three students whose chances of solving it are \(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}\). What is the probability in the
following cases?
(i) that the problem is solved
(ii) only one of them solves it correctly
(iii) at least one of them may solve it.
Solution:
Let A, B, C be three events when a problem in
Mathematics is solved by three students.
Given : P(A) = \(\frac{1}{2}\), P(B) = \(\frac{1}{3}\), P(C) = \(\frac{1}{4}\).
.-. P(A) = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\), P(B) = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
and P(C) = 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\).

(i) Probability that the problem is solved = Probability that the problem is solved by at least one student

(ii) Probability that only one solves it correctly = P(Aninc)+P(AnBnc)+P(AnBnc)

(iii) Probability that atleast one of them may solve the problem

Question 6.
In a set of 10 coins, 2 coins with heads on both sides. A coin is selected at random from this set and tossed five times. Of all the five times, the result was head, find the probability that the selected coin had heads on both sides. (A.I. C.B.S.E. 2015)
Solution:
Let the events be :
E₁ : Selecting a coin with two heads
E₂ : Selecting a normal coin and A : The coin falls head all the times.
Since E₁ and E₂ are mutually exclusive and by the data given in the problem, we have :
P(E₁) = \(\frac{2}{10}=\frac{1}{5}\)
P(A/E₂) = \(\frac{8}{10}=\frac{4}{5}\) = P(A/E₁) = 1
P(A/E₁) = 1
P(A/E₂) = \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{32}\)
Now P(A) = P(A ∩ E₁) + P(A ∩ E₂)
= P(E₁) P(A/ E₁) + P(E₂) P(A/ E₂)
= \(\frac{1}{5}+\frac{1}{40}=\frac{8+1}{40}=\frac{9}{40}\)

Question 7.
A person playsa game of tossing a coin thrice. For each head he is given XI by the organiser of the game and for each tail he has to give ₹1.50 to the organiser. Let ‘X’ denote the amount gained or lost by the person. Show that ‘X’ is a random variable and exhibit it as a function on the sample space of the experiment. (N.C.E.R.T.)
Solution:
Since ‘X’ is a number whose values are defined by the outcomes of the random experiment,
∴ ‘X’ is a random variable.
Now sample space is given by :
S = {HHH, HHT, HTH, THH, HIT, THT, TTH, ITT},
where H = Head and T = Tail.
Thus X (HHH) = 2 x 3 = ₹6
X (HHT) = X (HTH) = X (THH)
= 2×2-lxl.5 = ₹ 2.50
X (HTT) = X (THT) = X (TTH)
= lx2-2xl.5 = -₹l
and X (TTT) = -(3xl.5) = – ₹4.50.
Thus for each element of S, X takes a unique value.
∴ ‘X’ is a function on the sample space S having range = {6, 2.50, – 1, – 4.50}.

Question 8.
Two numbers are selected at random (without replacement) from first 7 natural numbers. If X denotes the smaller of the two numbers obtained, find the probability distribution of X. Also, find mean of the distribution.
(C.B.S.E. Sample Paper 2019-20)
Solution:
Let ‘X’ be the smaller of the two numbers obtained.
Thus, X takes values 1, 2, 3, 4, 5 and 6.

(ii) We have :

Question 9.
Two cards are drawn simultaneously (with-out replacement) from a well shuffled pack of 52 cards. Find the mean and variance of the number of red cards. (A.I.C.B.S.E. 2012)
Solution:
Here ‘X’ takes values 0, 1, 2

Question 10.
There is a group of 50 people who are patriotic out of which 20 believe in non-violence. Two persons are selected at random out of them, write the probability distribution for the selected persons who are non-violent. Also find the mean of the distribution.
Solution:
Let ‘X’ be the number of non-violent persons.
Here ‘X’ takes values 0, 1,2.
Thus no. of non-violent persons = 20
and no. of violent persons = 50 – 20 = 30.

Hence, probability distribution is given by :

Question 11.
Two groups are competing for the positions of the Board of Directors of a corporation. The probabilities that the first and second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group. (C.B.S.E. 2018 C)
Solution:
Let E₁ = First group wins, E₂ = Second group wins
H = Introduction of new product.
P(E₁) = 0.6, P(E₂) = 0.4
P(H/E₂) = 0.3 P(H/E₁) = 0.7
Now,
P(E₂/H) = \(\frac{\mathrm{P}\left(\mathrm{E}_{2}\right) \mathrm{P}\left(\mathrm{H} / \mathrm{E}_{2}\right)}{\mathrm{P}\left(\mathrm{E}_{2}\right) \mathrm{P}\left(\mathrm{H} / \mathrm{E}_{2}\right)+\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\mathrm{H} / \mathrm{E}_{1}\right)}\)

Question 12.
From a lot of 20 bulbs which include 5 defectives, a sample of 3 bulbs is drawn at random, one by one with replacement. Find the probability distribution of the number of defective bulbs. Also, find the mean of the distribution. (C.B.S.E. 2018 C)
Solution:
Let X denote the number of defective bulbs.
X = 0, 1, 2, 3.
P(X = 0) = \(\left(\frac{15}{20}\right)^{3}=\frac{27}{64}\)
P(X = 1) = \(3\left(\frac{5}{20}\right)\left(\frac{15}{20}\right)^{2}=\frac{27}{64}\)
P(X = 2) = \(3\left(\frac{5}{20}\right)^{2}\left(\frac{15}{20}\right)=\frac{9}{64}\)
P(X = 3 )=\(\left(\frac{5}{20}\right)^{3}=\frac{1}{64}\)
Mean = ΣXP(X) = \(\frac{27}{64}+\frac{18}{64}+\frac{3}{64}=\frac{3}{4}\)

Probability Important Extra Questions Long Answer Type 2

Question 1.
In a hockey match t wo teams A and B scored same number of goals upto the end of the game, so to decide the winner, the referee asked both the captains to throw a die alternately and decided that the team, whose captain gets a six first, will be declared the winner. If the captain of team A was asked to start, find their respective probabilities of winning the match. (A.I.C.B.S.E. 2013)
Solution:
P(A), probability of A’s getting a six = \(\frac { 1 }{ 6 }\)
\(\mathrm{P}(\overline{\mathrm{A}})\), probability of A’s not getting a six
= 1 – \(\frac{1}{6}=\frac{5}{6}\)
Thus, P(A) = \(\frac{1}{6}\) \(\mathrm{P}(\overline{\mathrm{A}})\) = \(\frac{5}{6}\).
Similarly, P(B) = \(\frac{1}{6}\), \(\mathrm{P}(\overline{\mathrm{B}})\). = \(\frac{5}{6}\)

Question 2.
If A and B are two independent events such that:
\(\mathbf{P}(\overline{\mathbf{A}} \cap \mathbf{B})=\frac{2}{\mathbf{1 5}}\) and \(\mathbf{P}(\mathbf{A} \cap \overline{\mathbf{B}})=\frac{1}{\mathbf{6}}\), then find P(A) and P(B). (C.B.S.E. 2015)
Solution:
Since A and B are independent events,
\(\overline{\mathrm{A}}\) and B, A and \(\overline{\mathrm{B}}\) are also independent events.
Now \(P(\bar{A} \cap B)=\frac{2}{15}\) => \(\mathrm{P}(\overline{\mathrm{A}}) \mathrm{P}(\mathrm{B})=\frac{2}{15}\)
(1 – P(A)P(B) = \(\frac{2}{15}\)
P(B) – P(A))P(B) = \(\frac{2}{15}\) ………… (1)
And P(A∩\(\overline{\mathrm{B}}\))
P(A) \(\overline{\mathrm{B}}\) = \(\frac{1}{6}\)
P(A) (1 — P(B)) = \(\frac{1}{6}\)
P(A)-P(A)P(B) = \(\frac{1}{6}\) ………… (2)
Putting P(A) = x and P(B) = y in (1) and (2), we get
y – xy = \(\frac{2}{15}\)
and x – xy = \(\frac{1}{6}\) ………..(2)
Subtracting (2) from (1) y = y – x = \(\frac{2}{15}\) –\(\frac{1}{6}\)
⇒ y – x = \(\frac{-1}{30}\)
⇒ y = x – \(\frac{1}{30}\) ………… (3)
Putting in (1)’, \(x-\frac{1}{30}-x\left(x-\frac{1}{30}\right)=\frac{2}{15}\)
⇒ \(x-\frac{1}{30}-x^{2}+\frac{x}{30}=\frac{2}{15}\)
⇒ 30x – 1 – 30x² + x = 4
⇒ 30x² – 31x + 5 = 0.

If the girl tossed a coin .three times and exactly
1 tail shown, then: „ n
{HTH, HHT, THH} = 3
∴ P(A/E₁) = \(\frac { 3 }{ 8 }\)

Let A be the event that the girl obtained ex-actly one tail. If the girl tossed a coin only once and exactly 1 tail.
∴ P(A/E₂) = \(\frac { 1 }{ 2 }\)
By Bayes’ Theorem,

Question 6.
A manufacturer has three machine operators A, B and C. The first operator A produces 1 % of defective items, whereas the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B on the job 30% of the time and C on the job for 20% of the time. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by A?
(Delhi 2019)
Solution:
Let the events be as below :
E₁ : Item produced by machine A.
E₂ : Item produced by machine B.
E₃ : Item produced by machine C. and D : Item drawn is defective.
We have to find P (E₁/D).
Now P(E₂) = \(\frac{50}{100}=\frac{1}{2}\) ,
P(E₂) = \(\frac{30}{100}=\frac{3}{10}\)
and P (E₃) = \(\frac{20}{100}=\frac{1}{5}\)
Also P(D/E₁) = \(\frac{1}{100}\)
P(D/E₂) = \(\frac{5}{100}\), P(D/E₃) = \(\frac{7}{100}\)
By Bayes’ Theorem,
P(E₁/D) =

Question 7.
Bag I contain 3 white and 4 black balls, while Bag II contains 5 white and 3 black balls. One ball is transferred at random from Bag I to Bag II and then a ball is drawn at random from Bag II. The ball is drawn is found to be white. Find the probability that the transferred ball is also white.
(C.B.S.E. 2019 C)
Solution:
Let the events be as below:
E₁ : 1 white ball is transferred from Bag I to Bag II
E₂ : 1 black ball is transferred from Bag I to Bag II
and A: 1 white ball is drawn from Bag II.

Question 8.
There are three coins. One is a two-headed coin (having head on both the faces), another is a biased coin that comes up heads 75% of the time and the third is an unbiased coin. One of the three coins is chosen at random and tossed. If it shows head, what is the probability that it was the two-headed coin? (C.B.S.E. Sample Paper 2018-19)
Solution:
Let the events be:
E₁ : coin is two headed
E₂ : coin is biased and
E₃ : coin is unbiased.
And, A : coin shows up head.
∴P(E₁) = P(E₂) = P(E₃) = \(\frac{1}{3}\)
and P(A/E₁) = 1, P(A/E₂) = \(\frac{75}{100}=\frac{3}{4}\)
and P(A/E₃) = \(\frac{1}{2}\)
By Bayes’ Theorem,
P(E₁/A) =

Hence, the required probability = \(\frac{4}{9}\)

Question 9.
Bag I contains 4 red and 2 green balls and Bag II contains 3 red and 5 green balls. One ball is transferred at random from Bag I to Bag II and then a ball is drawn at random from Bag II. The ball so drawn is found to be green in colour. Find the probability that the transferred ball is also green. (C.B.S.E. 2019 )
Solution:
Let the events be as below:
E₁ : 1 red ball is transferred from Bag I to Bag II
E₂ : 1 green ball is transferred from Bag I to Bag II
and A : 1 Green ball is drawn from Bag II.

Question 10.
There are three coins. One is a coin having tails on both faces, another is a biased coin that comes up tails 70% of the time and the third is an unbiased coin. One of the coins is chosen at random and tossed, it shows tail. Find the probability that it was a coin with tail on both the faces. (Outside Delhi 2019)
Solution:
Let the events be as below :
E₁ : Selected coin has tail on both faces
E₂ : Selected coin is biased
E₃ : Selected coin is unbiased and
A : Tail comes up.
Now P(E₁ ) = P(E₂ ) – P(E₃ ) = \(\frac{1}{3}\)

P(A/E₁) = 1,
P(A/E₂) = \(\frac{70}{100}=\frac{7}{10}\)
P(A/E₃) = \(\frac{1}{2}\)
By Bayes ’ Theorem :

Question 11.
Let a pair of dice be thrown and the random variable ‘X’ be the sum of the numbers that appear on the two dice. Find the mean (or expectation) of X. (N.C.E.R.T.)
Solution:
Clearly the sample space consists of 36 elementary events = {(x_i, y_i) ; x_i, y_i = 1, 2,…., 6}.
X, the random variable = Sum of the numbers on the two dice.
∴ ‘X’ takes values 2, 3,4, or 12.

Now P(X = 2) = P({1,1}) = \(\frac{1}{36}\)
P (X = 3) = P ({1,2}, {2,1}) =\(\frac{2}{36}\)
P (X = 4) = P {(1,3), (2,2), (3,1)} = \(\frac{3}{36}\)
P (X = 5) = P {(1,4),(2,3), (3,2),(4,1)} = \(\frac{4}{36}\)
P (X = 6) = P {(1,5), (2,4), (3,3), (4,2), (5,1)} = \(\frac{5}{36}\)
P (X = 7) = P {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)} = \(\frac{6}{36}\)
P (X = 8) = P {(2,6), (3,5), (4,4), (5, 3), (6,2)} = \(\frac{5}{36}\)
P (X = 9) = P {(3,6), (4,5), (5,4), (6,3)} = \(\frac{4}{36}\)
P (X = 10) = P {(4, 6), (5,5), (6,4)} = \(\frac{3}{36}\)
P (X = 11) = P {(5,6), (6,5)} = \(\frac{2}{36}\)
P (X = 12) = P {(6, 6} = \(\frac{1}{36}\)
Thus the probability distribution is:

= \(\frac{1}{36}\)(2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12) = \(\frac{1}{36}\) (252) = 7.
Hence, the required, mean = 7.

Question 12.
Find the mean and variance of the numbers obtained on a throw of an unbiased die.
(N.C.E.R.T.)
Solution:
Here sample space, S = {1, 2, 3, 4, 5, 6}.
Let ‘X’ denote the number obtained on the throw.
Thus ‘X’ takes values 1, 2, 3, 4, 5 or 6.
∴ P (1) = P (2) = = P(6) = \(\frac{1}{6}\).
∴ Probability distribution is :

Question 13.
Two cards are drawn simultaneously (or successively without replacement) from a well shuffled pack of 52 cards. Find the mean and variance of the number of kings. (C.B.S.E. Delhi 2019)
Solution:
Let X, number of kings = 0,1,2.
∴ P(X=0) = P (no king)
= \(\frac{48}{52} \times \frac{47}{51}=\frac{188}{221}\)
P(X = 1) = P (one king and one non-king)
= 2 x \(\frac{4}{52} \times \frac{48}{51}=\frac{32}{221}\)
P(X = 2) = P (two kings)
= \(\frac{4}{52} \times \frac{3}{51}=\frac{1}{221}\)
Hence, probability distribution is :

Here we are providing Class 12 Maths Important Extra Questions and Answers Chapter 12 Linear Programming. Class 12 Maths Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Maths Chapter 12 Important Extra Questions Linear Programming

Linear Programming Important Extra Questions Very Short Answer Type

Question 1.
Draw the graph of the following LPP:
5x + 2y ≤ 10, x ≥ 0,y ≥ 0.
Solution:
Draw the line AB : 5.v + 2y = 10 …(1),
which meets x-axis at A (2, 0) and y-axis at B (0,5).
Also x = 0 is y-axis and y = 0 is x-axis.
Hence, the graph of the given LPP is as shown (shaded):

Question 2.
Solve the system of linear inequations: x + 2y ≤ 10; 2x + y ≤ 8.
Solution:
Draw the st. lines x + 2y = 10 and 2x + y = 8.
These lines meet at E (2,4).
Hence, the solution of the given linear inequations is shown as shaded in the following figure :

Question 3.
Find the linear constraints for which the shaded area in the figure below is the solution set:

Solution:
From the above shaded portion, the linear constraints are :
2x + y ≥ 2,x – y ≤ 1,
x + 2y ≤ 8, x ≥ 0, y ≥ 0.

Question 4.
A small firm manufactures neclaces and bracelets. The total number of neclaces and bracelets that it can handle per day is at most 24. It takes one hour to make a bracelet and half an hour to make a neclace. The maximum number of hours available per day is 16. If the profit on a neclace is ₹100 and that on a bracelet is ₹300. Formulate an LPP for finding how many of each should be produced daily to maximize the profit ?
It is being given that at least one of each must be produced. (C.B.S.E. 2017)
Solution:
Let ‘x’ neclaces and ‘y’ bracelets be manufactured per day.
Then LPP problem is:
Maximize Z = 100x+300y
Subject to the constraints : x + y ≤ 24,
(1) (x) + \(\frac { 1 }{ 2 }\)y ≤ l6,
i.e. 2x + y ≤ 32
and x ≥ 1
and y ≥ 1
i.e. x – 1 ≥ 0
and y – 1 ≥ 0.

Question 5.
Old hens can be bought for ?2.00 each and young ones at ?5.00 each. The old hens lay 3 eggs per week and the young hens lay 5 eggs per week, each egg being worth 30 paise. A hen costs ₹1.00 per week to feed. A man has only ₹80 to spend for hens. Formulate the problem for maximum profit per week, assuming that he cannot house more than 20 hens.
Solution:
Let ‘x’ be the number of old hens and ‘y’ the number of young hens.
Profit = (3x + 5y) \(\frac { 30 }{ 100 }\) – (x + y) (1)
= \(\frac{9 x}{10}+\frac{3}{2}y\)x – y
= \(\frac{y}{2}-\frac{x}{10}=\frac{5 y-x}{10}\)
∴ LPP problem is:
Maximize Z = \(\frac{5 y-x}{10}\) subject to:
x ≥ 0,
y ≥ 0,
x + y ≤ 20 and
2x + 5y ≤ 80.

Linear Programming Important Extra Questions Very Long Answer Type 2

Question 1.
Maximize Z-5x + 3y
subject to the constraints:
3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0,y ≥ 0. (N.C.E.R.T)
Solution:
The system of constraints is :
3x + 5y ≤ 15 …(1)
5x + 2y ≤ 10 …(2)
and x ≥ 0, y≥ 0 …(3)
The shaded region in the following figure is the feasible region determined by the system of constraints (1) – (3):

It is observed that the feasible region OCEB is bounded. Thus we use Corner Point Method to determine the maximum value of Z, where :
Z = 5x + 3y …(4)

The co-ordinates of O, C, E and B are (0, 0), (2,0), \(\left(\frac{20}{19}, \frac{45}{19}\right)\) (Solving 3x + 5y = 15 and 5x + 2y – 10) and (0, 3) respectively.
We evaluate Z at each comer point:

Comer Point	Corresponding Value of Z
O: (0,0)	0
C: (2,0)	10
E(\(\left(\frac{20}{19}, \frac{45}{19}\right)\))	\(\frac{20}{19}\) (Maximum)
B(0.3)	9

Hence’ Z_max = at the Point \(\left(\frac{20}{19}, \frac{45}{19}\right)\)

Question 2.
Minimize Z = 3x + 2y subject to the constraints:
x +y ≥ 8, 3x + 5y ≤ 15, x ≥ 0, y ≥ 0. (N.C.E.R.T.)
Solution:
The system of constraints is :
x +y ≥ 8, , x ≥ 0, y ≥ 0…(1)
3x + 5y ≤ 15 …(2)
and x ≥ 0, y ≥ 0 …(3)

It is observed that there is no point, which satisfies all (1) – (3) simultaneously.
Thus there is no feasible region.
Hence, there is no feasible solution.

Question 3.
Determine graphically the minimum value of the objective function :
Z = – 50x + 20y
subject to the constraints:
2x-y ≥ – 5, 3x +y ≥ 3, 2x – 3y ≤ 12, x,y ≥ 0. (N.C.E.R.T.)
Sol. The system of constraints is :
2x-y ≥ – 5 …(1)
3x +y ≥ 3 …(2)
2x – 3y ≤ 12 …(3)
and x,y ≥ 0 …(4)
The shaded region in the following figure is the feasible region determined by the system of constraints (1) – (4).

It is observed that the feasible region is unbounded.
We evaluate Z = – 50x + 20y at the corner points :
A (1, 0), B (6, 0), C (0, 5) and D (0, 3) :

Corner Point	Corresponding Value of Z
A: (1,0)	-50
B : (6, 0)	– 300 (Minimum)
C : (0, 5)	100
D : (0, 3)	60

From the table, we observe that – 300 is the minimum value of Z.
But the feasible region is unbounded.
∴ – 300 may or may not be the minimum value of Z. ”

For this, we draw the graph of the inequality.
– 50x + 20y < – 300
i.e. – 5x + 2y < – 30.
Since the remaining half-plane has common points with the feasible region,
∴ Z = – 50x + 20y has no minimum value.

Question 4.
Minimize and Maximize Z = 5x + 2y subject to the following constraints : x – 2y ≤ 2, 3x + 2y < 12, -3x + 2y ≤ 3, x ≥ 0, y ≥ 0. (A.I.C.B.S.E. 2015)
Solution:
The given system of constraints is :
x – 2y ≤ 2 …(1)
3x + 2y < 12 …(2)
-3x + 2y ≤ 3 …(3)
and x ≥ 0, y ≥ 0.

The shaded region in the above figure is the feasible region determined by the system of constraints (1) – (4). It is observed that the feasible region OAHGF is bounded. Thus we use Corner Point Method to determine the maximum and minimum value of Z, where
Z = 5x + 2y …(5)

The co-ordinates of O, A, H, G and F are :
(0, 0). (2. 0), (\(\frac{7}{2}, \frac{3}{4}\)) and (\(\frac{3}{2}, \frac{15}{4}\)), \(\frac { 3 }{ 2 }\))
respectively. [Solving x

2y = 2 and 3x + 2y = 12 for
H and -3x + 2y = 3 and
3x + 2y = 12 for G]
We evaluate Z at each cormer point:

Corner Point	Corresponding value of Z
O: (0,0)	0 (Minimum)
A: (2,0)	10
H(\(\frac{7}{2}, \frac{3}{4}\))	19 (Maximum)
G(\(\frac{3}{2}, \frac{15}{4}\))	15
F: (0, \(\frac{3}{2}\)))	3

Hence, Z_max = 19 at (\(\frac{7}{2}, \frac{3}{4}\)) and
Z_max = 0at (0,0)

Question 5.
A dealer in rural area wishes to purchase a number of sewing machines. He has only ₹5760.00 to invest and has a space for at most 20 items. An electronic sewing machine costs him ₹360.00 and a manually operated sewing machine ₹240.00. He can sell electronic sewing machine at a profit of ₹22.00 and a manually operated sewing machine at a profit of ₹18.00. Assuming that he can sell all the items that he can buy, how should he invest his money in order to maximize his profit. Make it a linear programming problem and solve it graphically. (C.B.S.E. 2014)
Solution:
Let ‘x’ be the number of electronic operated machines and ‘y’ that of manually operated machines be purchased.
Then the LPP problem is as follows :
Maximize:
Z = 22 + 18y
Subject to:
x + y ≤ 20 …(1)
360x + 240y ≤ 5760
i.e. 3x + 2y ≤ 48 …(2)
and x ≥ 0, y ≥ 0 …(3)
The shaded region of the figure represents the feasible region OCEB, which is bounded.

Applying Corner Point Method, we have:

Corner Point	Corresponding value of Z
O: (0,0) C:(16,0) E: (8,12) B: (0,20)	0 352 392 (Maximum) 360

Thus, Z is maximum at E (8, 12).
Hence, the dealer should invest in 8 electronic and 12 manually operated machines.

Question 6.
A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of ₹ 35 per package of nuts and ₹ 14 per package of bolts. How many packages of each should be produced each day so as to maximise his profit, if he operates each machine for atmost 12 hours a day? Convert it into an LPP and solve graphically.
Solution:
Let ‘x’ and ‘y’ be the number of packages of nuts and bolts respectively.
We have the following constraints :
x ≥ 0 …(1)
y ≥ 0 …(2)
x + 3y ≤ 12 ….(3)
3x + y ≤ 12 …(4)
Now the profit,P= 35x+ 14y …..(5)
We are to maximize P subject to constraints (1) -(4).
Draw the line AB (x + 3y = 12)
Draw the line CD (3x + y = 12)
These meet at E (3, 3).
The shaded region in the figure represents the feasible region, which is bounded.

Applying Corner Point Method, we have :

Corner Point	P = 35x + 14y
O: (0,0)	0
C : (4, 0)	140
E: (3,3)	147 (Maximum)
B : (0,4)	56

Hence, max. profit is ₹ 147 and it is obtained when 3 packages each of Nuts and Bolts are produced daily.

Question 7.
Two tailors A and B earn ₹ 150 and ₹ 200 per day respectively. A can stitch 6 shirts and 4 pants per day, while B can stitch 10 shirts and 4 pants per day. Form a L.P.P. to minimize the labour cost to produce (stitch) at least 60 shirts and 32 pants and solve it graphically.
Solution:
Let the tailor A work for ‘x’ days and B for ‘y’ days.

Thus, we have the following constraints:
x ≥ 0 …(1)
y ≥ 0 …(2)
6x + 10y ≥ 60
i.e. 3x + 5y ≥ 30 …….(3)
4x + 4y ≥ 32
i.e. x + y ≥ 8 ……(4)
The objective function, or the cost Z is:
Z = 150x + 200 y ……..(5)
For the solution set, we draw the lines:
x = 0, y – 0, 3x + 5y = 30, x + y = 8

The feasible region (shaded) is unbounded. Let us evaluate Z at the comer points:
A (10,0), D(0, 8) and E(5, 3)
[Solving x + y – 8, Sx + 5y – 30; x – 5,y = 3]

Applying Comer Point Method, we have:

Corner Point	Z = 150 x + 200y
A : (10, 0) E: (5,3) D: (0, 8)	1500 1350 (Minimum) 1600

Hence, the tailor A should work for 5 days and B for 3 days.
To Check: Draw 150x + 200y < 1350 i.e. 3x + 4y < 27.
Since there is no region common with feasible region,
∴ Minimum value is ₹ 1350.

Question 8.
A dietician wishes to mix two types of food in such a way that the vitamin contents of the mixture contains at least 8 units of vitamin A and 10 units of vitamin C. Food I contains 2 units/kg of vitamin A and 1 unit/kg of vitamin C. It costs ₹50 per kg to produce food I. Food II contains 1 unit/kg of vitamin A and 2 units/kg of vitamin C and it costs ₹70 per kg to produce food n. Formulate this problem as a LPP to minimise the cost of a mixture that will produce the required diet. Also find the minimum cost
Solution:
Let the quantity of Food I = x kg
and the quantity of Food II = y kg.
Then the LPP problem is as below :
Z = 50x+70y …(1)
Subject to 2x + y ≥ 8 …(2)
x + 2y ≥ 10 ….(3)
and x ≥ 0, y ≥0 ……..(4)
For the solution, we draw the lines :
x = 0,y = 0,2x + y = 8 and x + 2y = 10

The feasible regin is as shown with vertices C(10,0), E(2,4) and B(0,8).
Applying Corner Point Method, we have

Corner Point	Z = 50x + 70y
C: (10,0) E:(2,4) B: (0,8)	500 380 (Minimum) 560

Thus minimum cost is ₹38Q when 2 kg of Food I and 4 kg of Food II are mixed.

Question 9.
A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given below:

Types of Toys	Machines
	I	II	III
A	20	10	10
B	10	20	30

The machines I, II and III are available for a maximum of 3 hours, 2 hours and 2 hours 30 minutes respectively. The profit on each toy of type A is ₹50 and that of type Bis ₹60. Formulate the above problem as a LPP and solve it graphically to maximize profit.
(C.B.S.E. Sample Paper 2018-19)
Solution:
Let ‘x’ and ‘y’ be the number of toys of type A and type B respectively.
Then maximize :
P = 50x + 60y …(1)
Subject to constraints :
20x +10v ≤ 180 …(2)
10x + 20y ≤ 120 …….(3)
10x + 30y ≤ 150 ……(4)
and x ≥ 0, y ≥ 0 …(5)

Applying Corner Point Method we have :

Corner point	P = 50jc + 60y
O: (0,0) F:(0,5) G: (6,3) H: (8,2) A: (9,0)	0 300 480 520 (Maximize) 450

Hence, maximum profit is ₹520 when x = 8 and y = 2. i.e., when 8 toys of type A and 2 toys of type B are made.

Question 10.
A factory manufactures two types of screws A and B, each type requiring the use of two machines, an automatic and a hand- operated. It takes 4 minutes on the automatic and 6 minutes on the hand operated machines to manufacture a packet of screws ‘A’ while it takes 6 minutes on the automatic and 3 minutes on the hand-operated machine to manufacture a packet of screws ‘B\ Each machine is available for at most 4 hours on any day. The manufactures can sell a packet of screws ‘A’ at a profit of 70 paise and screws ‘B’ at a profit of ₹ 1. Assuming that he can sell all the screws he manufactures, how many packets of each type should the factory owner produce in a day in order to maximize his profit? Formulate the above LPP and solve it graphically and find the maximum profit (C.B.S.E. 2018)
Solution:
Let the factory manufacture ‘x’ of type ‘A’ and ‘y’ of type ‘B’
Clearly x ≥ 0 …(1)
and y ≥ 0 …(2)
Since the machines can operate for at the most 4 hours a day,
4x + 6y ≤ 240
i.e., 2x + 3y ≤ 120 …(3)
and 6x + 3y ≤ 240
i.e., 2x + y ≤ 80 …(4)
The objective function or the profit, P, is:
P = 0.7 x + y …(5)

We drawn the lines :
x = 0,y = 0,
2x + 3y = 120
and 2x + y – 80

The feasible region is shown shaded OCPB is bounded, where O is (0, 0), C is (40, 0), B is (0, 40) and P is (30, 20).
[Solving 2x + y – 80 and 2x + 3y = 120; x = 30, y – 20]

Applying Corner Point Method, we have :

Comer point	P = 0.7x + y
O: (0, 0) C : (40, 0) P : (30, 20) B: (0,40)	0 28 41 (Maximum) 40

Hence, in order to maximize profit 30 packets of screw ‘A’ and 20 packets of screw ‘B’ should be manufactured and maximum profit = ₹ 41.

Question 11.
A small firm manufactures chairs and tables. Market demand and available resources indicate that the continued production of chairs and tables should not exceed 50 units per day. It takes 30 minutes to manufacture a chair and 1 hour to manufacture a table. A maximum of 40 man-hours per day are available. The profit on each chair is ₹ 40 and profit on each table is ₹ 60. Determine how many each of chairs and tables should be manufactured per day in order to maximize the profit. What is the maximum profit? Formulate LPPand solve graphically.
Solution:
Let ‘x’ and ‘y’ be the number of chairs and tables respectively.
We have: x ≥ 0 ………..(1)
y ≥ 0 ………(2)
x + y ≤ 50 ……(3)
and \(\frac{x}{2}\) + y ≤ 40
⇒ x + 2y < 80 …(4)
The objective function, or the profit, Z is
Z = 40 x + 60 y …(5)

We have to maximize Z subject to (1) – (4).
For solution set, we draw the lines:
x = 0, y – 0, x + y = 50 and x + 2y – 80.
The lines x + y = 50 and x + 2y = 80 meet at E (20, 30).

The shaded portion represents the feasible region, which is bounded.
Applying Corner Point Method, we have:

Corner Point	Z = 40x + 60y
0 =(0,0) A: (50, 0) E: (20,30) B: (0,40)	0 2000 2600 (Maximum) 2400

Hence, the maximum profit is ₹ 2600 when 20 chairs and 30 table are manufactured.

Studying from CBSE Class 12th Maths Revision Notes helps students to prepare for the exam in a well-structured and organised way. Making Class 12 Maths NCERT Notes saves students time during revision as they don’t have to go through the entire textbook. In CBSE Notes, students find the summary of the complete chapters in a short and concise way. Students can refer to the NCERT Solutions for Class 12 Maths, to get the answers to the exercise questions.

Class 12th Maths NCERT Notes | Maths Notes Class 12

1. Relations and Functions Class 12 Notes
2. Inverse Trigonometric Functions Class 12 Notes
3. Matrices Class 12 Notes
4. Determinants Class 12 Notes
5. Continuity and Differentiability Class 12 Notes
6. Application of Derivatives Class 12 Notes
7. Integrals Class 12 Notes
8. Application of Integrals Class 12 Notes
9. Differential Equations Class 12 Notes
10. Vector Algebra Class 12 Notes
11. Three Dimensional Geometry Class 12 Notes
12. Linear Programming Class 12 Notes
13. Probability Class 12 Notes

We hope students have found these Revision Notes for Class 12 Maths useful for their studies. If you have any queries related to Maths Notes Class 12, drop your questions below in the comment box.

By going through these CBSE Class 12 Maths Notes Chapter 11 Three Dimensional Geometry, students can recall all the concepts quickly.

Three Dimensional Geometry Notes Class 12 Maths Chapter 11

1. Direction cosines of a line:
Let AB be a line in space.
Through O, draw a line OP parallel to AB. Let OP makes angles α, β, and γ with OX, OY, and OZ respectively.

Cosines of the angles α, β, and γ,
i.e., cos α, cos β, and cos γ are known as the direction cosines of line AB.
Let l = cos α, m = cos β, n = cos γ
⇒ l, m, n are the direction cosines of the line AB Let us consider the ray BA. OQ is drawn parallel to BA. Now, OQ makes angles n – α, n – β, and n – γ with coordinates axes OX, OY, and OZ respectively.

∴ Direction cosines of BA are cos(π – α), cos (π – β) and cos (π – γ),
i.e., – cos α, – cos β, – cos γ or -l, -m, -n.

→ Relation between l, m, and n
AB is any ray having direction cosines l, m, n. Now OP is drawn parallel to AB, where P is the point (x, y, z). Let PM be drawn perpendicular to OY.

In ΔOPM OP = r (say). PM⊥OY.
∴ ∠PMO = 90°
Also, ∠POM = β.
∴ \(\frac{\mathrm{OM}}{\mathrm{OP}}\) = β
∴ \(\frac{y}{r}\) = m,
∴ y = rm

Similarly, x = rl and z = rn.
Now, OP² = x² + y² + z²
r² = (rl)² + (rm)² + (rn)²
or
r² = r²(l² + m² + n²)
⇒ l² + m² + n² = 1.

→ Direction Ratios of a line
Definition: The numbers which are proportional to the direction cosines of a line are known as direction ratios of the line.

Let Z, m, n be the direction cosines of a line.
Multiplying each by r, we ger rl, rm, rn the direction ratios.
Let rl = a,rm = b and rn – c.
Squaring and adding
r²l² + r²m² + r²n² = a² + b² + c²
∴ r²(l² + m² + n²) = a² + b² + c²
∵ l² + m² + n² = 1
∴ l = \(\frac{a}{r}=\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}\)

Similarly, m = \(\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}}\) and n = \(\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}\)

Thus, if a, b and c are the direction ratios of a line, the direction cosines are \(\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}\), \(\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}}\) and \(\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}\).

→ Direction cosines of the line passing through the points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂)
Direction ratios of the line PQ are x₂ – x₁, y₂ – y_1, z₂ – z₁
∴ PQ = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}\)
∴Direction cosines of PQ are \(\frac{x_{2}-x_{1}}{\mathrm{PQ}}\), \(\frac{y_{2}-y_{1}}{\mathrm{PQ}}\), \(\frac{z_{2}-z_{1}}{P Q}\)

2. Angle between the two lines
1. Let l_1, m₁, n₁, and l₂, m₂, n₂ be the direction cosines of the lines OP and OQ.

The angle θ between these lines is given by
cos θ = l₁l₂ + m₁m₂ + n₁n₂
and sin θ = \(\sqrt{\left(m_{1} n_{2}-m_{2} n_{1}\right)^{2}+\left(n_{1} l_{2}-n_{2} l_{1}\right)^{2}+\left(l_{1} m_{2}-l_{2} m_{1}\right)^{2}}\)

2. If a₁, b₁, c₁ and a₂, b₂, c₂ are the direction ratios of two lines, then

3. Two lines are perpendicular to each other,
if θ = \(\frac{π}{2}\) ⇒ cos θ = cos \(\frac{π}{2}\) = 0
⇒ l₁l₂ + m₁m₂ + n₁n₂ = 0
or
a₁a₂ + b₁b₂ + c₁c₂ = 0

4. When the lines are parallel
θ = 0 ⇒ sin θ = sin 0 = 0.

STRAIGHT LINE
3. Equation of a line through a given point:
(a) Let the line passes through \(\vec{a}\) and is parallel to vector \(\vec{b}\)

Then, the equation of the line is
\(\vec{r}\) – \(\vec{a}\) = λ\(\vec{b}\)
or
\(\vec{r}\) = \(\vec{a}\) + λ\(\vec{b}\)

(b) Let the point A be (x₁, y₁, z₁) and a, b, c are the direction ratios of the line. The equation of the line is
\(\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\)
If l, m, n are the direction cosines, then equation of the line is
\(\frac{x-x_{1}}{l}=\frac{y-y_{1}}{m}=\frac{z-z_{1}}{n}\)

4. Equation of the line passing through two points:
(a) Let \(\vec{a}_{1}\), \(\vec{a}_{2}\) be the position vectors of two points P and Q respectively.
⇒ \(\vec{b}\) = \(\vec{a}_{1}\) – \(\vec{a}_{2}\)

∴ Equation of PQ is
\(\vec{r}\) = \(\vec{a}\) + λ(\(\vec{a}_{2}\) – \(\vec{a}_{1}\))

(b) Direction ratios of the line passing through P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) are x₂ – x₁, y₂ – y₁ and z₂ – z₁
∴ Equation of the line PQ is
\(\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}\)

5. Angle between two straight lines:

(a) Let the two lines be
\(\vec{r}\) = \(\vec{a}_{1}\) + λ\(\vec{b}_{1}\)
\(\vec{r}\) = \(\vec{a}_{2}\) + λ\(\vec{b}_{2}\)
If θ be the angle between them, then
cos θ = \(\frac{\vec{b}_{1} \vec{b}_{2}}{\left|\vec{b}_{1}\right|\left|\vec{b}_{2}\right|}\)

(b) Let the lines be
\(\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}\)
and \(\frac{x-x_{2}}{a_{2}}=\frac{y-y_{2}}{b_{2}}=\frac{z-z_{2}}{c_{2}}\)
i.e., a1, b1, c1 and a2, b2, c2 are the direction ratios of the lines.
∴ cos θ = \(\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\)
If l1, m1, n1 and l₂, m₂, n₂ are the direction cosines, then
cos θ = l₁l₂ + m1m₂ +n1n₂

6. Shortest Distance:
(a)Let \(\vec{r}\) = \(\vec{a}_{1}\) + λ\(\vec{b}_{1}\) and
\(\vec{r}\) = \(\vec{a}_{2}\) + λ\(\vec{b}_{2}\) be the two non-intersecting lines.
The shortest distance between the given lines = \(\left|\frac{\left(\vec{b}_{1} \times \vec{b}_{2}\right) \cdot\left(\vec{a}_{2}-\vec{a}_{1}\right)}{\left|\vec{b}_{1} \times \vec{b}_{2}\right|}\right|\)

(b) Let the lines be

If the lines are intersecting, then lines are coplanar.
⇒ S.D. = 0
⇒ (\(\vec{b}_{1} \times \vec{b}_{2}\)) – (\(\vec{a}_{1} \times \vec{a}_{2}\)) = 0
or
\(\left|\begin{array}{ccc}
x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\
a_{1} & b_{1} & c_{1} \\
a_{2} & b_{2} & c_{2}
\end{array}\right|\) = 0

7. Distance between parallel lines:
Let the parallel lines be
\(\vec{r}\) = \(\vec{a}_{1}\) + λ\(\vec{b}_{1}\) and
\(\vec{r}\) = \(\vec{a}_{2}\) + μ\(\vec{b}_{2}\)

Shortest distance d between these lines is given by,
d = \(\left|\frac{\vec{b} \times\left(\vec{a}_{1}-\vec{a}_{2}\right)}{|\vec{b}|}\right|\)

PLANES
8. Different forms of equations of a plane:
1. Normal Form.
(a) Let the plane ABC be at a distance d from the origin. ON is the normal to the plane in the directon n̂. Equation of the plane is \(\vec{r}\). n̂ = d.

(b) If l, m, n are the direction cosines of the normal to the plane which is at distance d from the origin. The equation of the plane is lx + my + nz = d.
However, general form of the equation of a plane are \(\vec{r}\) .\(\overrightarrow{\mathrm{N}}\) = D and Ax + By + Cz + D = 0.

2. (a) Let the plane passes through a point A and let it perpendicular to the vector \(\overrightarrow{\mathrm{N}}\)
∴ Equation of the plane
(\(\vec{r}\) – \(\vec{a}\)).\(\overrightarrow{\mathrm{N}}\) = 0.

(b) If a plane passes through (x₁, y₁, z₁) and perpendicular to the line with direction ratios a, b, c the equation of the plane is
a(x – x₁) + b(y – y₁)+c(z – z₁) =0

3. Equation of the plane passing through three points.
(a) Let the three points be A, B and C whose position vectors be, \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\)
The equation of the plane is
\((\vec{r}-\vec{a}) \cdot[(\vec{b}-\vec{a}) \times(\vec{c}-\vec{a})]\) = 0

(b) Let the three points through which the plane is passing be A(x₁, y₁, z₁), B(x₂, y₂, z₂), and C(x₃, y₃, z₃).
Then, equation of the plane is
\(\left|\begin{array}{llll}
x & y & z & 1 \\
x_{1} & y_{1} & z_{1} & 1 \\
x_{2} & y_{2} & z_{2} & 1 \\
x_{3} & y_{3} & z_{3} & 1
\end{array}\right|\) = 0
or
\(\left|\begin{array}{ccc}
x-x_{1} & y-y_{1} & z-z_{1} \\
x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\
x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1}
\end{array}\right|\) = 0

4. Intercepts form of the equation of the plane
Let the plane make the intercepts a, b, and c on coordinate axes OX, OY, and OZ respectively.

Then, the equation of the plane is
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\) = 1

9. Plane passing through the intersection of the two planes:
Let the equation of two planes be
\(\vec{r}\). \(\vec{n}_{1}\) = d1 and
\(\vec{r}\) .\(\vec{n}_{2}\) = d2

The equation of the plane passing through the line of intersection of the given planes is
(\(\vec{r}\). \(\vec{n}_{1}\) – d1) + λ(\(\vec{r}\) .\(\vec{n}_{2}\) – d2) = 0
or
\(\vec{r}\) .(\(\vec{n}_{1}\) + λ\(\vec{n}_{2}\)) = (d1 + λd2)

(b) Equation of the plane passing through the line of intersection of the planes
a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 is (a₁x + b₁ y + c₁z + d₁ + λ(a₂x + b₂y + c₂z + d₂) = 0
or
(a₁ + λa₂)x + (b₁ + λb₂)y + (c₁ + λc₂)z + d₁ + λd₂ = 0,
where λ is determined according to the given condition.

10. Coplanarity of two lines:

(a) The two lines
\(\vec{r}\) = \(\vec{a}_{1}\) + λ\(\vec{b}_{1}\) and
\(\vec{r}\) = \(\vec{a}_{2}\) + λ\(\vec{b}_{2}\) intersect each other, if \(\left(\vec{a}_{2}-\vec{a}_{1}\right) \cdot\left(\vec{b}_{2} \times \vec{b}_{1}\right)\) = 0

11. The angle between the planes:
Definition : The angle between the two planes is the angle between their normals. The angle 0 between the planes
\(\vec{r}\) = \(\vec{n}_{1}\) = d1 and \(\vec{r}\) = \(\vec{n}_{1}\) = d2 is given by cos θ = \(\frac{\left|\vec{n}_{1} \cdot \vec{n}_{2}\right|}{\left|\vec{n}_{1}\right|\left|\vec{n}_{2}\right|}\)

(b) If the planes are
a₁x + b₁y + C₁z + d₁ = 0
a₂x + b₂y + c₂z + d₂ = 0, then the angle θ between these planes is given by
cos θ = \(\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\)

(c) The planes are perpendicular to each other, if θ = \(\frac{π}{2}\)
∴ cos \(\frac{π}{2}\) = 0 ⇒ \(\vec{n}_{1}\).\(\vec{n}_{2}\) = 0
or
a₁a₂ + b₁b₂ + c₁c₂ = 0

(d) The planes a₁x + b₁y +c₁z + d₁ = 0
and a₂x + b₂y + c₂z + d₂ = O
are parallel, if \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\).

(e) Equation of the plane parallel to
\(\vec{r} \cdot \vec{n}\) = d is \(\vec{r} \cdot \vec{n}\) = λ
Plane parallel to
ax + by + cz + d = 0 is ax + by + cz + λ = 0.

12. Distance of a point from th plane:
(a) Let the plane be \(\vec{r} \cdot \vec{n}\) = d.
∴ Perpendicular distance of the point \(\vec{a}\) from the plane
=|d – \(\vec{a} \cdot \hat{n}\)|

(b) Let the equation of the plane
be Ax + By + Cz + D= O.
The distance of the point (x1, y1, z1) from this plane
= \(\left|\frac{A x_{1}+B y_{1}+C z_{1}+D}{\sqrt{A^{2}+B^{2}+C^{2}}}\right|\)

13. Angle between a line and a plane:
Definition: The angle between a line and plane is said to be the complement of the angle between the line and the normal to the plane.

(a) Let the line and plane be
\(\vec{r}\) = \(\vec{a}\) + λ\(\vec{b}\) and \(\vec{r} \cdot \vec{n}\) = d.

If θ be the angle between the plane and the line, then
sin θ = \(\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|}\)

(b) ¡et the line and the plane be \(\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\) and Ax + By + Cz + D = O.
The angle Φ between them is given by
sin Φ = \(\frac{a \cdot \mathrm{A}+b \cdot \mathrm{B}+c \cdot \mathrm{C}}{\sqrt{a^{2}+b^{2}+c^{2}} \sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}+\mathrm{C}^{2}}}\).

1. IMPORTANT RESULT

l² + m² + n² = 1, where < l, m, n > are direction-cosines of a st. line.

2. DIRECTION-RATIOS

The direction-ratios of the line joining of the points (x₁, y₁, z₁) and (x₂, y₂, z₂) are :
<x₂-x₁, y₂-y₁ ,z₂-z₁ >.

3. (i) Angle between two lines. The angle between two lines having direction-cosines
< l₁ m₁ n₁ > and < l₂, m₂, n₂ > is given by :
cos θ = |l₁l₂+ m₁m₂ + n₁n₂|.

(ii) The lines are:
(a) perpendicular l₁l₂+ m₁m₂ + n₁n₂ = 0.
(b) parallel iff l₁ = l_2, m₁ = m₂, + n₁ = n₂ 

4. SHORTEST DISTANCE

The shortest distance between two lines :
\(\vec{r}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}} \text { and } \vec{r}=\overrightarrow{a_{2}}+\mu \overrightarrow{b_{2}} \text { is }\left|\frac{\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right) \cdot\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)}{\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|}\right|\)

5. EQUATIONS OF PLANES

(i) Equation of a plane, which is at a distance ‘p’ from the origin and perpendicular to the unit
vector \(\hat{n}\) is \(\vec{r} \cdot \hat{n}=p\)

(ii) General Form. The general equation of first degree i.e.ax + by + cz + d- 0 represents a plane,

(iii) One-point Form. The equation of a plane through (*,, y,, z,) and having <a,b,c> as direction-
ratios of the normal is a (x – x₁) + b (y – y₁) + c (z – z₁) = 0.

(iv) Three-point Form. The equation of the plane through (x₁, y₁, z₁), (x₂, y₂, z₂) and (x₃, y₃, z₃) is:
\(\left|\begin{array}{ccc}
x-x_{1} & y-y_{1} & z-z_{1} \\
x-x_{2} & y-y_{2} & z-z_{2} \\
x-x_{3} & y-y_{3} & z-z_{3}
\end{array}\right|=0\)

(v) Intercept Form. Equation of the plane, which cuts off intercepts a, b, c on the axes, is :
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\) = 1

6. ANGLE BETWEEN TW O PLANES
The angle between the planes :
a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 is given by :
cos θ = \(\frac{\left|a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}\right|}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\)

7. BISECTING PLANES
The equations of the planes bisecting the planes :
a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 are:
\(\frac{a_{1} x+b_{1} y+c_{1} z+d_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}}=\pm \frac{a_{2} x+b_{2} y+c_{2} z+d_{2}}{\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\)

By going through these CBSE Class 12 Maths Notes Chapter 10 Vector Algebra, students can recall all the concepts quickly.

Vector Algebra Notes Class 12 Maths Chapter 10

Some Basic Concepts:
1. Vector (Definition): A quantity that has magnitude, as well as direction, is called a vector.
\(\overrightarrow{\mathrm{AB}}\) is a directed line segment.

It is a vector \(\overrightarrow{\mathrm{AB}}\) and its direction is from A to B.
→ Initial Points: The point A wherefrom the vector \(\overrightarrow{\mathrm{AB}}\) starts is known as the initial point.

→ Terminal point: Point B, where it ends is said to be the terminal point.

→ Magnitude: The distance between initial point and terminal point of a vector is the magnitude (or length) of the vector \(\overrightarrow{\mathrm{AB}}\) . It is denoted by |\(\overrightarrow{\mathrm{AB}}\)| or |\(\overrightarrow{\mathrm{a}}\)| or simply as a.

2. Position Vector: Consider a point F(x, y, z) in space. The vector \(\overrightarrow{\mathrm{OP}}\) with an initial point as origin O and terminal point P, is called the position vector of P. It may be denoted by \(\overrightarrow{\mathrm{p}}\).

3. Direction Cosines: Let OX, OY, OZ be the coordinates axes and F(x, y, z) be any point in the space. Let OP makes angles α, β, γ with coordinates axes OX, OY, OZ respectively.

The angles α, β, γ are known as direction angles. The cosine of these angles, i.e., cos α, cos β, cos γ is called direction cosines of line OP. These direction cosines are denoted by Z, m, n, i.e., Z = cos α, m = cos β, and n = cos γ.

4. Relation between l, m, n, and Direction Ratios: The perpendiculars PA, PB, PC are drawn on coordinate axes OX, OY, OZ respectively. Let OP = r,
In ΔOAP, ∠A = 90°, cos α = \(\frac{x}{r}\) = l ∴ x = Ir
In ΔABP, ∠B = 90°, cos β = \(\frac{y}{r}\) = m ∴ y = mr
In ΔOCP, ∠C = 90°, cos γ = \(\frac{z}{r}\) = n ∴ z = nr

Thus, the coordinates of P may be expressed as (lr, mr, nr).
Also, OP² = x² + y² + z² = r² = (lr)² + (mr)² + (nr)²
⇒ l² + m² + n² = 1.
When l, m, n are multiplied by a scalar k, we get the direction ratios, i.e., lk, mk, nk are the direction ratios of OA.

Types of vectors:
1. Zero Vector or Null Vector: A vector whose initial and terminal points coincide is known as zero vector (\(\overrightarrow{0}\)).

2. Unit Vector: A vector whose magnitude is unity is said to be a unit vector. It is denoted as â, so that | â | = 1.

3. Co-initial Vectors: Two or more vectors having the same initial point are called co-initial vectors.

4. Collinear Vectors: If two or more vectors are parallel to the same line, such vectors are known as collinear vectors.

5. Equal Vectors: If two vectors \(\vec{a}\) and \(\vec{b}\) have the same magnitude and direction, regardless of the positions of their initial points, such vectors are said to be equal, i.e., \(\vec{a}\) = \(\vec{b}\) .

6. Negative of a Vector: A vector whose magnitude is the same as that of a given vector (say \(\overrightarrow{\mathrm{AB}}\)), but the direction is opposite to that of it, is known as negative of vector \(\overrightarrow{\mathrm{AB}}\). i.e., \(\overrightarrow{\mathrm{BA}}\) = – \(\overrightarrow{\mathrm{AB}}\).

6. Sum of Vectors:
1. Sum of Vectors \(\vec{a}\) and \(\vec{b}\):
Let the vectors \(\vec{a}\) and \(\vec{b}\) be so positioned that initial point of one coincides with terminal point of the other i.e., let \(\vec{a}\) = \(\overrightarrow{\mathrm{AB}}\) and \(\vec{b}\) = \(\overrightarrow{\mathrm{BA}}\) .

Then, the vector \(\vec{a}+\vec{b}\) is represented by the third side of ΔABC
i. e.,
\(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=\overline{\mathrm{AC}}\) ……..(1)
or
\(\overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{AB}}=\overline{\mathrm{BC}}\)
and
\(\overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{BC}}=\overline{\mathrm{AB}}\)

This is known as the triangle law of addition.
Further, \(\overrightarrow{\mathrm{AC}}\) = – \(\overrightarrow{\mathrm{CA}}\)
So, (1) may be written as
\(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}\) = – \(\overrightarrow{\mathrm{CA}}\)

∴ \(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}\) = \(\overrightarrow{0}\)

When sides of a triangle ABC are taken in order, i.e., initial and terminal points coincides, then
\(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}\) = \(\overrightarrow{0}\)

2. Parallelogram law of Addition:
If the two vectors \(\vec{a}\) and \(\vec{b}\) are represented by the two adjacent sides OA and OB of a parallelogram OACB, then their sum \(\vec{a}\) + \(\vec{b}\) is represented in magnitude and direction by the diagonal OC of the parallelogram through their common point O.

i.e., \(\overline{\mathrm{OA}}+\overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{OC}}\)

Properties of Vector Addition:
Property 1: For two vectors \(\vec{a}\) and \(\vec{b}\), the sum is commutative i.e.,
\(\vec{a}+\vec{b}\) = \(\vec{b}+\vec{a}\)

Property 2: For three vectors \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) the sum of vectors is associative,- i.e.,
\((\vec{a}+\vec{b})+\vec{c}\) =\(\vec{a}+(\vec{b}+\vec{c})\)

7. Multiplication of a vector by a scalar:
Let \(\vec{a}\) be the given vector and λ be a scalar.
Product of λ and \(\vec{a}\) = λ\(\vec{a}\).
(i) When λ is +ve, \(\vec{a}\) and λ \(\vec{a}\) have the same sense of direction.
(ii) When λ is -ve, \(\vec{a}\) and λ\(\vec{a}\) are of the opposite directions.
Also, |λ\(\vec{a}\)| = |λ| |\(\vec{a}\)|

→ Additive Inverse of Vector \(\vec{a}\)
If there exists a vector – \(\vec{a}\) such that
\(\vec{a}\) + (- \(\vec{a}\)) = \(\vec{a}\) – \(\vec{a}\) = \(\vec{0}\), then – \(\vec{a}\) is called the additive inverse of \(\vec{a}\).

→ Unit Vector \(\vec{a}\)
Let λ = \(\frac{1}{|\vec{a}|}\) . So, | λ\(\vec{a}\)| = |λ| |\(\vec{a}\)|
= \(\frac{1}{|\vec{a}|}\) × |\(\vec{a}\)| = 1
∴ \(\frac{1}{|\vec{a}|}\) × |\(\vec{a}\)| = a, where |\(\vec{a}\)| ≠ 0 and \(\vec{a}\) is the unit vector.

8. Components of Vector:
Let us take the points A(1, 0, 0), B(0,1, 0) and 0(0, 0,1) on the co-ordinate axes OX, OY and OZ respectively.
Obviously |\(\overrightarrow{\mathrm{OA}}\)| = 1, |\(\overrightarrow{\mathrm{OB}}\)| = 1 and |\(\overrightarrow{\mathrm{OC}}\)| = 1.

Vectors \(\overrightarrow{\mathrm{OA}}\), \(\overrightarrow{\mathrm{OB}}\) and \(\overrightarrow{\mathrm{OC}}\) each having magnitude 1 are known as unit vectors. They are denoted by î, ĵ and k̂ respectively.

Consider the vector \(\overrightarrow{\mathrm{OP}}\), where P is the point (x, y, z). Now, OQ, OR, OS are the projections of OP on co-ordinates axes.

∴ OQ = x, OR = y, OS = z
\(\overrightarrow{\mathrm{OQ}}\) = xî, \(\overrightarrow{\mathrm{OR}}\) =yĵ, \(\overrightarrow{\mathrm{OS}}\) = zk̂.
⇒ \(\overrightarrow{\mathrm{OP}}\) – xî + yĵ +zk̂.
Also, |\(\overrightarrow{\mathrm{OP}}\)| = \(\sqrt{x^{2}+y^{2}+z^{2}}\) = |\(\vec{r}\)|

x, y, z are called the scalar components and xî, yĵ, and zk̂ are the vector components of vector \(\overrightarrow{\mathrm{OP}}\)

Some properties:
Let \(\vec{a}\) =a₁î + a₂ĵ + a₃k̂ and \(\vec{b}\) =b₁î +b₂ĵ +b₃k̂
1. \(\vec{a}\) + \(\vec{b}\) =(a₁ î + a₂ĵ + a₃k̂) + (b₁ î +b₂ĵ +b₃k̂)
⇒ (a₁ + b₁)î + (a₂ + b₂)ĵ + (a₃ + b₃)k̂

2. \(\vec{a}\) – \(\vec{b}\) or (a₁î + a₂ĵ +a₃k̂) = (b₁î + b₂ĵ + b₃k̂)

3. λ\(\vec{a}\) = λ(a₁î + a₂ĵ + a₃k̂)
⇒ (λa₁)î + (λa₂)ĵ+ (λa₃)k̂

4. \(\vec{a}\) and \(\vec{b}\) are collinear, if and only if there exists a non-zero scalar X such that \(\vec{b}\) = λ\(\vec{a}\), i.e.,

9. Vector joining two points:
Let P₁(x₁, y₁, z₁) and P₂(x₂, y₂, z₂) be the two points. Then, the vector joining the points P₁ and P₂ is \(\overline{\mathrm{P}_{1} \mathrm{P}_{2}}\)
Join P₁ and P₂ with O.

Now,

10. Dividing a line segment in a given ratio:
1. A line segment PQ is divided by a point R in the ratio m: n internally.

i.e., \(\frac{\mathrm{PR}}{\mathrm{RQ}}=\frac{m}{n}\)
If \(\vec{a}\) and \(\vec{b}\) are the position vectors of P and Q then the position vector \(\vec{r}\) of R is given by
\(\vec{r}\) = \(\frac{m \vec{b}+n \vec{a}}{m+n}\)

2. When R divides PQ externally,
\(\frac{\mathrm{PR}}{\mathrm{RQ}}=\frac{m}{n}\)

Replacing n by -n,
\(\vec{r}\) = \(\frac{m \vec{b}-n \vec{a}}{m-n}\)

Product of two vectors (Dot Product)
1. Scalar (or Dot) Product:
Scalar product of two vectors \(\vec{a}\) and \(\vec{b}\) is defined as \(\vec{a}\) . \(\vec{b}\) = | \(\vec{a}\) | \(\vec{b}\) | cos 0, where 0 is the angle between a and b (0,< θ < π).

2. From definition we derive:
(a) \(\vec{a}\).\(\vec{b}\) is a scalar quantity.

(b) When θ = 0, \(\vec{a}\).\(\vec{b}\) = |\(\vec{a}\)| |\(\vec{a}\).\(\vec{b}\)|.
Also, \(\vec{a}\). \(\vec{a}\) = |\(\vec{a}\)|2 = |\(\vec{a}\)|.

(c) When θ = \(\frac{π}{2}\)
⇒ When \(\vec{a}\) ⊥ \(\vec{a}\), \(\vec{a}\). \(\vec{b}\) = 0.

(d) When either \(\vec{a}\) = 0 or \(\vec{b}\) = 0, \(\vec{a}\). \(\vec{b}\) = 0.

(e) cos θ = \(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\)

3. Properties of scalar Product
(a) Scalar product is commutative, i.e.,
\(\vec{a}\) . \(\vec{b}\) = \(\vec{b}\).\(\vec{a}\)

(b) If a is scalar, then
(α \(\vec{a}\) ). \(\vec{b}\) = α(\(\vec{a}\).\(\vec{b}\)) – \(\vec{a}\). (α\(\vec{b}\))

11. Projection of vector along a directed line:
Let the vector \(\overrightarrow{\mathrm{AB}}\) makes ar angle θ with directed line l.
Projection of AB on l = |\(\overrightarrow{\mathrm{AB}}\)| cos θ = \(\overrightarrow{\mathrm{AC}}\) = \(\overrightarrow{\mathrm{P}}\).

The vector is \(\overrightarrow{\mathrm{P}}\) is called the projection vector. Its magnitude is
|\(\overrightarrow{\mathrm{P}}\)|, which is known as projection of vector \(\overrightarrow{\mathrm{AB}}\).

The angle θ between \(\overrightarrow{\mathrm{AB}}\) and \(\overrightarrow{\mathrm{AC}}\) is given by
cos θ = \(\frac{\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{AC}}}{|\overline{\mathrm{AB}}||\overline{\mathrm{AC}}|}\)

Now, projection AC = \(\overrightarrow{\mathrm{AB}}\) cos θ = cos θ \(\frac{\overline{\mathrm{AB}} \cdot \overrightarrow{\mathrm{AC}}}{|\mathrm{AC}|}\)
= \(\overrightarrow{\mathrm{AB}}\).(\(\frac{\overrightarrow{\mathrm{AC}}}{|\overrightarrow{\mathrm{AC}}|}\))

If, \(\overrightarrow{\mathrm{AB}}\) = \(\vec{a}\) and \(\overrightarrow{\mathrm{AC}}\) = \(\vec{p}\), then
AC = \(\vec{a}\)(\(\frac{\vec{p}}{|\vec{p}|}\)) = \(\vec{a}\).\(\vec{p}\)

Thus, the projection of \(\vec{a}\) on \(\vec{b}\)
= \(\vec{a}\)(\(\frac{\vec{b}}{|\vec{b}|}\)) = \(\vec{a}\).\(\vec{b}\).

Note: If α, β, γ are the direction angles of the vector
\(\vec{a}\) = (a₁î + a₂ĵ + a₃k̂), the direction cosines of a are given as

cos α = \(\frac{\vec{a} \cdot \hat{i}}{|\vec{a}||\hat{i}|}=\frac{a_{1}}{|\vec{a}|}\)
cos β = \(\frac{a_{2}}{|\vec{a}|}\)
cos γ = \(\frac{a_{3}}{|\vec{a}|}\)

12. Vector Product of two vectors:
1. Definition: The vector product of two non-zero vectors \(\vec{a}\) and \(\vec{b}\), denoted by \(\vec{a}\) × \(\vec{b}\) is defined as
\(\vec{a}\) × \(\vec{b}\) = |\(\vec{a}\) ||\(\vec{b}\) |sin θ.n̂
where θ is the angle between \(\vec{a}\) and \(\vec{b}\), 0 ≤ θ ≤ π. Unit vector n̂ is perpendicular to both vectors \(\vec{a}\) and \(\vec{b}\), such that \(\vec{a}\), \(\vec{b}\) and n̂ form a right handed system.

2. Note:
(a) \(\vec{a}\) × \(\vec{b}\) = | \(\vec{a}\) | | \(\vec{b}\) | sin θ. n̂

1. If \(\vec{a}\) = 0 or \(\vec{b}\) = 0, \(\vec{a}\) × \(\vec{a}\) = \(\vec{0}\)
2. \(\vec{a}\) || \(\vec{b}\), \(\vec{a}\) × \(\vec{b}\) = 0

(b) \(\vec{a}\) × \(\vec{b}\) = – \(\vec{b}\) × \(\vec{a}\)
\(\vec{a}\) × \(\vec{b}\) ≠ \(\vec{b}\) × \(\vec{a}\)
⇒ vector product is not commutative.

(c) When θ = \(\frac{π}{2}\), \(\vec{a}\) × \(\vec{b}\) = |\(\vec{a}\)| × |\(\vec{b}\)| × n̂
or
| \(\vec{a}\) × \(\vec{b}\) | = |\(\vec{a}\)||\(\vec{b}\)|

(d) î × î = ĵ × ĵ = k̂ × k̂ = 0
and î × ĵ = k̂, ĵ × k̂ = ĵ, k̂ × î = ĵ.
⇒ĵ × î = – k̂, k̂ × ĵ = -î , î × k̂ = – ĵ.

(e) sin θ = \(\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}\)

(f) If \(\vec{a}\) and \(\vec{b}\) represent adjacent sides of a parallelogram, then its area |\(\vec{a}\) × \(\vec{b}\)|.

(g) If \(\vec{a}\) and \(\vec{b}\) represent adjacent sides of a triangle, then its area = \(\frac{1}{2}\) |\(\vec{a}\) × \(\vec{b}\)|

(h) Distributive Property
\(\vec{a}\) × (\(\vec{b}\) + \(\vec{c}\)) = \(\vec{a}\) × \(\vec{b}\) + \(\vec{b}\) x× \(\vec{c}\)
1. Let α be a scalar, then
α(\(\vec{a}\) × \(\vec{b}\)) = (α\(\vec{a}\)) × \(\vec{b}\) = \(\vec{a}\) × (α\(\vec{b}\))

2. \(\vec{a}\) = (a₁ î + a₂ ĵ + a₃ k̂) and
\(\vec{b}\) = (b₁ î + b₂ ĵ + b₃ k̂), then

\(\vec{a}\) × \(\vec{a}\) = \(\left|\begin{array}{lll}
\hat{i} & \hat{j} & \hat{k} \\
a_{1} & a_{2} & a_{3} \\
b_{1} & b_{2} & b_{3}
\end{array}\right|\)

1. PROPERTIES OF \TXHORS UNDER ADDITION

(i) Commutative Law. \(\vec{a}+\vec{b}=\vec{b}+\vec{a}\)
(ii) Associative Law, \(\vec{a}+(\vec{b}+\vec{c})=(\vec{a}+\vec{b})+\vec{c}\)
(iii) Additive Identity. \(\vec{a}+\overrightarrow{0}=\vec{a}=\overrightarrow{0}+\vec{a}\)
(iv) Additive Inverse. \(\vec{a}+(-\vec{a})=\overrightarrow{0}=(-\vec{a})+\vec{a}\)

2. \(\vec{AB}\) = (POSITION VECTOR OF B)-(POSITION VECTOR OF A).

3. SCALAR PRODUCT

(i) Def. The scalar product of \(\vec{a}\) and \(\vec{b}\) is defined as \(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta\), where ‘θ’ is the angle between \(\vec{a}\) and \(\vec{b}\)
(ii) Condition of Perpendicularity. If \(\vec{a}\) and \(\vec{b}\) are perpendicular, then \(\vec{a} \cdot \vec{b}\) = 0.
(iii) If \(\hat{i}, \hat{j}, \hat{k}\) are unit vectors, which are perpendicular to each other, then :
\(\hat{i} \cdot \hat{j}=\hat{j} \cdot \hat{k}=\hat{k} \cdot \hat{i}=0\) and \(\hat{i}^{2}=\hat{j}^{2}=\hat{k^{2}}\)

(iv) Properties:
(a) CommutativeLaw. \(\vec{a}\cdot \vec{b}=\vec{b} \cdot \vec{a}\)

(b) Associative Law, does not hold.

(c) Distributive Laws, \(\vec{a} \cdot(\vec{b}+\vec{c})=\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c}\) and \((\vec{b}+\vec{c}) =\vec{b} \cdot \vec{a}+\vec{c} \cdot \vec{a}\)

(d) \((m\vec{a}) \cdot \vec{b}=m(\vec{a} \cdot \vec{b})=\vec{a} \cdot(m \vec{b})\), m being a scalar.

(e) \(\vec{a} \cdot \vec{a}=a^{2}\)

4. VECTOR PRODUCT

(i) Definition. The vector product of \(\vec{a}\) and \(\vec{b}\) is given by: \({a} \times \vec{b}=|\vec{a} \| \vec{b}| \sin \theta \hat{n}\) , where ‘θ’ the angle between \(\vec{a}\) and \(\vec{b}\) and \(\hat{n}\) is a unit vector perpendicular to the plane of \(\vec{a}\) and \(\vec{b}\) .

(ii) When two vectors are parallel, then \(\vec{a} \times \vec{b}=\overrightarrow{0}\)

(iii) If \(\hat{i}, \hat{j}, \hat{k}\) are unit vectors, which are perpendicular to each other, then :
\(\hat{i} \times \hat{j}=\hat{k}=-\hat{j} \times \hat{i} ; \text { etc. } \hat{i} \times \hat{i}=\hat{j} \times \hat{j}=\hat{k} \times \hat{k}=\overrightarrow{0}\)

(iv) Properties :

(a) Commutative Law does not hold.

(b) \((m \vec{a}) \times \vec{b}=m(\vec{a} \times \vec{b})=\vec{a} \times(m b)\), m being a scalar.

(c) If \(\vec{a}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}, \vec{b}=b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}\), then \(\vec{a} \times \vec{b}\) = \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
a_{1} & a_{2} & a_{3} \\
b_{1} & b_{2} & b_{3}
\end{array}\right|\)

(d) Distributive Law. \(\vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}\)

5. SCALAR TRIPLE PRODUCT

(i) Def. If \(\vec{a}, \vec{b}, \vec{c}\) are three vectors, then the scalar product of \(\vec{a} \times \vec{b}\) with \(\vec{c}\) is called scalar triple product

(ii) Properties:
(a) Condition for three vectors to be coplanar. If \(\vec{a}, \vec{b}, \vec{c}\) are all coplanar vectors, then \(\left[\begin{array}{lll}
\rightarrow & \rightarrow & \rightarrow \\
a & b & c
\end{array}\right]\) = 0

(b) If any two of the three vectors \(\vec{a}, \vec{b}, \vec{c}\) are equal, then \([\vec{a} \vec{b} \vec{c}]\) = 0

(c) If any two of the three vectors \(\vec{a}, \vec{b}, \vec{c}\) are parallel, then \(\left[\begin{array}{ccc}
\rightarrow & \rightarrow & \rightarrow \\
a & b & c
\end{array}\right]\) = 0

(d) In the scalar triple product, the position of dot and cross can be interchanged at pleasure provided the cyclic order of the vectors is maintained.

(e) In the scalar triple product, the change in the cyclic order of die vectors changes the sign of the product.

Here we are providing Class 12 Maths Important Extra Questions and Answers Chapter 2 Inverse Trigonometric Functions. Class 12 Maths Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Maths Chapter 2 Important Extra Questions Inverse Trigonometric Functions

Inverse Trigonometric Functions Important Extra Questions Very Short Answer Type

Question 1.
Find the principal value of sin^-1 ( \(\frac { 1 }{ 2 }\) )
Solution:
\(\sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)
Hence, the principal value of sin^-1 ( \(\frac { 1 }{ 2 }\) ) is \(\frac{\pi}{6}\)

Question 2.
What is the principal value of:
cos^-1 (cos \(\frac{2 \pi}{3}\) + sin^-1 (sin \(\frac{2 \pi}{3}\) ) ?
Solution:

Question 3.
Find the principal value of:
tan^-1 (√3)- sec^-1 (-2). (A.I.C.B.S.E. 2012)
Solution:

Question 4.
Evaluate : tan ^-1 ( 2 cos (2 sin^-1 ( \(\frac{1}{2}\) )))
Solution:

Question 5.
Find the value of tan^-1(√3) – cot^-1(-√3). (C.B.S.E. 2018)
Solution:
tan^-1(√3) – cot^-1(—√3)
\(=\frac{\pi}{3}-\left(\pi-\frac{\pi}{6}\right)=-\frac{\pi}{2}\)

Question 6.
If sin^-1 ( \(\frac { 1 }{ 3 }\) ) + cos^-1 x = \(\frac{\pi}{2}\), then find x.(C.B.S.E. 2010C)
Solution:
sin^-1 ( \(\frac { 1 }{ 3 }\) ) + cos^-1 x = \(\frac{\pi}{2}\)
⇒ x = 1/3
[sin^-1 x + cos^-1 x = \(\frac{\pi}{2}\)

Question 7.
If sec^-1 (2) + cosec^-1 (y) = \(\frac{\pi}{2}\) , then find y.
Solution:
sec^-1 (2) + cosec^-1 (y) = \(\frac{\pi}{2}\)
⇒ y = 2 [∵ sec^-1 x + cosec^-1 x= \(\frac{\pi}{2}\) ]

Question 8.
Write the value of sin [ \(\frac{\pi}{3}\) – sin ^-1 ( \(\frac { -1 }{ 2 }\) ) ]
Solution:
sin [ \(\frac{\pi}{3}\) – sin^-1 ( \(\frac { -1 }{ 2 }\) ) ]
= sin [ \(\frac{\pi}{3}\) + sin^-1 ( \(\frac { 1 }{ 2 }\) ) ]
[∵ sin^-1 (-x) = -sin^-1x]
= sin ( \(\frac{\pi}{3}+\frac{\pi}{6}\) ) = sin \(\frac{\pi}{2}\) = 1

Question 9.
Prove the following:

Solution:

Question 10.
If tan^-1 x + tan^-1y = \(\frac{\pi}{4}\) , xy < 1, then write the value of the x + y + xy (A.I.C.B.S.E. 2014)
Solution:
We have tan^-1 x + tan^-1y = \(\frac{\pi}{4}\)

Hence x + y + xy = 1.

Question 11.
Prove that: 3 sin^-1 x = sin^-1(3x – 4x²);
x ∈ [\(\frac { -1 }{ 2 }\) , \(\frac { 1 }{ 2 }\)] (C.B.S.E 2018)
Solution:
To prove: 3 sin^-1 x = sin^-1(3x – 4x²)
Put sin^-1 x = θ
so that x = sin θ.
RHS = sin^-1 (3 sin θ – 4 sin³ θ)
= sin^-1 (sin 3θ) = 3θ = 3 sin^-1x = LHS.

Inverse Trigonometric Functions Important Extra Questions Short Answer Type

Question 1.
Express sin^-1 ( \(\frac{\sin x+\cos x}{\sqrt{2}}\) )
Where \(-\frac{\pi}{4}\) < x < \(\frac{\pi}{4}\), in the simples form.
Solution:

Question 2.
Prove that :
\(\cos ^{-1} \frac{12}{13}+\sin ^{-1} \frac{3}{5}=\sin ^{-1} \frac{56}{65}\)   (A.I.C.B.S.E. 2019; C.B.S.E. 2010)
Solution:

Question 3.
Prove that :
sin ^-1 \(\frac{8}{17}\) + cos ^-1 \(\frac{4}{5}\) = cos^-1 \(\frac{36}{77}\) (A.I.C.B.S.E. 2019)
Solution:
L.H.S = sin^-1 \(\frac{8}{17}\) + cos^-1 \(\frac{4}{5}\)
= tan^-1 \(\frac{8}{15}\) + tan^-1\(\frac{3}{4}\)

= R.H.S

Question 4.
Solve the following equation:
\(\tan ^{-1}\left(\frac{x+1}{x-1}\right)+\tan ^{-1}\left(\frac{x-1}{x}\right)=\tan ^{-1}(-7)\) (A.I.C.B.S.E. 2019; C)
Solution:

⇒ 2x² – 8x + 8 = 0
⇒ x² – 4x + 4 = 0
⇒ (x – 2)² = 0.
Hence, x = 2.

Question 5.
Solve the following equation:
2 tan^-1(sin x) = tan^-1 (2 sec x), x ≠ \(\frac{\pi}{2}\)   (C.B.S.E. (F) 2012)
Solution:
2 tan^-1(sinx) = tan^-1 (2 secx)

⇒ tan ^-1 (2sec x tan x) = tan^-1 (2sec x)
⇒ 2 sec x tan x = 2 sec x
tan x = 1 [∵ sec x ≠ 0 ]
Hence, x= \(\frac{\pi}{2}\)

Question 6.
Solve the following equation
cos (tan^-1 x) = sin ( cot ^-1\(\frac { 3 }{ 4 }\) )
(A.I.C.B.S.E. 2013)
Solution:
We have:
cos (tan^-1 x) = sin ( cot ^-1\(\frac { 3 }{ 4 }\) )
cos (tan^-1 x) = sin ( sin ^-1\(\frac { 4 }{ 5 }\) )
cos (tan^-1 x) = \(\frac { 4 }{ 5 }\)
tan^-1 x = cos^-1\(\frac { 4 }{ 5 }\)
⇒ tan^-1 x = tan ^-1\(\frac { 3 }{ 4 }\)
Hence x = \(\frac { 3 }{ 4 }\)

Question 7.
Prove that
3cos^-1 x = cos^-1 (4x³ – 3x), x ∈ [ \(\frac { 1 }{ 2 }\) , 1 ]
Solution:
Put x = cos θ in RHS
As 1/2 ≤ x ≤ 1
RHS = cos^-1 (4cos³ θ – 3cos θ),
= cos^-1 (cos 3θ) = 3θ = 3cos^-1 x = L.H.S

Inverse Trigonometric Functions Important Extra Questions Long Answer Type 1

Question 1.
prove that \(\frac { 1 }{ 2 }\) ≤ x ≤ 1, then
\(\cos ^{-1} x+\cos ^{-1}\left[\frac{x}{2}+\frac{\sqrt{3}-3 x^{2}}{2}\right]=\frac{\pi}{3}\) (CBSE. Sample paper 2017 – 18)
Solution:

Question 2.
Find the value of :
\(\tan ^{-1}\left(\frac{x}{y}\right)-\tan ^{-1}\left(\frac{x-y}{x+y}\right)\)
Solution:

Question 3.
Prove that :
tan ^-1( \(\frac{1}{2}\) ) + tan ^-1 ( \(\frac{1}{5}\) ) + tan ^-1( \(\frac{1}{8}\) ) = \(\frac{\pi}{4}\) (C.B.S.E. 2013: A.I.C.B.S.E. 2011)
Solution:

Question 4.

Solution:

Question 5.

Solution:
Put cos ^-1 (a/b) = θ so that cos θ = a/b.

Question 6.

Solution:

Question 7.

( A.I.C.B.S.E. 2016)
Solution:

Question 8.
\(\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{8}=\frac{\pi}{4}\) (C.B.S.E. 2016)
Solution:

Question 9.
Solve for x : 2 tan^-1 (cos x ) = tan^-1(2 cosec x). (C.B.S.E. 2016)
Solution:
2 tan^-1 (cos x ) = tan^-1 (cos x) + tan^-1 (cos x)


= tan^-1 (2cot x cosec x) …(1)
Now 2 tan^-1(cos x) = tan^-1(2 cosec x)
⇒ tan^-1(2 cot x cosec x) = tan^-1 (2cosec x)
[Using (1)]
⇒ 2 cot x cosec x = 2 cosec x
⇒ cotx cosecx = cosec x
⇒  sin x = tan x sin x
⇒either sin x = 0 or tan x = 1.
Hence, x = nπ ∀ n ∈ Z or x
= πm + \(\frac{\pi}{4}\) ∀ m ∈ Z

Question 10.
If tan^-1 \(\left(\frac{x-2}{x-4}\right)\) + tan^-1 \(\left(\frac{x+2}{x+4}\right)\) = \(\frac{\pi}{4}\) find the value of ‘x’   (A.I.C.B.S.E. 2014)
Solution:
We have

⇒ 2x² – 16 = -12
⇒ 2x² = 16 – 12
⇒ 2x² = 4
⇒ x² = 2
Hence x = ±√2

Question 11.
If tan^-1 \(\left(\frac{x-3}{x-4}\right)\) + tan^-1 \(\left(\frac{x+3}{x+4}\right)\) = \(\frac{\pi}{4}\) find the value of ‘x’ (A.I.C.B.S.E. 2017)
Solution:

⇒ 2x² – 24 = -7
⇒ 2x² = 17
⇒ 2x² = 4
⇒ x² = 17/2
Hence x = \(\pm \sqrt{\frac{17}{2}}\)

Question 12.
Prove that : \(\tan ^{-1} \sqrt{x}=\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right)\)
x∈ [0,1] (C.B.S.E. 2019C)
Solution:
Let tan ^-1 √x = θ
So that √x = tan θ i.e. x = tan²θ

Question 13.
Solve for x : tan ^-1 (2x) + tan ^-1 (3x) = \(\frac{\pi}{4}\) (C.B.S.E. 2019)
Solution:
The given equation is
tan ^-1 (2x) + tan ^-1 (3x) = \(\frac{\pi}{4}\) …. (1)

⇒ 5x = 1 – 6x²
⇒ 6x² + 5x – 1 = 0
⇒ x = -1 or x = \(\frac{1}{6}\)
Hence, x = \(\frac{1}{6}\)
[∵ x = -1 does not satisfy (1)]

Question 14.
Solve : tan^-1 4x + tan^-1 6x = \(\frac{\pi}{4}\) (C.B.S.E. 2019)
Solution:
We have tan^-1 4x + tan^-1 6x = \(\frac{\pi}{4}\)

⇒ 10x = 1 – 24x²
⇒ 24x² + 10x – 1 = 0
⇒ 24x²+ 12x – 2x – 1 = 0
⇒ 12x(2x + 1) – 1 (2x + 1) = 0
⇒ (2x+1)(12x – 1) = 0
⇒ 2x + 1 = 0 or 12x – 1 = 0
x = \(-\frac{1}{2}\) or x = \(\frac{1}{2}\).
Hence, x = \(-\frac{1}{2}\) or \(\frac{1}{12}\)
As x = \(-\frac{1}{2}\) does not satisfy the given equation.
Hence x = \(-\frac{1}{12}\)

Question 15.
If (tan^-1 x )² + (cot^-1 x )² = \(\frac{5 \pi^{2}}{8}\) , then find ‘x’ (C.B.S.E. 2015)
Solution:
We have
(tan^-1 x )² + (cot^-1 x )² = \(\frac{5 \pi^{2}}{8}\)
Put tan^-1 x = t so that cot^-1x = \(\frac{\pi}{2}\) – t
∴ (1) become : t² + (\(\frac{\pi}{2}\) – t)² = \(\frac{5 \pi^{2}}{8}\)

When tan^-1x = \(\frac{3 \pi}{4}\)
then x = tan \(\frac{3 \pi}{4}\) = -1.
When tan^-1 x = \(-\frac{\pi}{4}\),
then x = tan(\(-\frac{\pi}{4}\)) = -1.
Hence, x = -1.

Question 16.
Write : tan ^-1\(\frac{1}{\sqrt{x^{2}-1}}\) , |x| > 1 in the simplest form.
Solution:
Put x = sec θ

Question 17.
Prove that

(C.B.S.E. 2012)
Solution:

Question 18.
Prove that :
tan^-1x + tan^-1 \(\frac{2 x}{1-x^{2}}\) = tan ^-1\(\frac{3 x-x^{3}}{1-3 x^{2}}\) (N.C.E.R.T
Solution:
Put x = tan θ so that θ = tan ^-1 x

= 3θ = 3 tan^-1x = tan^-1x + 2tan^-1x
= tan^-1x + tan^-1\(\frac{2 x}{1-x^{2}}\)
= L.H.S.

Question 19.
Simplify : tan^-1 [ \(\left[\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right]\)]
If \(\frac{a}{b}\) tan x > -1    (N.C.E.R.T)
Solution:

Question 20.
Prove that tan ^-1 [ \(\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\) ]
= \(\frac{\pi}{4}+\frac{1}{2}\) cos^-1 x ; \(-\frac{1}{\sqrt{2}}\) ≤ x ≤ 1. (C.B.S.E. 2019C)
Solution:

Question 21.
Show that

(N.C.E.R.T.A.I. CBSE 2014)
Solution:

Question 22.
cos[tan^-1{sin (cot^-1x)}] = \(\sqrt{\frac{1+x^{2}}{2+x^{2}}}\)
(A.I. C.B.S.E. 2010)
Solution:
Put cot^-1 = θ so that x = cot θ
∴ sin θ = \(\frac{1}{\sqrt{1+x^{2}}}\)

Question 23.
Find the value of sin (2 tan^-1 1/4) + cos (tan^-1 2√2)   (C.B.S.E. Sample Paper 2018 – 2019)
Solution:
Put tan^-1 1/4 = θ so that θ = 1/4
Now, sin 2θ = \(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\)

To evaluate cos(tan^-1 2√2) :
Put tan^-12√2 = Φ
So that tan Φ = 2√2
cos Φ = 1/3
Hence, sin (2tan^-1(1/4)) + cos (tan^-12√2)
= sin 2θ + cos Φ = \(\frac{8}{17}+\frac{1}{3}=\frac{41}{51}\)
[Using (1) and (2)]

Question 24.
Solve for x:
tan^-1(x – 1) + tan^-1x + tan^-1(x + 1) = tan^-1(3x).    (A.I.C.B.S.E. 2016)
The given equation is:
tan^-1(x —1) + tan^-1(x) + tan^-1(x + 1) = tan^-13x
tan^-1(x— 1)+tan^-1(x+ 1) = tan^-13x – tan^-1


⇒ 1 + 3x² = 2 – x²
⇒ 4x² = 1
⇒ x² = 1/4
⇒ x = \(\pm \frac{1}{2}\)
Hence, x = 0, \(\pm \frac{1}{2}\)

Here we are providing Class 12 Maths Important Extra Questions and Answers Chapter 3 Matrices. Class 12 Maths Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Maths Chapter 3 Important Extra Questions Matrices

Matrices Important Extra Questions Short Answer Type

Question 1.
Write the element a₂₃ of a 3 x 3 matrix A = [a_ij] whose elements atj are given by : \(\frac{|i-j|}{2}\)
Solution:
We have [a_ij] = \(\frac{|i-j|}{2}\)
∴ a₂₃ = \(\frac{|2-3|}{2}=\frac{|-1|}{2}=\frac{1}{2}\)

Question 2.
For what value of x is
\(\left[\begin{array}{lll}
1 & 2 & 1
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 0 \\
2 & 0 & 1 \\
1 & 0 & 2
\end{array}\right]\left[\begin{array}{l}
0 \\
2 \\
x
\end{array}\right]=0\) ? (C.B.S.E. 2019(C))
Answer:
We have
\(\left[\begin{array}{lll}
1 & 2 & 1
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 0 \\
2 & 0 & 1 \\
1 & 0 & 2
\end{array}\right]\left[\begin{array}{l}
0 \\
2 \\
x
\end{array}\right]=0\)

[1 + 4 + 1 2 + 0 + 0 0 + 2 + 2] \(\left[\begin{array}{l}
0 \\
2 \\
x
\end{array}\right]\) = 0
[6 2 4 ]\(\left[\begin{array}{l}
0 \\
2 \\
x
\end{array}\right]\) = 0
⇒ [0 + 4 + 4x] =0
⇒ [4 + 4x] = [0]
⇒ 4 + 4x = 0.
Hence, x = -1.

Question 3.
Find a matrix A such that 2A – 3B + 5C = 0,
Where B = \(\left[\begin{array}{ccc}
-2 & 2 & 0 \\
3 & 1 & 4
\end{array}\right]\) and C = \(\left[\begin{array}{ccc}
2 & 0 & -2 \\
7 & 1 & 6
\end{array}\right]\)
Solution:
Here, 2A – 3B + 5C = 0
⇒ 2A = 3B – 5C

Question 4.
If A = \(\left(\begin{array}{l}
\cos \alpha-\sin \alpha \\
\sin \alpha-\cos \alpha
\end{array}\right)\) , then for what value of ‘α’ is A an identity matrix? (C.B.S.E. 2010)
Solution:
Here A = \(\left(\begin{array}{l}
\cos \alpha-\sin \alpha \\
\sin \alpha-\cos \alpha
\end{array}\right)\)
Now A = I = \(\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)\) when
cos α = 1 and sin α = 0.
Hence, α = 0.

Question 5.
Find the values of x, y, z and t, if:
\(2\left[\begin{array}{ll}
x & z \\
y & t
\end{array}\right]+3\left[\begin{array}{rr}
1 & -1 \\
0 & 2
\end{array}\right]=3\left[\begin{array}{ll}
3 & 5 \\
4 & 6
\end{array}\right]\)
Solution:
We have :

⇒ 2x + 3 = 9 …………. (1)
2z – 3 = 15 …………. (2)
2y = 12 …………. (3)
2t + 6 = 18 …………. (4)

From (1), ⇒ 2x = 9 – 3
⇒ 2x = 6
⇒ x = 3.

From (3) 2y = 12
⇒ y = 6.

From (2), ⇒ 2z – 3 = 15
⇒ 2z = 18
⇒ z = 9.

From (4), 2t + 6 = 18
⇒ 2t = 12
⇒ t = 6.
Hence, x = 3,y = 6, z = 9 and t = 6.

Question 6.
If A = \(\left[\begin{array}{rrr}
2 & 0 & 1 \\
2 & 1 & 3 \\
1 & -1 & 0
\end{array}\right]\), then find (A² – 5A). (CBSE 2019)
Solution:
We have A = \(\left[\begin{array}{rrr}
2 & 0 & 1 \\
2 & 1 & 3 \\
1 & -1 & 0
\end{array}\right]\)
Then A² = AA

Question 7.
If A = \(\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\) and I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) , find k so that A² = 5A + b kI (C.B.S.E. Sample Paper 2018-2019)
Solution:

⇒ 8 = 15 + k and 3 = 10 + k
⇒ k = -1 and k = -7.
Hence, k – -7.

Question 8.
If A and B are symmetric matrices, such that AB and BA are both defined, then prove that AB – BA is a skew symmetric matrix. (A.I.C.B.S.E. 2019)
Solution:
Since A and B are symmetric matrices,
∴ A’ = A and B’ = B …(1)
Now,(AB – BA)’= (AB)’ – (BA)’
= B’A’ – A’B’
= BA – AB [Using (1)]
= – (AB – BA).
Hence, AB – BA is a skew-symmetric matrix.

Question 9.
For the matrix A = \(\left[\begin{array}{ll}
2 & 3 \\
5 & 7
\end{array}\right]\) find (A + A’) and show that it is a symmetric matrix. (A.I.C.B.S.E. 2019)
Solution:

Now (A + A’)’ = \(\left[\begin{array}{cc}
4 & 8 \\
8 & 14
\end{array}\right]\) = (A + A’)
Hence (A + A’) is symmetric

Question 10.
If the matrix A = \(\left[\begin{array}{ccc}
0 & a & -3 \\
2 & 0 & -1 \\
b & 1 & 0
\end{array}\right]\) is skew symmetric, find the values of ‘a’ and ‘b’: (C.B.S.E. 2018)
Solution:

Comparing, 2 = -a ⇒ a = -2
and – 3 = -b ⇒ b = 3.
Hence, a = -2 and b = 3.

Matrices Important Extra Questions Long Answer Type 1

Question 1.
Find the values of a, b, c and d from the following equation :
\(\left[\begin{array}{cc}
2 a+b & a-2 b \\
5 c-d & 4 c+3 d
\end{array}\right]=\left[\begin{array}{cc}
4 & -3 \\
11 & 24
\end{array}\right]\) (N.C.E.R.T)
Solution:
We have
\(\left[\begin{array}{cc}
2 a+b & a-2 b \\
5 c-d & 4 c+3 d
\end{array}\right]=\left[\begin{array}{cc}
4 & -3 \\
11 & 24
\end{array}\right]\)
Comparing the corresponding elements of two given matrices, we get:
2a + b = 4 …(1)
a-2b = – 3 …(2)
5c-d = 11 …(3)
4c + 3d = 24 …(4)
Solving (1) and (2):
From (1),
b = 4 – 2a …(5)
Putting in (2), a – 2 (4 – 2a) = – 3
⇒ a – 8 + 4a = -3
⇒ 5a = 5
⇒ a = 1.
Putting in (5),
b = 4 – 2(1) = 4 – 2 = 2.
Solving (3) and (4):
From (3),
d = 5c- 11 …(6)
Putting in (4),
4c+ 3 (5c- 11) = 24
⇒ 4c + 15c – 33 = 24
⇒ 19c = 57
⇒ c = 3.
Putting in (6),
d = 5 (3) – 11 = 15 – 11 = 4.
Hence, a = 1, b = 2, c = 3 and d = 4.

Question 2.
If \(\left[\begin{array}{rrr}
9 & -1 & 4 \\
-2 & 1 & 3
\end{array}\right]\) = A + \(\left[\begin{array}{rrr}
1 & 2 & -1 \\
0 & 4 & 9
\end{array}\right]\) then find the matrix A. (C.B.S.E. 2013)
Solution:

Comparing:
9 = a₁₁ + 1 – 1 = a₁₂ + 2,
4 = 1₁₃ – 1, -2 = a₂₁
1 = a₂₂ + 4, and 3 = a₂₃ + 9
a₁₁ = 8, a₁₂ = – 3,
a₁₃ = 5, a₂₁ = -2
a₂₂ = – 3, and a₂₃ = – 6.
Hence, A = \(\left[\begin{array}{rrr}
8 & -3 & 5 \\
-2 & -3 & -6
\end{array}\right]\)

Question 3.
If A = \(=\left[\begin{array}{rr}
2 & 2 \\
-3 & 1 \\
4 & 0
\end{array}\right]\) B = \(\left[\begin{array}{ll}
6 & 2 \\
1 & 3 \\
0 & 4
\end{array}\right]\)find the matrix C such that A + B + C is a zero matrix.
Solution:

Comparing :
8 + c₁₁ = 0 ⇒ c₁₁ = -8,
4 + C₁₂ = 0 ⇒ C₁₂ = -4,
– 2 + C₂₁ = 0 ⇒ C₂₁ = 2
4 + C₂₂ = 0 ⇒ C₂₂ =- 4,
4 + c₃₁ = 0 ⇒ C₃₁ = -4
and 4 + c₃₂ = 0 ⇒ C₃₂ = -4.
Hence, C = \(\left[\begin{array}{rr}
-8 & -4 \\
2 & -4 \\
-4 & -4
\end{array}\right]\)

Question 4.
If A = \(\left[\begin{array}{rr}
8 & 0 \\
4 & -2 \\
3 & 6
\end{array}\right]\) B = \(\left[\begin{array}{rr}
2 & -2 \\
4 & 2 \\
-5 & 1
\end{array}\right]\) then find the matrix ‘X’, of order 3 x 2, such that 2A + 3X = 5B. (N.C.E.R.T.)
Solution:
We have : 2A + 3X = 5B
⇒ 2A + 3X-2A = 5B-2A
⇒ 2A-2A + 3X = 5B-2A
⇒ (2A – 2A) + 3X = 5B – 2A
⇒ O + 3X = 5B – 2A
[ ∵ – 2A is the inverse of2A]
⇒ 3X = 5B – 2A.
[ ∵ O is the additive identity]
Hence, X = \(\frac{1}{3}\)(5B – 2A)

Question 5.
If A is a square matrix such that A2 = A, then write the value of 7A – (I + A)3, where I is an identity matrix. (A.I.C.B.S.E. 2014)
Solution:
(I + A)² = (I + A) (I + A)
= II + IA + AI + AA
= I + A + A + A²
= I + 2A + A [∵ A² = A]
= I + 3A …(1)
∴ (I + A)³ = (I + A)² (I + A)
= (I + 3A) (I + A) [Using (1)]
= II + IA + 3AI + 3AA
= I + A + 3A + 3A²
= I + A + 3A + 3A [∵ A² = A]
= I + 7 A …(2)
Hence, 7A – (I + A)³ = 7A – (I + 7A)
[Using (2)]
= -I.

Question 6.
Compute the indicated product:
\(\left[\begin{array}{lll}
2 & 3 & 4 \\
3 & 4 & 5 \\
4 & 5 & 6
\end{array}\right]\left[\begin{array}{rrr}
1 & -3 & 5 \\
0 & 2 & 4 \\
3 & 0 & 5
\end{array}\right]\) (NCERT)
Solution:

Number of columns of A = Number of rows of B = 3.
Thus AB is defined and is a 3 x 3 matrix.


Question 7.
IF A = \(\), show that A² – 6A² + 7A + 2I = 0   (N.C.E.R.T)
Solution:

Question 8.
Prove the following by the principle of Mathematical Induction:
If A = \(\left[\begin{array}{ll}
3 & -4 \\
1 & -1
\end{array}\right]\)
than An = \(\left[\begin{array}{cc}
1+2 n & -4 n \\
n & 1-2 n
\end{array}\right]\) (N.C.E.R.T)
Solution:
Set I. When n = 1

Thus the result is true when n = 1.
Step II. Let us assume that the result is true for any natural number m, where 1 ≤ m ≤ n

Thus the result is true when n = m+ 1.
Hence, by Mathematical Induction, the required result is true for all n ∈ N.

Question 9.
If A = \(\left[\begin{array}{r}
-2 \\
4 \\
5
\end{array}\right]\), B = [1 3 -6], then verify that (AB)’ = B’A’ (N.C.E.R.T.)
Solution:
We have :
A = \(\left[\begin{array}{r}
-2 \\
4 \\
5
\end{array}\right]\), B = [1 3 -6]

From (1) and (2), (AB)’ = B’A’
which verifies the result.

Question 10.
By using elementary transformations, find the inverse of the matrix A = \(\left[\begin{array}{ll}
1 & 3 \\
2 & 7
\end{array}\right]\)
(N.C.E.R.T.)
Solution:
(By Elementary Row Transformations)
We know A = I₂A

Matrices Important Extra Questions Long Answer Type 2

Question 1.
Two farmers Ram Kishan and Gurucharan Singh cultivate only three varieties of rice namely Basmati, Permal and Naura. The sale (in ?) of these varieties of rice by both the farmers in the month of September and October are given by the following matrices A and B :

Find :
(i) What were combined sales in September and October for each farmer in each variety?
(ii) What was the change in sales from September to October?
(iii) If both farmers receive 2% profit on gross sales, compute the profit for each farmer and for each variety in October. (N.C.E.R.T.)
Solution:
(i) Combined sales in September and October :

(ii) Change in sales from September to October:

(iii) 2% of B = \(\frac{2}{100}\) x B = 0.02 B

Hence, Ram Kishan receives ₹ 100, ₹ 200 and ₹ 120 as profit and Gurucharan Singh receives ₹ 400, ₹200 and ₹ 200 in each variety of rice in the month of October.

Question 2.
Three schools A, B and C organised a mela for collecting funds for helping the rehabilitation of flood victims. They sold handmade fans, mats and plates from recycled meterial at a cost of ₹25, ₹100 and ₹50 each. The number of articles sold are given below :

Find the funds collected by each school separately by selling the above articles. Also, find the total fund collected for the purpose.
(C.B.S.E. 2015)
Solution:
Quantity matrix, A = \(\left[\begin{array}{lll}
40 & 25 & 35 \\
50 & 40 & 50 \\
20 & 30 & 40
\end{array}\right]\)
Cost Matrix B = [laatex]\left[\begin{array}{c}
25 \\
100 \\
50
\end{array}\right][/latex]
Total fund = Quantity matrix x Cost matrix

Fund collected by school A = ₹5,250
Fund collected by school B = ₹7,750
Fund collected by school C = ₹5,500 and total fund collected = 5250 + 7750 + 5500
= ₹18,500.

Question 3.
If A = \(\left(\begin{array}{rr}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right)\), find α satisfing 0 < α < when A + A^T = √2I₂
Where A^T is transpose of A. (A.I.C.B.S.E. 2016)
Solution:

Question 4.
If \(\left(\begin{array}{cc}
a+b & 2 \\
5 & b
\end{array}\right)\) = \(\left(\begin{array}{ll}
6 & 5 \\
2 & 2
\end{array}\right)^{\prime}\) (C.B.S.E. 2010C)
Solution:

Comparing:
a + b – 6 ………….. (1)
and b = 2 ………….(2)
Putting the value of b from (2) in (1),
a + 2 = 6.
Hence, a = 4.

Question 5.
Express the matrix A as the sum of a symmetric and a skew-symmetric matrix, where :
A = \(\left[\begin{array}{rrr}
3 & -2 & -4 \\
3 & -2 & -5 \\
-1 & 1 & 2
\end{array}\right]\) (A.I.C.B.S.E. 2010)
Solution:
We have

Question 6.
Find the inverse of the following matrix, using elementary operations :
A = \(\left[\begin{array}{ccc}
1 & 2 & -2 \\
-1 & 3 & 0 \\
0 & -2 & 1
\end{array}\right]\)  (C.B.S.E. 2019)
Solution:
We know that A + I₃A\

Question 7.
Find the inverse of the following matrix, using elementary transformation:
\(\left[\begin{array}{rrr}
2 & -1 & 3 \\
-5 & 3 & 1 \\
-3 & 2 & 3
\end{array}\right]\)  (C.B.S.E. Sample Paper 2017-18)
Solution:
We know that A = IA

Question 8.
If A = \(\left[\begin{array}{rrr}
2 & 1 & 1 \\
1 & 0 & 1 \\
0 & 2 & -1
\end{array}\right]\) , find the inverse of A, using elementary row transformations and hence solve the following equation:
XA = [1 0 1 ] (C.B.S.E. Sample Paper 2017-18)
Solution:
(i) We know that A = I₃ A

Here, x = 0, y = 1, z = 0

Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

Class 12 Physics Chapter 14 Important Extra Questions Semiconductor Electronics: Materials, Devices and Simple Circuits

Semiconductor Electronics Important Extra Questions Very Short Answer Type

Question 1.
Draw the energy band diagram for a p-type semiconductor.
Answer:
The energy level diagram is shown below.

Question 2.
Draw the voltage-current characteristic of a p-n junction diode in forwarding bias and reverse bias.
Answer:
The characteristics are as shown.

Question 3.
Draw the voltage-current characteristic for a Zener diode.
Answer:
The V-l characteristic of the Zener diode is as shown.

Question 4.
Draw the energy band diagram for n-type semiconductor.
Answer:
The diagram is as shown.

Question 5.
An ac input signal of frequency 60 Hz is rectified by an
(i) Half wave and an
Answer:
The output frequency remains the same in a half-wave rectifier, i.e. 60 Hz.

(ii) Full-wave rectifier. Write the output frequency in each case.
Answer:
The output frequency becomes twice the input frequency in the case of the full-wave rectifier, i.e. 120 Hz.

Question 6.
Give the ratio of the number of holes and the number of conduction electrons in an intrinsic semiconductor.
Answer:
The ratio is one.

Question 7.
What is the depletion region in a p-n junction?
Answer:
It is a thin layer between p and n sections of the p-n junction which is devoid of free electrons and holes.

Question 8.
Name an impurity which when added to pure silicon makes it a
(i) p-type semiconductor
Answer:
Boron, aluminum, etc.

(ii) n-type semiconductor.
Answer:
Phosphorous, antimony, etc.

Question 9.
Which type of biasing gives a semiconductor diode very high resistance?
Answer:
Reverse biasing.

Question 10.
Identify the biasing in the figure given below.

Answer:
Forward biasing.

Question 11.
Draw the circuit symbol of (a) photodiode, and (b) light-emitting diode.
Answer:
The circuit symbols are as shown below.

Question 12.
What is the function of a photodiode? (CBSE AI 2013C)
Answer:
It functions as a detector of optical signals.

Question 13.
When a p-n junction diode is forward biased, how will its barrier potential be affected? (CBSEAI 2019)
Answer:
Potential barrier decreases in forwarding bias.

Question 14.
Name the junction diode whose l-V characteristics are drawn below: (CBSE Delhi 2017)

Answer:
Solar cell.

Question 15.
How does the width of the depletion region of a p-n junction vary if the reverse bias applied to it decreases?
Answer:
With the increase in the reverse bias, the depletion layer increases.

Question 16.
How does the width of the depletion region of a p-n junction vary if the reverse bias applied to it decreases?
Answer:
If the reverse bias decreases, the width of the depletion layer also decreases.

Question 17.
Why is the conductivity of n-type semiconductors greater than that of p-type semiconductors even when both of these have the same level of doping?
Answer:
It is because in n-type the majority carriers are electrons, whereas in p-type they are holes. Electrons have greater mobility than holes.

Question 18.
How does the conductance of a semiconducting material change with rising in temperature?
Answer:
Increases with an increase in temperature.

Question 19.
How is a sample of an n-type semiconductor electrically neutral though it has an excess of negative charge carriers?
Answer:
It is because it contains an equal number of electrons and protons and is made by doping with a neutral impurity.

Question 20.
How is the bandgap, Eg, of a photodiode related to the maximum wavelength, λ_m, that can be detected by it?
Answer:
E_g = \(\frac{h c}{\lambda_{m}}\)

Question 21.
Zener diodes have higher dopant densities as compared to ordinary p-n junction diodes. How does it affect the
(i) Width of the depletion layer
Answer:
Junction width will be small and

(ii) Junction field?
Answer:
The junction field will be high.

Question 22.
Can the potential barrier across a p-n junction be measured by simply connecting a voltmeter across the junction? (NCERT Exemplar)
Answer:
No, because the voltmeter must have a resistance very high compared to the junction resistance, the latter being nearly infinite.

Semiconductor Electronics Important Extra Questions Short Answer Type

Question 1.
Draw a labeled circuit diagram of a full-wave rectifier using a p-n junction.
Answer:
The diagram is as shown.

Question 2.
What is a solar cell? How does it work? Give one of its uses.
Answer:
It is a p-n junction used to convert light into electrical energy. In such a diode, one region either the p-type or the n-type is made so thin that light falling on the diode is not absorbed appreciably before reaching the junction. The thin region in the solar cell is called the emitter and the other is called the base. The magnitude of current depends upon the intensity of light reaching the junction. A solar cell can be used to charge storage batteries during the daytime, which can be used during the night.

These are used as power supplies for satellites and space vehicles.

Question 3.
Draw the output signal in a p-n junction diode when a square input signal of 10 V as shown in the figure is applied across it. (CBSE AI 2019)

Answer:
The diode will conduct only when it is forward biased. Therefore, till the input voltage is + 5 V, we will get an output across R, accordingly the output waveform shown in the figure.

Question 4.
The following diagrams, indicate which of the diodes are forward biased and which are reverse biased.

Answer:
(a) Forward biased.
(b) Reverse biased.
(c) Forward biased,
(d) Reverse biased.

Question 5.
Mention the important considerations required while fabricating a p-n junction diode to be used as Light-Emitting Diode (LED). What should be the order of bandgap of an LED if it is required to emit light in the visible range? (CBSE Delhi 2013)
Answer:
The important considerations are

• It should be heavily doped.
• The diode should be encapsulated with a transparent cover so that emitted light can come out.

The semiconductor used for the fabrication of visible LEDs must at least have a bandgap of 1.8 eV.

Question 6.
In the given circuit diagram shown below, two p-n junction diodes D₁ and D₂ are connected with a resistance R and a dc battery E as shown. Redraw the diagram and indicate the direction of flow of appreciable current in the circuit. Justify your answer.

Answer:
The redrawn diagram showing the flow of appreciable current is shown below.

Here diode D₂ is forward biased, hence it conducts. Therefore appreciable current will pass through it. However, diode 0, is reverse biased, hence negligible current will flow through it.

Question 7.
The diagram below shows a piece of pure semiconductor S in series with a variable resistor R and a source of constant voltage V. Would you increase or decrease the value of R to keep the reading of ammeter (A) constant when semiconductor S is heated? Give reason.

Answer:
When a semiconductor is heated, its resistance decreases. As a result, the total resistance of the circuit will decrease. In order to maintain constant current flow, the total resistance of the circuit must remain constant. Hence, the external resistance has to be increased to compensate for the decrease of resistance of the semiconductor.

Question 8.
Two semiconductor materials X and Y showed in the figure are made by doping germanium crystal with indium and arsenic respectively. The two are joined end to end and connected to a battery as shown,
(i) Will the junction be forward or reverse biased?
(ii) Sketch a V-l graph for this arrangement.

Answer:
Material X is p-type and material Y is n-type.
(i) The junction is reverse biased.
(ii) For the V-l graph
The characteristics are as shown.

Question 9.
Draw the output waveform across the resistor (figure). (NCERT)

Answer:
It is a half-wave rectifier, therefore only the positive cycle will be rectified. Thus the output waveform is as shown.

Question 10.
(i) Name the type of a diode whose characteristics are shown in figure
(a) and figure (b).
(ii) What does point P in figure (a) represent?
(iii) What do the points P and Q in figure (b) represent? (NCERT Exemplar)

Answer:
(i) Zener junction diode and solar cell.
(ii) Zener breakdown voltage
(iii) P-open circuit voltage.
Q-short circuit current

Question 11.
A Zener of power rating 1 W is to be used as a voltage regulator. If Zener has a breakdown of 5 V and it has to regulate voltage which fluctuated between 3 V and 7 V, what should be the value of R, for safe operation (figure)? (NCERT Exemplar)

Answer:
Given P = 1 W, Vz = 5V, Vs = 7 V, Rs = ?
We know that

Question 12.
Two material bars A and B of the equal area of the cross-section are connected in series to a dc supply. A is made of usual resistance wire and B of an n-type semiconductor.
(i) In which bar is the drift speed of free electrons greater?
(ii) If the same constant current continues to flow for a long time, how will the voltage drop across A and B be affected? Justify each answer. (CBSE Sample Paper 2018-19)
Answer:
(i) Drift speed in B (n-type semiconductor) is higher.
Reason: Since the two bars A and B are connected in series, the current through each is the same.
Now l = neAv_d
Or
v_d = \(\frac{1}{n e A}\) ⇒ v_d ∝ \(\frac{1}{n}\) (As l and A are same).

As n is much lower in semiconductors, drift velocity will be more.

Question 13.
Explain how the width of the depletion layer in a p-n junction diode changes when the junction is (i) forward biased and (ii) reverse biased. (CBSE Delhi 2018C)
Answer:
The width of the depletion region in a p-n junction diode decreases when it is forward biased because the majority of charge carriers flow towards the junction. While it increases when the junction diode is reverse biased because the majority of charge carriers move away from the junction.

Semiconductor Electronics Important Extra Questions Long Answer Type

Question 1.
Define the terms ‘potential barrier’ and ‘depletion region’ for a p – n junction diode. State how the thickness of the depletion region will change when the p-n junction diode is (i) forward biased and (ii) reverse biased.
Answer:
Potential barrier: The potential barrier is the fictitious battery, which seems to be connected across the p-n junction with its positive terminal in the n-region and the negative terminal in the p-region.

Depletion region: The region around the junction, which is devoid of any mobile charge carriers, is called the depletion layer or region.

1. When the p-n junction is forward biased, there is a decrease in the depletion region.
2. When the p-n junction is reverse biased, there is an increase in the depletion region.

Question 2.
Explain (i) forward biasing and (ii) reverse biasing of a p-n junction diode.
Answer:
(i) A p-n junction is said to be forward-biased if its p-type is connected to the positive terminal and its n-type is connected to the negative terminal of a battery.
(ii) A p-n junction is said to be reverse-biased if its n-type is connected to the positive terminal and its p-type is connected to the negative terminal of a battery. The diagrams are as shown.

Question 3.
Draw V-l characteristics of a p-n junction diode. Answer the following questions, giving reasons:
(i) Why is the current under reverse bias almost independent of the applied potential up to a critical voltage?
(ii) Why does the reverse current show a sudden increase at the critical voltage? Name any semiconductor device which operates under the reverse bias in the breakdown region. (CBSEAI 2013)
Answer:
The characteristics are as shown.

(i) This is because even a small voltage is sufficient to sweep the minority carriers from one side of the junction to the other side of the junction.

(ii) As the reverse bias voltage is increased, the electric field at the junction becomes significant. When the reverse bias voltage V = V_z critical voltage, then the electric field strength is high enough to pull valence electrons from the host atoms on the p-side which are accelerated to the n-side. These electrons account for the high current observed at the breakdown.

Zener diode operates under the reverse bias in the breakdown region.

Question 4.
Draw the energy band diagrams of (i) n-type and (ii) p-type semiconductor at temperature T > 0 K.
In the case of n-type Si semiconductors, the donor energy level is slightly below the bottom of the conduction band, whereas in p-type semiconductors, the acceptor energy level is slightly above the top of the valence band. Explain what role do these energy levels play in conduction and valence bands. (CBSE AI 2015 C)
Answer:
For energy bands
(i) The energy level diagram is shown below.

(ii) The diagram is shown as

In the energy band diagram of n-type Si semiconductor, the donor energy level E_A is slightly below the bottom Ec of the conduction band and electrons from this level move into the conduction band with a very small supply of energy. At room temperature, most of the donor atoms get ionized but very few (-10-12) atoms of Si get ionized. So the conduction band will have most electrons coming from the donor impurities.

Similarly, for p-type semiconductors, the acceptor energy level E_A is slightly above the top E_v of the valence band. With the very small supply of energy, an electron from the valence band can jump to the level E_A and ionize the acceptor negatively. Alternately, we can also say that with a very small supply of energy, the hole from level EA sinks down into the valence band. Electrons rise up and holes fall down when they gain external energy.

Question 5.
Give reasons for the following:
(i) High reverse voltage does not appear across an LED.
Answer:
It is because the reverse breakdown voltage of LED is very low, i.e. nearly 5 V.

(ii) Sunlight is not always required for the working of a solar cell.
Answer:
Because solar cells can work with any light whose photon energy is more than the bandgap energy.

(iii) The electric field, of the junction of a Zener diode, is very high even for a small reverse bias voltage of about 5 V. (CBSE Delhi 2016C)
Answer:
The heavy doping of p and n sides of the p-n junction makes the depletion region very thin, hence for a small reverse bias voltage, the electric field is very high.

Question 6.
State the reason why the photodiode is always operated under reverse bias. Write the working principle of operation of a photodiode. The semiconducting material used to fabricate a photodiode has an energy gap of 1.2 eV. Using calculations, show whether it can detect light of a wavelength of 400 nm incident on it. (CBSE Al 2017C)
Answer:
It is easier to observe the change in the current with the change in the light intensity if a reverse bias is applied. Thus photodiode is used in the reverse bias mode even when the current in the forward bias is more the energy gap (E_g) of the semiconductor, then electron-hole pairs are generated due to the absorption of photons.

Due to the electric field of the junction, electrons and holes are separated before they recombine. The direction of the electric field is such that electrons reach the n-side and holes reach the p-side. Electrons are collected on the n-side and holes are collected on the p-side giving rise to an emf. When an external load is connected, current flows.

Given λ = 400 nm,
Energy of photon
E = \(\frac{h c}{\lambda}=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{400 \times 10^{-9} \times 1.6 \times 10^{-19}}\) = 3.105 eV

Since the bandgap is lesser than this energy, therefore it will be able to detect the wavelength.

Question 7.
Explain the two processes involved in the formation of a p-n junction diode. Hence define the term ‘barrier potential’. (CBSE Delhi 2017C)
Answer:
The two processes are
(i) Diffusion and
(ii) Drift

• Diffusion: The holes diffuse from the p-side to the n-side and electrons diffuse from the n-side to the p-side.
• Drift: The motion of charge carriers due to the applied electric field which results in the drifting of holes along the electric field and of electrons opposite to the electric field.

The potential barrier is the fictitious battery that seems to be connected across the junction with its positive end on the n-type and the negative end on the p-type.

Question 8.
Explain briefly how a photodiode operates.
Answer:
A Photodiode is again a special purpose p-n junction diode fabricated with a transparent window to allow light to fall on the diode. It is operated under reverse bias. When the photodiode is illuminated with light (photons) with energy (hv) greater than the energy gap (E_g) of the semiconductor, then electron-hole pairs are generated due to the absorption of photons. The diode is fabricated such that the generation of e-h pairs takes place in or near the depletion region of the diode.

Question 9.
Name the p-n junction diode which emits spontaneous radiation when forward biased. How do we choose the semiconductor, to be used in these diodes, if the emitted radiation is to be in the visible region?
Answer:
The p-n junction diode, which emits spontaneous radiation when forward biased, is the “light-emitting diode” or LED.

The visible tight is from 400 nm to 700 nm and the corresponding energy is between 2.8 eV to 1.8 eV. Therefore, the energy gap of the semiconductor to be used in LED, in order to have the emitted radiation be in the visible region, should be 1.8 eV. Phosphorous doped gallium arsenide and gallium phosphide are two such suitable semiconductors.

Question 10.
The figure shows the V-l characteristic of a semiconductor diode designed to operate under reverse bias. (CBSE Al 2019)

(a) Identify the semiconductor diode used.
Answer:
The diode used is Zener diode

(b) Draw the circuit diagram to obtain the given characteristics of this device.
Answer:
The circuit diagram is as shown.

(c) Briefly explain one use of this device.
Answer:
The Zener diode can be used as a voltage regulator in its breakdown region. The Zener voltage remains constant even when the current through the Zener diode changes.

Question 11.
With the help of a diagram, show the biasing of a light-emitting diode (LED). Give its two advantages over conventional incandescent lamps.
Answer:
The biasing of a light-emitting diode (LED), has been shown below.

Two main advantages of LED over conventional incandescent lamps are as follows:

1. Low operational voltage and less power consumption.
2. Fast action and no warm-up time required.
3. The bandwidth of emitted light is 100 A to 500 A or in other words, it is nearly (but not exactly) monochromatic.
4. Long life and ruggedness.
5. Fast on-off switching capability.

Question 12.
(a) Writetheprincipleofasemiconductor device which is used as a voltage regulator.
Answer:
(a) Zener diode is used as a voltage regulator
Principle: It is based on the Principle that when breakdown voltage V₂ takes place, there is a large change in the reverse current even with the insignificant change in the reverse bias voltage.

(b) With the help of a circuit diagram explain its working.
Answer:

Working: If the reverse voltage across a Zener diode is increased beyond the breakdown voltage V_z, the current increases sharply and large current l_z flows through the Zener diode and the voltage drop across Rs increases maintaining the voltage drop across RL at constant value V_o = V_z.

On the other hand, if we keep the input voltage constant and decrease the load resistance RL, the current across the load will increase. The extra current cannot come from the source because drop-in Rs will not change as the Zener is within its regulating range. The additional load current will pass through the Zener diode and is known as Zener current l_z so that the total current (l_L + l_z) remains constant.

(c) Draw its l-V characteristics. (CBSE 2019C)
Answer:
l-V Characteristics of Zener diode

Question 13.
With what considerations in view, a photodiode is fabricated? State its working with the help of a suitable diagram.
Even though the current in the forward bias is known to be more than in the reverse bias, yet the photodiode works in reverse bias. What is the reason? (CBSE Delhi 2015)
Answer:
It is fabricated with a transparent window to allow light to fall on the diode. It is fabricated such that the generation of e-h pairs takes place in or near the depletion region of the diode.

When the photodiode is illuminated with light (photons) with energy (h_w) greater than the energy gap (E_g) of the semiconductor, then electron-hole pairs are generated due to the absorption of photons. The diode is fabricated such that the generation of e-h pairs takes place in or near the depletion region of the diode. Due to the electric field of the junction, electrons and holes are separated before they recombine.

The direction of the electric field is such that electrons reach the n-side and holes reach the p-side. Electrons are collected on the n-side and holes are collected on the p-side giving rise to an emf. When an external load is connected, current flows. The magnitude of the photocurrent depends on the intensity of incident light (photocurrent is proportional to incident light intensity). The diagram is as shown.

It is easier to observe the change in the current with a change in the light intensity if a reverse bias is applied. Thus photodiode is used in the reverse bias mode even when the current in the forward bias is more.

Question 14.
Draw the circuit diagram of a full-wave rectifier and explain its working. Also, give the input and output waveforms. (CBSE Delhi 2019)
Answer:
The circuit diagram is as shown.

The two ends S₁ and S₂  of a center-tapped secondary of a transformer are connected to the P sections of the two diodes D₁ and D₂ respectively. The n-sections of the two diodes are joined together and their com¬mon junction is connected to the central tap C of the secondary winding through a load resistance R_L. The input is applied across the primary and the output is ob¬tained across the load resistance R_L. The arrows show the direction of the current.

Assume that the end A of the secondary is positive during the first half cycle of the supply voltage. This makes diode D₁ forward biased and diode D₂ reverse biased. Thus diode D₁ conducts and an output is obtained across the load R_L.

During the second half cycle of the supply voltage, the polarities of the secondary windings reverse. A becomes negative and B becomes positive with respect to the central terminal C. This makes diode D₂ forward biased. Hence it conducts and an output is obtained across R_L.

The input-output waveforms are as shown.

Question 14.
Draw the circuit diagram to show the use of a p-n junction diode as a half-wave rectifier. Also show the input and the output voltages, graphically. Explain its working.
Answer:
The diagram is as shown.

The input and output waveforms are as shown.

A p-n junction diode is used as a half-wave rectifier. Its work is based on the fact that the resistance of the p-n junction becomes low when forward biased and becomes high when reversing biased. These characteristics of a diode are used in rectification.

Question 15.
Distinguish between conductors, insulators, and semiconductors on the basis of the band theory of solids.
Answer:
The diagrams are as shown.

Metals: A distinguishing character of all conductors, including metals, is that the valence band is partially filled or the conduction and the valence band overlap. Electrons in states near the top of the filled portion of the band have many adjacent unoccupied states available, and they can easily gain or lose small amounts of energy in response to an applied electric field. Therefore these electrons are mobile and can contribute to electrical and thermal conductivity. Metallic crystals always have partially filled bands figure (ii).

Insulators: In the case of insulators, there is a large energy gap of approximately 6 eV depending upon the nature of the crystal. Electrons, however, heated, find it difficult or practically impossible to jump this gap and thus never reach the conduction band. Thus electrical conduction is not possible through an insulator figure (iii).

Semiconductors: There is a separation between the valence band and the conduction band. The energy gap is of the order of 1 eV (0.67 eV for germanium and 1.12 eV for silicon). At absolute zero the electrons cannot gain this energy. But at room temperature, these electrons gain energy and move into the conduction band where they are free to move even under the effect of a weak electric field figure (i).

Question 16.
What is a Zener diode? How is it symbolically represented? With the help of a circuit diagram, explain the use of the Zener diode as a voltage stabilizer.
Answer:
It is a special diode made to work only in the reverse breakdown region.
Symbol:

The figure below shows the use of the Zener diode in providing a constant voltage supply.

This use of the Zener diode is based on the fact that in the reverse breakdown (or Zener) region, a very small change in voltage across the Zener diode produces a very large change in the current through the circuit. If voltage is increased beyond Zener voltage, the resistance of the Zener diode drops considerably. Consider that the Zener diode and a resistor R, called dropping resistor, are connected to a fluctuating voltage supply, such that the Zener diode is reverse biased.

Whenever voltage across the diode tends to increase, the current through the diode rises out of proportion and causes a sufficient increase in the voltage drop across the dropping resistor. As a result, the output voltage lowers back to the normal value. Similarly, when the voltage across the diode tends to decrease, the current through the diode goes down out of proportion, so that the voltage drop across the dropping resistor is much less and now the output voltage is raised to normal.

Question 17.
Explain briefly with the help of a circuit diagram how V-l characteristics of a p-n junction diode are obtained in (i) forward bias and (ii) reverse bias.
Answer:
Forward biased characteristics: A p-n junction is said to be forward-biased if its p-type is connected to the positive terminal and its n-type is connected to the negative terminal of a battery shows a circuit diagram that is used to study the forward characteristics of a p-n junction. The p-n junction is forward biased. Different readings are taken by changing the voltage and noting the corresponding milliammeter current.

Practically no current is obtained till the applied voltage becomes greater than the barrier potential. Above the potential barrier voltage, even a small change in potential causes a large change in current.

Reverse biased characteristics: In reverse biased characteristics, instead of a milliammeter, a microammeter is used. The voltage across the p-n junction is increased and the corresponding current is noted.

In the reverse bias, the diode current is very small. As the voltage has increased the current also increases. At a certain voltage, the current at once becomes very large. This voltage is called Breakdown voltage or Zener voltage. At this voltage, a large number of covalent bonds break releasing a large number of electrons and holes. Hence a large current is obtained. The characteristics are as shown.

Question 18.
Explain how the heavy doping of the p and n sides of a p-n junction diode helps in internal field emission (or Zener breakdown), even with a reverse bias voltage of a few volts only. Draw the general shape of the V-I characteristics of a Zener diode. Discuss how the nature of these characteristics led to the use of a Zener diode as a voltage regulator.
Answer:
Consider a p-n junction where both p- and n-sides are heavily doped. Due to the high dopant densities, the depletion layer junction width is small and the junction field will be high. Under large reverse bias, the energy bands near the junction and the junction width decrease. Since the junction width is < 10^-7 m, even a small voltage (say 4 V) may give a field as large as 4 × 10^-7 Vm^-1. The high junction field may strip an electron from the valence band which can tunnel to the n-side through the thin depletion layer. Such a mechanism of emission of electrons after a certain critical field or applied voltage V is termed as internal field emission which gives rise to a high reverse, current, or breakdown current.

The general shape of the V-l characteristics of a Zener diode is as shown in the figure.

Suppose an unregulated dc input voltage (V_i) is applied to the Zener diode (whose breakdown voltage is V_z as shown in the figure. If the applied voltage V_i > V_z, then the Zener diode is in the breakdown condition. As a result of a wide range of values of load (R_L), the current in the circuit or through the Zener diode may change but the voltage across it remains unaffected by the load. Thus, the output voltage across the Zener is a regulated voltage.

Question 19.
(i) Describe briefly with the help of a necessary circuit diagram, the working principle of a solar cell.
Answer:
Solar Cell: A solar cell is a junction diode that converts solar energy into electrical energy. In a solar cell, the n-region is very thin and transparent so that most of the incident light reaches the junction. The thin region is called the emitter and the other base. When light is incident on it, it passes through the crystal onto the junction. The electrons and holes are generated due to light (with h_v > E_g). The electrons are kicked to the n-side and holes to the p-side due to the electric field of the depletion region. Thus p-side becomes positive and the n-side becomes negative giving rise to a photo-voltage. Thus it behaves as a cell.

(ii) Why are Si and GaAs preferred materials for solar cells? Explain. (CBSE AI 2011C)
Answer:
Si and GaAs are preferred for solar cell fabrication due to the fact that their bandgap is ideal. Further, they have high electrical conductivity and high optical absorption.

Question 20.
Explain, with the help of a circuit diagram, the working of a photo-diode. Write briefly how it is used to detect optical signals. (CBSE Delhi 2013)
Answer:
When the photodiode is illuminated with light (photons) with energy (h_v) greater than the energy gap (E_g) of the semiconductor, then – electron-hole pairs are generated due to the absorption of photons. The diode is fabricated such that the generation of e-h pairs takes place in or near the depletion region of the diode. Due to the electric field of the junction, electrons and holes are separated before they recombine. The direction of the electric field is such that electrons reach the n-side and holes reach the p-side. Electrons are collected on the n-side and holes are collected on the p-side giving rise to an emf. When an external load is connected, current flows.

The magnitude of the photocurrent depends on the intensity of incident light incident on it. This helps in detecting optical signals.

Question 21.
(i) Explain with the help of a diagram, how depletion region and potential barrier are formed in a junction diode.
(ii) If a small voltage is applied to a p-n junction diode, how will the barrier potential be affected when it is (i) forward biased and (ii) reverse biased? (CBSE AI2015)
Answer:
(i) Since there is an excess of electrons in the n-type and excess of holes in the p-type, on the formation of a p-n junction the electrons from the n-type diffuse into the p-region, and the holes in the p-type diffuse into the n-region.

The accumulation of electric charge of opposite polarities in the two regions across the junction establishes a potential difference between the two regions. This is called the potential barrier or junction barrier. The potential barrier developed across the junction opposes the further diffusion of the charge carriers from p to n and vice versa. There is a region on either side of the junction where there is a depletion of mobile charges and has only immobile charges. The region around the junction, which is devoid of any mobile charge carriers, is called the depletion layer or region.

(ii) (a) In forwarding bias the potential barrier decreases.
(b) In reverse bias the potential barrier increases.

Question 22.
Write the two processes that take place in the formation of a p-n junction. Explain with the help of a diagram, the formation of depletion region and barrier potential in a p-n junction. (CBSE Delhi 2017)
Answer:

1. Drift and
2. diffusion.

The n-type has an excess of electrons and the p-type has an excess of holes. When a p-n junction has formed the electrons from the n-type diffuse into the p-region and the holes in the p-type diffuse into the n-region. These diffusing electrons and holes combine near the junction. Each combination eliminates an electron and a hole. This results in the n-region near the junction becoming positively charged by losing its electrons and the p-region near the junction becoming negatively charged by losing its holes.

This accumulation of electric charge of opposite polarities in the two regions across the junction establishes a potential difference between the two regions. This is called the potential barrier or junction barrier.
The potential barrier developed across the junction opposes the further diffusion of the charge carriers from p to n and vice versa. As a result, a region develops on either side of the junction where there is a depletion of mobile charges and has only immobile charges. The region around the junction which is devoid of any mobile charge carriers is called the depletion layer or region.

Question 23.
(i) State briefly the processes involved in the formation of the p-n junction explaining clearly how the depletion region is formed.
Answer:
As we know that n-type semi-conductor has more concentration of electrons than that of a hole and a p-type semi-conductor has more concentration of holes than an electron. Due to the difference in concentration of charge carriers in the two regions of the p-n junction, the holes diffuse from p-side to n-side, and electrons diffuse from n-side to p-side. When an electron diffuses from n to p, it leaves behind it an ionized donor on the n-side. The ionized donor (+ve charge) is immobile as it is bound by the surrounding atoms. Therefore, a layer of positive charge is developed on the n-side of the junction. Similarly, a layer of negative charge is developed on the p-side.

Hence, a space-charge region is formed on either side of the junction, which has immobile ions and is devoid of any charge carrier, called depletion layer or depletion region.

The potential barrier is the fictitious battery which seems to be connected across the junction with its positive end on the n-type and the negative end on the p-type.

(ii) Using the necessary circuit diagrams, show how the V – l characteristics of a p-n junction are obtained in
(a) Forward biasing
(b) Reverse biasing
How these characteristics are made use of in rectification? (CBSE Delhi 2014)
Answer:
(a) p-n junction diode under forwarding bias: The p-side is connected to the positive terminal and the n-side to the negative terminal. Applied voltage drops across the depletion region. Electron in n-region moves towards the p-n junction and holes in the p-region move towards the junction. The width of the depletion layer decreases and hence, it offers less resistance. Diffusion of majority carriers takes place across the junction. This leads to the forward current.

(b) p-n junction diode under reverse bias: Positive terminal of the battery is connected to the n-side and negative terminal to p-side. Reverse bias supports the potential barrier. Therefore, the barrier height increases and the width of the depletion region also increases. Due to the majority of carriers, there is no conduction across the junction. A few minority carriers across the junction after being accelerated by the high reverse bias voltage. This constitutes a current that flows in opposite direction, which is called reverse current.

For V-l curves

A p-n junction diode is used as a half-wave rectifier. Its work is based on the fact that the resistance of the p-n junction becomes low when forward biased and becomes high when reversing biased. These characteristics of the diode are used in rectification.

Question 24.
(i) Explain with the help of a suitable diagram, the two processes which occur during the formations of a p-n junction diode. Hence define the terms (i) depletion region and (ii) potential barrier.
(ii) Draw a circuit diagram of a p-n junction diode under forwarding bias ‘ and explain its working. (CBSE 2018C)
Answer:
(i) The two important processes are diffusion and drift.
Due to the concentration gradient, the electrons diffuse from the n-side to the p side, and holes diffuse from the n-side to the n side.

Due to the diffusion, an electric field develops across the junction. Due to the field, an electron moves from the p-side to the n-side; a hole moves from the n-side to the p-side. The flow of the charge carriers due to the electric field is called drift.

(a) Depletion region: It is the space charge region on either side of the junction that gets depleted of free charges is known as the depletion region.
(b) Potential Barrier: The potential difference that gets developed across the junction and opposes the diffusion of charge carriers and brings about a condition of equilibrium, which is known as the barrier potential.

(ii) The circuit diagram is as shown

Working:
In the forward bias condition, the direction of the applied voltage is opposite to the barrier potential. This reduces the width of the depletion layer as well as the height of the barrier. A current can, therefore, flow through the circuit. This current increases (non¬linearly) with the increase in the applied voltage.

Numerical Problems

Question 1.
(i) Three photodiodes D₁, D₂, and D₃ are made of semiconductors having band gaps of 2.5 eV, 2 eV, and 3 eV respectively. Which of them will not be able to detect light of wavelength 600 nm? (CBSE Delhi 2019)
Answer:
λ = 600 nm
The energy of a photon of wavelength λ
E = \(\frac{h c}{\lambda}=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{600 \times 10^{-9} \times 1.6 \times 10^{-19}}\) = 2.08 eV

The bandgap energy of diode D₂ (= 2eV) is less than the energy of the photon. Hence diode D₂ will not be able to detect light of wavelength 600 nm.

(ii) Why photodiodes are required to operate in reverse bias? Explain.
Answer:
A photodiode when operated in reverse bias can measure the fractional change in minority carrier dominated reverse bias current with greater ease than when forward biased.

Question 2.
In half-wave rectification, what is the output frequency if the input frequency is 50 Hz? What is the output frequency of a full-wave rectifier for the same input frequency? (NCERT)
Answer:
The output frequency of the half-wave rectifier is the same as the input frequency, while that of the full-wave rectifier is double that of the input. Therefore the frequency is 50 Hz for half-wave and 100 Hz for full-wave.

Question 3.
A p-n photodiode is fabricated from a semiconductor with a bandgap of 2.8 eV. Can it detect a wavelength of 6000 nm? (NCERT)
Answer:
Given λ = 6000 nm,
Energy of photon

Since the bandwidth is greater than this energy, it will not be able to detect the wavelength.

Question 4.
Three photodiodes D₁, D_2, and D₃ are made of semiconductors having band gaps of 2.5 eV, 2 eV, and 3 eV, respectively. Which ones will be able to detect light of wavelength 600 nm? (NCERT Exemplar)
Answer:
The energy of an incident light photon is given by

For the incident radiation to be detected by the photodiode, the energy of the incident radiation photon should be greater than the bandgap. This is true only for D₂. Therefore, only D₂ will detect this radiation.

Question 5.
If each diode in the figure has a forward bias resistance of 25 Ω and infinite resistance in reverse bias, what will be the values of the current l₁ l₂, l₃, and l₄? (NCERT Exemplar)

nswer:
Current l₃ is zero as the diode in that branch is reverse biased.

Resistance in the branches AB and EF is each (125 +25) Ω = 150 Ω
As AB and EF are identical paraLleL branches, their effective resistance is 150/2 = 75 Ω

Therefore net resistance in the circuit Is
Rnet =75 + 25 = 100 Ω

Therefore current l1 is
l₁ = 5/100 = o.05 A

As resistance of branches AB and EF is equal, the current l1 will be equally shared by the two, hence
l₂ = l₄ = 0.05/2 = 0.025A
Hence l₁ = 0.05 A, l₂ = l₄ = 0.025 A, l₃ = 0

Question 6.
Assuming the ideal diode, draw the output waveform for the circuit given in the figure. ExplaIn the waveform. (NCERT Exemplar)

Answer:
When input voltage Will is greater than 5 V, the diode wilt becomes forward biased and will conduct. When the input is Less than 5 V, the diode will be reverse biased and will not conduct, i.e. open circuit, hence the output is as shown.

Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 10 Wave Optics. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

Class 12 Physics Chapter 10 Important Extra Questions Wave Optics

Wave Optics Important Extra Questions Very Short Answer Type

Question 1.
Sketch the refracted wavefront emerging from convex tens, If a plane wavefront is an incident normally on it.
Answer:
The figure is as shown.

Question 2.
How would you explain the propagation of light on the basis of Huygen’s wave theory?
Answer:
To explain the propagation of light we have to draw a wavefront at a later instant when a wavefront at an earlier instant is known. This can be drawn by the use of Huygen’s principle.

Question 3.
Draw the shape of the reflected wavefront when a plane wavefront is an incident on a concave mirror.
Answer:
The reflected wavefront is as shown.

Question 4.
Draw the shape of the refracted wavefront when a plane wavefront is an incident on a prism.
Answer:
The shape of the wavefront is as shown.

Question 5.
Draw the type of wavefront that corresponds to a beam of light diverging from a point source.
Answer:
The wavefront formed by the light coming from a very far off source is a plane and for a beam of light diverging from a point, a wavefront is a number of concentric circles.

Question 6.
Draw the type of wavefront that corresponds to a beam of light coming from a very far off source.
Answer:
The wavefront is as shown.

Question 7.
Name two phenomena that establish the wave nature of light.
Answer:
Interference and diffraction of light.

Question 8.
State the conditions which must be satisfied for two light sources to be coherent.
Answer:
(a) Two sources must emit light of the same wavelength (or frequency).
(b) The two light sources must be either in-phase or have a constant phase difference.

Question 9.
Draw an intensity distribution graph for diffraction due to a single-slit.
Answer:
The intensity distribution for a single-slit diffraction pattern is as shown.

Question 10.
Name one device for producing plane polarised light. Draw the graph showing the variation of intensity of polarised light transmitted by an analyser.
Answer:
Nicol prism can be used to produce plane polarised light. The graph is as shown.

Question 11.
State Huygens’ principle of diffraction of light. (CBSE AI 2011C)
Answer:
Huygens principle states that
(a) Each point on a wavefront is a source of secondary waves which travel out with the same velocity as the original waves.
(b) The new wavefront is given by the forward locus of the secondary wavelets.

Question 12.
In what way is a plane polarised tight different from an unpolarised light? (CBSE AI 2012C)
Answer:
Plane polarized light vibrates 1n only one plane.

Question 13.
Which of the following waves can be polarised: (i) Heatwaves (ii) Sound waves? Give a reason to support your answer. (CBSE Delhi 2013)
Answer:
Heatwaves are transverse In nature.

Question 14.
Define the term ‘wavefront’. (CBSE AI 2014C)
Answer:
It Is defined as the locus of all points In a medium vibrating in the same phase.

Question 15.
Define the term ‘coherent sources’ which are required to produce interference pattern in Young’s double-slit experiment. (CBSE Delhi 2014C)
Answer:
Two sources that are In phase or have a constant phase difference are called coherent sources.

Question 16.
What change would you expect if the whole of Young’s double-slit apparatus were dipped into the water?
Answer:
The wavelength λ, of light In water, is less than that in air. Since the fringe width β is directly proportional to the wavelength of light, therefore, the fringe width will decrease.

Question 17.
When light travels from a rarer to a denser medium, it loses some speed. Does the reduction in speed Imply a reduction in the energy carried by the light wave?
Answer:
No, the energy carried by a wave depends upon the amplitude of the wave and not on Its speed of propagation.

Question 18.
If one of the slits say S₁, is covered then what changes occur in the Intensity of light at the centre of the screen?
Answer:
The intensity 1s decreased four times because l ∝ 4a² where a is the amplitude of each wave.

Question 19.
How does the angular separation between fringes in a single-slit diffraction experiment change when the distance of separation between the slit and screen is doubled? (CBSE AI 2012)
Answer:
No change.

Question 20.
What is the effect on the interference fringes in Young’s double-slit experiment If the separation between the screen and slits Is Increased?
Answer:
The fringe width Increases.

Question 21.
How does the Intensity of the central maximum change If the width of the slit Is halved in a single-slit diffraction experiment?
Answer:
The width of the central maxima is doubled and the intensity is reduced to one-fourth of Its original value.

Question 22.
The polarising angle of a medium Is 60°, what is the refractive index of the medium? (CBSE Delhi 2019)
Answer:
Using the expression
μ = tan i_p = tan 60° = 1.732.

Question 23.
In the wave picture of light intensity of light is determined by the square of the amplitude of the wave. What determines the intensity of light in the photon picture of light?
Answer:
For a given frequency intensity of light in the photon, the picture is determined by the number of photons crossing a unit area per unit time.

Question 24.
How does the Intensity of the central maximum change if the width of the slit is halved in the single-slit diffraction experiment?
Answer:
The width of the central maximum is doubled and the intensity is reduced to one-fourth of its original value.

Question 25.
What would happen If the path difference between the interfering beams that is S₂P – S₁P became very large?
Answer:
If the path difference becomes very large it may exceed the coherent length. Thus the coherence of the waves reaching P is lost and no interference takes place.

Question 26.
In Young’s double-slit experiment, what would happen to the intensity of the maxima and the minima if the size of the hole illuminating the two coherent holes were gradually Increased?
Answer:
The fringe width will decrease and finally, there will be general illumination on the screen.

Question 27.
What is the Brewster angle for air to glass transition? (Refractive index of glass =1.5.) (NCERT)
Answer:
Given μ = 1.5, i_P = ?
Using the relation μ = tan i_P we have
l_p = tan^-1 (1.5) = 56.3°

Question 28.
Is Huygen’s principle valid for longitudinal sound waves? (NCERT Exemplar)
Answer:
Yes

Question 29.
Consider a point at the focal point of a convergent lens. Another convergent lens of the short focal length is placed on the other side. What is the nature of the wavefronts emerging from the final image? (NCERT Exemplar)
Answer:
Spherical.

Question 30.
What is the shape of the wavefront on earth for sunlight? (NCERTExemplar)
Answer:
Spherical with a huge radius as compared to the earth’s radius so that it is almost a plane.

Question 31.
Draw a graph showing the intensity distribution of fringes due to diffraction at a single-slit. (CBSE 2018C)
Answer:

Wave Optics Important Extra Questions Short Answer Type

Question 1.
How can one distinguish between an unpolarised and linearly polarised light beam using polaroid? (CBSE Delhi 2019)
Answer:
The two lights will be allowed to pass through a polariser. When the polarizer is rotated in the path of these two light beams, the intensity of light remains the same in all the orientations of the polariser, then the light is unpolarised. But if the intensity of light varies from maximum to minimum then the light beam is a polarised light beam.

Question 2.
What is meant by plane polarised light? What type of waves shows the property of polarisation? Describe a method of producing a beam of plane polarised light?
Answer:

1. The light that has its vibrations restricted in only one plane is called plane polarised light.
2. Transverse waves show the phenomenon of polarization.
3. Light is allowed to pass through a polaroid. The polaroid absorbs those vibrations which are not parallel to its axis and allows only those vibrations to pass which are parallel to its axis.

Question 3.
Write the Important characteristic features by which the Interference can be distinguished from the observed diffraction pattern. (CBSE AI 2015)
Answer:
(a) In the interference pattern the bright fringes are of the same width, whereas in the diffraction pattern they are not of the same width.
(b) In interference all bright fringes are equally bright while in diffraction they are not equally bright.

Question 4.
State Brewster’s law. The value of Brewster’s angle for the transparent medium is different for the light of different colours. Give reason. (CBSE Delhi 2016)
Answer:
When the reflected ray and the refracted ray are perpendicular then μ = tani_p where i_p is the polarising angle or Brewster angle.

Brewster’s angle depends upon the refractive index of the two media in contact. The refractive index in turn depends upon the wavelength of light used (different colours) hence Brewster’s angle is different for different colours.

Question 5.
Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids.
Answer:
Let l_o be the intensity of polarised light after passing through the first polarizer P₁. Then the intensity of light after passing through the second polarizer P2 will be l = l_ocos 2θ, where θ is the angle between pass axes of P₁ and P₂. Since P₁ and P₃ are crossed the angle between the pass axes of P₂ and P₃ will be (π/2 – θ). Hence the intensity of light emerging from P₃ will be
l = l_o cos² θ cos² (90° – θ) = l_o cos² θ sin² θ = (l_o /4)sin² 2θ

Therefore, the transmitted intensity will be maximum when θ = π/4

Question 6.
Is energy conserved in interference? Explain.
Answer:
Yes, energy is conserved in interference. Energy from the dark fringes is accumulated in the bright fringes. If we take
l = 4a²cos²\(\frac{\phi}{2}\), then intensity at bright points is l_max = 4a² and intensity at the minima l_min = 0. Hence average intensity in the pattern of the fringes produced due to interference is given by
Ī = \(\frac{I_{\max }+I_{\min }}{2}=\frac{4 a^{2}+0}{2}\) = 2a²

But if there is no interference then total intensity at every point on the screen will be l = a² + a² = 2a², which is the same as the average intensity in the interference pattern.

Question 7.
An incident beam of light of intensity lo is made to fall on a polaroid A. Another polaroid B is so oriented with respect to A that there is no light emerging out of B. A third polaroid C is now introduced midway between A and B and is so oriented that its axis bisects the angle between the axes of A and B. What is the intensity of light now between (i) A and C (ii) C and B? Give reasons for your answers.
Answer:
Polaroids A and B are oriented at an angle of 90°, so no light is emerging out of B. On placing polaroid C between A and B such that its axis bisects the angle between axes of A and B, then the angle between axes of polaroids A and B is 45° and that of C and B also 45°.
(a) Intensity of light on passing through Polaroid A or between A and C is l₁ = \(\frac{l_{0}}{2}\)
(b) On passing through polaroid C, intensity of light between C and B becomes
l₂ = l₁ cos² θ = \(\frac{l_{0}}{2}\) × cos² 45° = \(\frac{l_{0}}{4}\)

Question 8.
One of the slits of Young’s double-slit experiment is covered with a semi¬transparent paper so that it transmits lesser light. What will be the effect on the interference pattern?
Answer:
There will be an interference pattern whose fringe width is the same as that of the original. But there will be a decrease in the contrast between the maxima and the minima, i.e. the maxima will become less bright and the minima will become brighter.

Question 9.
Light from a sodium lamp is passed through two polaroid sheets P₁ and P₂ kept one after the other. Keeping P₁, fixed, P₂ is rotated so that its ‘pass axis can be at different angles, θ, with respect to the pass-axis of P₁.
An experimentalist records the following data for the intensity of light coming out of P₂ as a function of the angle θ.

I = Intensity of beam falling on P₁
(a) One of these observations is not in agreement with the expected theoretical variation of I, identify this observation and write the correct expression.
(b) Define the Brewster angle and write the expression for It in terms of the refractive index of the medium.
Answer:
(a) The observation \(\frac{1}{\sqrt{2}}\) is not correct. It should be 1/2.
(b) It is the angle of incidence at which the refracted and the reflected rays are perpendicular to each other. It is related to the refractive index as tan i_p = μ

Question 10.
How will the interference pattern in Young’s double-slit experiment get affected, when
(a) distance between the slits S₁, and S₂ reduced and
(b) the entire set-up is immersed in water? Justify your answer in each case. (CBSE Delhi 2011C)
Answer:
We know that fringe width β of the dark or bright fringes is given by β = \(\frac{D \lambda}{d}\) where d is the distance between the slits.
(a) When the distance between the slits, i. e. d is reduced then p will increase. The interference pattern will thus become broader.
(b) When the entire set up is immersed in water, the pattern will become narrow due to the decrease in the wavelength of light. The new wavelength λ’ = λ/n, hence β’= β/n

Question 11.
Discuss the intensity of transmitted light when a Polaroid sheet is rotated between two crossed polaroids? (NCERT)
Answer:
Let l0 be the intensity of polarised light after passing through the first polarizer P₁. Then the intensity of light after passing through the second polarizer P₂ will be l = l_o cos² θ

where θ is the angle between pass axes of P₁ and P₂. Since P₁ and P₃ are crossed the angle between the pass axes of P₂ and P₃ will be (π/2 – θ). Hence the intensity of light emerging from P₃ will be

Therefore, the transmitted intensity will be maximum when θ = π/4

Question 12.
A Polaroid (I) is placed In front of a monochromatic source. Another Polaroid (II) is placed in front of this Polaroid (I) and rotated till no light passes. A third Polaroid (III) is now placed in between (I) and (II). In this case, will light emerge from (II)? Explain. (NCERT Exemplar)
Answer:
Only in the special case when the pass axis of (III) is parallel to fill or (II) there shall be no light emerging. In all other cases, there shall be light emerging because the pass axis of (II) is no longer perpendicular to the pass axis of (III).

Question 13.
(a) Good quality sunglasses made of polaroids are preferred over ordinary coloured glasses. Explain why.
(CBSE AI2019)
Answer:
Polaroid sunglasses are preferred over coloured sunglasses because they reduce the intensity of light.

(b) How is plane polarised light defined?
Answer:
Plane polarised light: Light in which vibrations of electric field vector are restricted to one plane containing the direction of propagation.

(c) A beam of plane polarised light is passed through a polaroid. Show graphically, a variation of the intensity of the transmitted light with the angle of rotation of the Polaroid.
Answer:
The graphical variation is as shown.

Wave Optics Important Extra Questions Long Answer Type

Question 1.
Define the term wavefront. Using Huygen’s wave theory, verify the law of reflection.
Or
Define the term, “refractive index” of a medium. Verify Snell’s law of refraction when a plane wavefront is propagating from a denser to a rarer medium. (CBSE Delhi 2019)
Answer:
The wavefront is a locus of points that oscillate in the same phase.

Consider a plane wavefront AB incident obliquely on a plane reflecting surface MM^–. Let us consider the situation when one end A of was front strikes the mirror at an angle i but the other end B has still to cover distance BC. The time required for this will be t = BC/c.

According to Huygen’s principle, point A starts emitting secondary wavelets and in time t, these will cover a distance c t = BC and spread. Hence, with point A as centre and BC as radius, draw a circular arc. Draw tangent CD on this arc from point C. Obviously, the CD is the reflected wavefront inclined at an angle ‘r’. As incident wavefront and reflected wavefront, both are in the plane of the paper, the 1^st law of reflection is proved.

To prove the second law of reflection, consider ΔABC and ΔADC. BC = AD (by construction),
∠ABC = ∠ADC = 90° and AC is common.

Therefore, the two triangles are congruent and, hence, ∠BAC = ∠DCA or ∠i = ∠r, i.e.the angle of reflection is equal to the angle of incidence, which is the second law of reflection.
Or
The refractive index of medium 2, w.r.t. medium 1 equals the ratio of the sine of the angle of incidence (in medium 1) to the sine of the angle of refraction (in medium 2), The diagram is as shown.

From the diagram

Question 2.
(a) Sketch the refracted wavefront for the incident plane wavefront of the light from a distant object passing through a convex lens.
Answer:

(b) Using Huygens’s principle, verify the laws of refraction when light from a denser medium is incident on a rarer medium.
Answer:
Refraction from denser to the rarer medium: Let XY be plane refracting surface separating two media of refractive index μ₁ and μ₂ (μ₁ > μ₂)

Let a plane wavefront AB incident at an angle i. According to Huygen’s principle, each point on the wavefront becomes a source of secondary wavelets and

Time is taken by wavelets from B to C = Time taken by wavelets from A to D


(c) For yellow light of wavelength 590 nm incident on a glass slab, the refractive index of glass Is 1.5. Estimate the speed and wavelength of yellow light Inside the glass slab. (CBSE 2019C)
Answer:
Given λ = 590 nm, μ = 1.5
Velocity of light inside glass slab.
∴ v = \(\frac{C}{\mu}=\frac{3 \times 10^{8}}{1.5}\) = 2 × 10⁸ ms^-1

Wavelength of yellow light inside the glass slab.
λ₁ = \(\frac{\lambda}{\mu}=\frac{290}{1.5}\) = 393.33 nm

Question 3.
(a) State the postulates of Huygen’s wave theory.
Answer:
The postulates are
All points on a given wavefront are taken as point sources for the production of spherical secondary waves, called wavelets, which propagate outward with speed characteristic of waves in that medium.

(b) Draw the type of wavefront that corresponds to a beam of light (i) coming from a very far off the source and (ii) diverging from a point source.
Answer:
After some time has elapsed, the new position of the wavefront is the surface tangent to the wavelets or the envelope of the wavelets in the forward direction.


Question 4.
What is meant by the diffraction of light? Obtain an expression for the first minimum of diffraction.
Answer:
The divergence of light from its initial line of travel when it passes through an opening or an obstacle is called diffraction or the phenomenon of bending of light around the sharp corners and spreading into the regions of the geometrical shadow is called diffraction.

Consider that a monochromatic source of light S, emitting light waves of wavelength λ, is placed at the principal focus of the convex lens L₁. A parallel beam of light, i.e. a plane wavefront, gets incident on a narrow slit AB of width ‘a’ as shown in the figure.

The diffraction pattern is obtained on a screen Lying at a distance D from the slit and at the focal plane of the convex lens L₂.

Consider a point P on the screen at which wavelets travelLing in a direction making angle O with CO are brought to focus by the lens. The wavelets from different parts of the slit do not reach point P in phase, although they are initially in phase. It is because they cover unequal distances in reaching point R The waveLets from points A and B will have a path difference equal to BN.

From the right angLed ANB, we have
BN = AB sin θ or BN = a sin θ …(1)

Suppose that the point P on the screen is at such a distance from the centre of the screen that BN = λ. and the angle θ = θ₁.

Then, equation 1 gives
λ = a sin θ₁ or sin θ₁ = \(\frac{λ}{a}\)

Such a point one screen will be the position of the first secondary minimum.

Question 5.
Describe an experiment to show that light waves are transverse in nature.
Answer:
Light is a transverse wave. This can be shown with the help of this simple experiment. The figure shows an unpolarized light beam incident on the first polarising sheet, called the polariser where the transmission axis is indicated by the straight line on the polariser.

The polariser can be a thin sheet of tourmaline (a complex boro-silicate). The Light, which is passing through this sheet, is polarised vertically as shown, where the transmitted electric vector is E_o. A second polarising sheet called the analyser intercepts this beam with its transmission axis at an angle θ to the axis of the polariser. As the axis of the analyser is rotated slowly, the intensity of Light received beyond It goes on decreasing.

When the transmitting axis of the analyser becomes perpendicular to the transmission axis of the polariser, no beam is obtained beyond the analyser. This means that the anaLyser has further polarised the beam coming from the polariser. Since only transverse waves can be polarised, therefore, this shows that light waves are transverse In nature.

Question 6.
Derive an expression for the width of the central maxima for diffraction of light at a single-slit. How does this width change with an increase in the width of the slit?
Answer:
The width of the central maxima for diffraction of light at a single-slit is the distance from the 1st diffraction minima on one side of the central maxima to the 1st diffraction minima on another side of the central maxima.

If a be the width of slit then for 1st diffraction minima, we have
sin θ = ± λ
or
sin θ = θ = ± \(\frac{λ}{a}\)

Angular width of central maxima
= 2θ = ± \(\frac{2λ}{a}\)

If D be the distance between the slit and the screen, then linear width ‘x’ of the central maxima is given by
x = d × 2θ = D × \(\frac{2λ}{a}\) = \(\frac{2Dλ}{a}\)

As the width (a) of the slit is increased, the linear width of central maxima goes on decreasing because x ∝ \(\frac{1}{a}\)

Question 7.
(a) Sketchthegraphshowingthevariation of the intensity of transmitted light on the angle of rotation between a polarizer and an analyser.
Answer:
The graph is as shown below

(b) A ray of light is incident at an angle of incidence i_p on the surface of separation between air and a medium of refractive index µ, such that the angle between the reflected and refracted ray is 90°. Obtain the relation between i_p and µ.
Answer:
Suppose an unpolarised light beam is an incident on a surface as shown in the figure below. The beam can be described by two electric field components, one parallel to the surface (the dots) and the other perpendicular to the first and to the direction of propagation (the arrows). It is found that the parallel components reflect more strongly than the other component, this results in a partially polarised beam. Furthermore, the refracted ray is also partially polarized.

Now suppose the angle of incidence, i is varied until the angle between the reflected and the refracted beam is 90°. At this particular angle of incidence, the reflected beam is completely polarised with its electric vector parallel to the surface, while the refracted beam is partially polarised. The angle of incidence at which this occurs is called the polarising angle i_p.

An expression can be obtained relating the polarising angle to the index of refraction n, of the reflecting surface. From the figure, we see that at the polarising angle
i_p + 90° + r = 180°
or
r = 90° – i_p.

Using Snell’s law we have n = \(\frac{\sin i}{\sin r}=\frac{\sin i_{p}}{\sin r}\)

Now sin r = sin (90° – i_p) = cos i_p, therefore the above expression becomes
n = \(\frac{\sin i_{p}}{\cos i_{p}}\) = tan ip

Question 8.
Describe Young’s double-slit experiment to produce an interference pattern due to a monochromatic source of light. Deduce the expression for the fringe width. (CBSE Delhi 2011)
Answer:
Consider a point P on the viewing screen; the screen is Located at a perpendicular distance D from the screen containing the slits S₁ and S₂, which are separated by a distance d as shown in the figure. Let us assume that the source of tight is monochromatic having a waveLength X. Under these conditions, the waves emerging from S and S have the same frequency and amplitude and are In phase.

If δ is the path difference between the waves from slits S₂ and S₁ at point P then
δ = r₂ – r₁ = d sin θ …(1)

where it is assumed that r1 and r2 are parallel, which is approximately true because D is much greater than d.

If this path difference is either zero or some Integral multiple of the wavelength, the waves are in phase at P and constructive Interference results. Therefore, the condition for bright fringes, or constructive interference, at P is given by
δ = d sin θ = m λ …(2)

Here m is called the order number. In order to find the position of maxima measured vertically from O to P, i.e. y we assume that D >> d, that is, the distance from the slits to the screen is much larger than the distance between the two slits. (In practice D = 1 m and d = a few millimetre) Under these conditions, θ is small, and so the approximation sin θ ≈ tan θ. Therefore, from triangle POQ in the figure, we find that
sin θ = tan θ = \(\frac{Y}{D}\) …(3)

Substituting this in equation (2) we have
d \(\frac{Y}{D}\) = m λ …(2)

rewriting the above equation we have
y = \(\frac{mDλ}{d}\) …(4)
This gives the distance of the mth maxima from the centre 0 of the screen. For m = 0, 1, 2, 3, 4, ……. the values of y are
y_o = 0 …….. position of the central bright fringe

y₁ = \(\frac{Dλ}{d}\) ……. position of the first bright fringe

y₂ = \(\frac{2Dλ}{d}\) ……. position of the second bright fringe

y_m = \(\frac{mDλ}{d}\) …….position of the mth bright fringe

The spacing between any two bright fringes gives the width of a dark fringe. Therefore,

This gives the width of the dark fringe. Similarly, when the path difference is an odd multiple of λ/2, the waves arriving at P will be 180° out of phase and will give rise to destructive interference. There¬fore, the condition for dark fringes, or de¬structive interference, at P is given by

This gives the distance of the mth minima from the centre O of the screen. For m is equal to 0, 1, 2, 3, ……. we have
y’_o = \(\frac{Dλ}{2d}\), ……. position of the first dark fringe,
y’₁ = \(\frac{3Dλ}{2d}\) ……. position of the second dark fringe
y’₂ = \(\frac{5Dλ}{2d}\) ……. position of the third dark fringe and,
y’_m = \(\frac{(2 m+1) D \lambda}{2 d}\) ……. position of the mth dark fringe

Now the spacing between two consecutive dark fringes gives the width of the bright fringe, therefore, a width of the bright fringe is

From equations 5 and 9 we find that the fringe width of the bright and the dark fringe is the same, therefore, the fringe width is given by
β = \(\frac{Dλ}{d}\) …(10)

Question 9.
(a) Explain, with the help of a diagram, how plane polarised light Is obtained by scattering.
Answer:
It is found that when unpolarised sunlight is incident on a small dust particle or air molecules it is scattered in all directions. The light scattered in a direction perpendicular to the incident light is found to be completely polarised.

The electric vector of incident light has components both in the plane of the paper as well as perpendicular to the plane of the paper. The electrons under the influence of the electric vector acquire motion in both directions. However, radiation scattered by the molecules in perpendicular direction contains only components represented by dots (.), i.e. polarised perpendicular to the plane of the paper.

(b) Between two polaroids placed In crossed position a third Polaroid is introduced. The axis of the third Polaroid makes an angle of 30° with the axis of the first polaroid. Find the intensity of transmitted light from the system assuming / to be the intensity of polarised light obtained from the first polaroid. (CBSE A! 2011C)
Answer:
Let l_o be the intensity of light passing through the first polaroid.

The intensity of light passing through the middle polaroid whose axes are inclined at 30° to the first polaroid by Malus law is
l’ = l_o cos² 30° = l_o × 3/4 = 3l_o/4

The intensity of light passing through the system is, therefore, (for the second crystal θ = 60°)
l” = l’cos² 60° = 3l_o/4 × 1 /4
or
l” = 3l_o/16

Question 10.
(a) Why are coherent sources necessary to produce a sustained interference pattern?
Answer:
Interference will be sustained if there is a constant phase difference between the two interfering waves. This is possible if the two waves are coherent.

(b) In Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference Is λ, is K units. Find out the intensity of light at a point where the path difference is λ/3. (CBSE Delhi 2012)
Answer:
Intensity at any point on the screen
l = l₁ + l₂ + 2\(\sqrt{1_{1}1_{2}}\) cos Φ

Let l0 be the intensity of either source, then l₁= l₂ = l_o
When p = λ, Φ = 2π

Question 11.
Use Huygen’s principle to explain the formation of the diffraction pattern due to a single-slit illuminated by a monochromatic source of light.
When the width of the slit is made double the original width, how would this affect the size and intensity of the central diffraction band? (CBSE Delhi 2012)
Answer:
(a) The arrangement is shown below in the figure.

When a plane wavefront WW’ falls on a single-slit AB, each point on the unblocked portion ADB of wavefront sends out secondary, wavelets in all the directions. For secondary waves meeting at point O, the path difference between waves is zero and hence the secondary waves reinforce each other giving rise to the central maximum at the symmetrical point O.

Consider secondary waves travelling in a direction making an angle θ with DO and reaching the screen at point P. Obviously, path difference between the extreme secondary waves reaching point P from A and B.
= BC = AB sin θ = a sin θ.

If this path difference a sin θ = λ, then point P will be minimum Intensity. In this situation, wavefront may be supposed to consist of two equal halves AD and BD and for every point on AD, there will be a corresponding point on DB having a path difference λ/2. Consequently, they nullify the effect of each other and point P behaves as the first secondary minimum. In general, if path difference a sin θ = nλ where n = 1, 2, 3, …. then we have secondary minima corresponding to that angle of diffraction θ_n.

However, if for some point P₁ on the screen secondary waves BP₁ and AP₁ differ In path by 3λ/2 then point P₁ will be the position of the first secondary maxima. Because in this situation, wavefront AB may be divided into three equal parts such that path difference between corresponding points on the first and second will be λ/2 and they will nullify. But secondary waves from the third part remain as such and give rise to the first secondary maxima, whose Intensity will be much less than that of central maxima.

In general, if path difference a sin θ = (2n + 1)θ/2, where n = 1, 2, 3, …… then we have nth secondary maxima corresponding to these angles.

The width of the central maxima is the distance between the first secondary minima on either side of the centre of the screen. The width of the central maxima is twice the angle θ subtended by the first minima on either side of the central maxima. Now sin θ = \(\frac{\lambda}{a}\). Since θ is small a therefore it can be replaced by tan θ, hence sin θ = \(\frac{\lambda}{a}=\frac{y}{L} or y = [latex]\frac{L \lambda}{\mathrm{a}}\).

This gives the distance of the first secondary minima on both sides of the centre of the screen. Therefore, the width of the central maxima is 2y, hence
2y = \(\frac{2L \lambda}{\mathrm{a}}\)

(b) The size reduces by half according to the relation: size = X/d. Intensity becomes twice the original intensity.

Question 12.
(a) Using the phenomenon of polarization, show how the transverse nature of light can be demonstrated.
Answer:
Light is a transverse wave. This can be shown with the help of this simple experiment. The figure shows an unpolarized light beam incident on the first polarising sheet, called the polariser where the transmission axis is indicated by the straight line on the polariser.

The polariser can be a thin sheet of tourmaline (a complex boro-silicate). The Light, which is passing through this sheet, is polarised vertically as shown, where the transmitted electric vector is E_o. A second polarising sheet called the analyser intercepts this beam with its transmission axis at an angle θ to the axis of the polariser. As the axis of the analyser is rotated slowly, the intensity of Light received beyond It goes on decreasing.

When the transmitting axis of the analyser becomes perpendicular to the transmission axis of the polariser, no beam is obtained beyond the analyser. This means that the anaLyser has further polarised the beam coming from the polariser. Since only transverse waves can be polarised, therefore, this shows that light waves are transverse in nature.

(b) Two polaroids P₁ and P₂ are placed with their pass axes perpendicular to each other. Unpolarised light of intensity l0 is Incident on P₁. A third polaroid P₃ is kept in between P₁ and P₂ such that its pass axis makes an angle of 30° with that of P₁. Determine the intensity of light transmitted through P₁ P₂ and P₃. (CBSE AI 2014)
Answer:
The intensity of light passing through P₁ is half of the light falling on it. Therefore, the light coming out of P₁ is l0_o/2 Now light coming out of P₃ is

and light coming out of polariser P₂ is

Question 13.
What does a Polaroid consist of? Show using a simple Polaroid that light waves are transverse in nature. The intensity of light coming out of a Polaroid does not change irrespective of the orientation of the pass axis of the polaroid. Explain why. (CBSE AI 2015)
Answer:
A polaroid consists of long-chain molecules aligned in a particular direction.

Let the light from an ordinary source (like a sodium lamps pass through a polaroid sheet P₁ it is observed that its intensity is reduced by half. Now, let an identical piece of polaroid P₂ be placed before P₁ As expected, the light from the lamp is reduced in intensity on passing through P₂2 alone. But now rotating P₁ has a dramatic effect on the light coming from P₂. In one position, the intensity transmitted by P₂ followed by P₁ is nearly zero. When turned by 90° from this position, P₁ transmits nearly the full intensity emerging from P₂ shown in the figure.

Since only transverse waves can be polarised, therefore, this experiment shows that light waves are transverse in nature. The electric vectors (associated with the propagating light wave) along the direction of the aligned molecules get absorbed. Thus, if an unpolarized light wave is an incident on such a polaroid then the light wave will get linearly polarized with the electric vector oscillating along a direction perpendicular to the aligned molecules; this direction is known as the pass-axis of the Polaroid.

Thus, if the light from an ordinary source (like a sodium lamp) passes through a polaroid sheet P it is observed that its intensity is reduced by half. Rotating P has no effect on the transmitted beam and transmitted intensity remains constant as there is always an electric vector that oscillates in a direction perpendicular to the direction of the aligned molecules.

Question 14.
Find an expression for intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids. In which position of the polaroid sheet will the transmitted Intensity be maximum? (CBSE Delhi 2015)
Answer:
Let l0 be the intensity of polarised light after passing through the first polarizer P₁. Then the intensity of light after passing through the second polarizer P₂ will be
l = l_o cos² θ

where θ is the angle between pass axes of P₁ and P₂. Since P₁ and P₃ are crossed the angle between the pass axes of P₂ and P₃ will be (π/2 – θ). Hence the intensity of light emerging from P₃ will be

The transmitted intensity will be maximum when 2θ = π/2 or θ = π/4

Question 15.
Distinguish between unpolarised and linearly polarised light. Describe with the help of a diagram how unpolarised light gets linearly polarised by scattering. (CBSE Delhi 2015)
Answer:
The two lights will be allowed to pass through a polariser. When the polarizer is rotated in the path of these two light beams, the intensity of light remains the same in all the orientations of the polariser, then the light is unpolarised. But if the intensity of light varies from maximum to minimum then the light beam is a polarised light beam.

It is found that when unpolarised sunlight is incident on a small dust particle or air molecules it is scattered in all directions. The light scattered in a direction perpendicular to the incident light is found to be completely polarised.

The electric vector of incident light has components both in the plane of the paper as well as perpendicular to the plane of the paper. The electrons under the influence of the electric vector acquire motion in both directions. However, radiation scattered by the molecules in perpendicular direction contains only components represented by dots (.), i.e. polarised perpendicular to the plane of the paper.

Question 16.
(a) Explain how a diffraction pattern is formed due to interference of secondary wavelets of light waves from a slit.
Answer:
(a) Diffraction at single slit: Consider monochromatic plane wavefront WW’ incident on the slit AB as shown in Fig. Imagine the slit to be divided into a large number of very narrow strips of equal width parallel to the slit. When the wavefront reaches the slit, each narrow strip parallel to the slit can be considered to be a source of Huygen’s secondary wavelets.

Central Maxima. Let us first consider the effect of all the wavelets at point O. All wavelets cover the same distance and reach 0 in the same phase. The wavelets superimpose constructively and give maximum intensity at O.

Position of secondary minima. Consider the intensity at a point P at an angle θ with the original direction.
Draw AL ⊥ BP.
In ΔABL


or
sin θ = \(\frac{λ}{a}\)

For second minima, BL = 2A
or a sin θ = 2λ

Similarly, for nth minima a sin θ = nλ
or sin θ = \(\frac{nλ}{a}\) , where n = 1, 2, ……

Position of secondary maxima: If path difference BL = \(\frac{3λ}{2}\), then slit AB can be divided into three equal parts and the path difference between wavelets from corresponding points in the first two parts will be \(\frac{λ}{2}\) and hence cancels each other’s effect and produces destructive Interference and the third path will produce its maxima at P called first secondary maxima.

Similarly, if BL = \(\frac{5λ}{2}\), we get second secondary maxima and so on.

For nth maxima
a sin θ = (2n + 1)\(\frac{λ}{a}\)
or
sin θ = \(\frac{(2 n+1) \lambda}{2 a}\)

Thus we find that the intensity of the central fringe is maximum whereas that of other fringes fall off rapidly in either direction from the centre of the fringe pattern.

(b) Sodium light consists of two wavelengths, 5900 Å and 5960 Å. If a slit of width 2 × 10^-4 m is Illuminated by sodium light, find the separation between the first secondary maxima of the diffraction pattern of the two wavelengths on a screen placed 1.5m away. (CBSE 2019C)
Answer:
The separation between the first secondary maxima of the diffraction pattern of two wavelengths is:

Question 17.
(a) Derive Snell’s law on the basis of Huygen’s wave theory when light Is travelling from a denser to a rarer medium.
Answer:
The ray diagram is as shown.


(b) Draw the sketches to differentiate between plane wavefront and spherical wavefront. (CBSE AI 2016)
Answer:
Spherical wavefront and plane wavefront

Question 18.
The figure drawn here shows the geometry of path differences for diffraction by a single-slit of width a.

Give appropriate ‘reasoning’ to explain why the intensity of light is
(a) Maximum at the central point C on the screen.
(b) (Nearly) zero for point P on the screen when θ = λ / a.
Hence write an expression for the total linear width of the central maxima on a screen kept at a distance D from the plane of the slit. (CBSE Delhi 2016C)
Answer:
(a) At central point C, the angle is zero, all path differences are zero. Hence all the parts of the slit contribute to the same phase. This gives the maximum intensity at point C.
(b)

From figure
NP – LP = NQ= a sin θ = aθ When θ = λ/a

Then path difference NP – LP = aθ = λ.
Hence MP – LP = NP – MP = λ/2

It implies that the contribution from corresponding points in two halves of the slit has a phase difference of π. Therefore, contributions from two halves cancel each other in pairs, resulting in a zero net intensity at point P on the screen. Half angular width of central maxima = λ/a Half linear width = λD/a

Linear width of central maxima = 2λD/a

Question 19.
Two polaroids, P₁ and P₂, are ‘set-up’ so that their ‘pass axis is ‘crossed’ with respect to each other. AQ third Polaroid, P₃ is now introduced between these two so that its ‘pass axis makes an angle with the ‘pass axis of P_1. A beam of unpolarized light, of Intensity I, Is Incident on P₁ If the Intensity of light that gets transmitted through this combination of three polaroids, Is I’, find the ratio \(\left(\frac{I^{\prime}}{I}\right)\) when θ equals: (i) 30°, (ii) 45° (CBSE Delhi 2016C)
Answer:
The diagram is as shown.

The intensity of unpolarized light is given as l.
It becomes half after passing through Polaroid P₁ (l/2)
Using Malus law for the intensity of light passing through P3 we have
l₁ = \(\left(\frac{l}{2}\right)\) cos²θ

This intensity l₁ is Incident on P₂, hence the intensity of light coming out of P₂ will be

Question 20.
What is the effect on the Interference pattern observed In Young’s double-slit experiment In the following cases:
(a) Screen is moved away from the plane of the slits,
(b) Separation between the slits is Increased and
(c) Widths of the slits are doubled. Give the reason for your answer.
Answer:
The fringe width is given by the expression
β = \(\frac{Dλ}{d}\)

(a) When D Is Increased, the fringe width Increases.
(b) When d 1s Increased the fringe width decreases.
(c) If the width w of the slits Is changed then Interference occurs only If \(\frac{1}{w}\) > \(\frac{1}{d}\) remains satisfied, where d is the distance between the slits.

Question 21.
Two slits In Young’s double-slit experiment are illuminated by two different lamps emitting light? Will you observe the Interference pattern? Justify your answer. Find the ratio of Intensities at two points on a screen In Young’s double-slit experiment, when waves from two slits have a path difference of (i) 0 and (ii) λ/4.
Answer:
The two sources act as Independent sources of light and hence can never be coherent. In a light, source light is produced by billions of atoms under proper excitation condition and each atom acts independently of the other atoms. Thus there Is no coherence ‘ between these two Independent sources hence no interference.

The phase difference corresponding to the two paths are Φ = 0 and Φ = π/2

Now intensity at the screen when the phase difference is Φ = 0 is
l_A = l₁ +l₂ + 2\(\sqrt{l_{1} l_{2}}\)cos Φ
or
l_A = l + l + 2\(\sqrt{l l}\)cos Φ = 4l

Now intensity at the screen when the phase difference is Φ = 90° Is
l_B = l₁ +l₂ + 2\(\sqrt{l_{1} l_{2}}\)cos Φ
or
l_B = l + l + 2\(\sqrt{l l}\)cos 90° = 2l
Therefore, ratio of intensities is
\(\frac{l_{A}}{l_{B}}=\frac{4l}{2l}\) = 2

Question 22.
(a) In a single-slit diffraction pattern, how does the angular width of the central maximum vary, when
(i) the aperture of the slit Is Increased?
(ii) distance between the slit and the screen is decreased?
Justify your answer In each case.
Answer:
The angular width of the central maxima in a single-slit diffraction pattern is given by 2θ = \(\frac{2λ}{a}\) where λ is the wavelength of light and ‘a’ the slit width.
(i) When the aperture of the slit is increased the angular width decreases.
(ii) When the distance between the slit and the screen is decreased, the angular width will remain the same but the linear width will increase.

(b) How Is the diffraction pattern different from the interference pattern obtained In Young’s double¬slit experiment? (CBSE Delhi 2011C)
Answer:
The difference is shown in the table:

Diffraction	Interference
1. It is due to interference between the wavelets starting from two parts of the same wavefront.	1. It is a superposition of the two waves starting from two coherent sources.
2. The intensity of the consecutive bands goes on decreasing.	2. All bright fringes have the same intensity.
3. Fringes have poor contrast.	3. Fringes have good contrast.
4. Diffraction fringes are not of the same width.	4. Fringes may or may not be of the same width.

Question 23.
(a) Can two independent monochromatic light sources be used to obtain a steady interference pattern? Justify your answer. (CBSE 2019C)
Answer:
No, because the phase difference between the light waves from two independent sources keeps on changing continuously and for a steady interference pattern, the phase difference between the waves should remain constant with time. Hence two independent monochromatic light sources cannot produce a steady interference pattern.

(b) In Young’s double-slit experiment, explain the formation of interference fringes and obtain an expression for the fringe width.
Answer:
Interference of light

Let S₁, S₂ be the two fine slits illuminated by a monochromatic source S of wavelength λ.

The intensity of light at any point P on the screen at a distance D from the slit depends upon the path difference between S₂P and S₁P.
∴ Path difference = S₂P – S₁P
= (S₂A + AP) – S₁P

Path difference = S₂A = d sin θ
Since θ is small, sin θ can be replaced by tan θ.
∴ Path difference = d tan θ = d\(\frac{y}{D}\)

Constructive interference [Bright Fringes]. For bright fringes, the path difference should be equal to integral multiple of A.
∴ \(\frac{dy}{D}\) = nλ
or
y = n\(\frac{λD}{d}\)

For nth fringe, let us write y as y_n
so y_n = n\(\frac{λD}{d}\)

The spacing between two consecutive bright fringes is equal to the width of a dark fringe.
∴ Width of the dark fringe.
β = y_n – y_n-1 = \(\frac{nλD}{d}\) – (n – 1)\(\frac{λD}{d}\) = \(\frac{λD}{d}\) …(i)

Destructive interference. [Dark Fringes].
For dark fringes, the path difference should be an odd multiple of λ/2.
∴ \(\frac{d}{D}\)y = (2n + 1)λ/2
or
y’ = \(\frac{D}{d}\)(2n + 1)\(\frac{λ}{2}\) = \(\frac{(2 n+1) \lambda D}{2 d}\)

For nth fringe, writing y as yn, we get
y’_n = \(\frac{(2 n+1) \lambda D}{2 d}\).

The spacing between two consecutive dark fringes is equal to the width of a bright fringe.
∴ Width of the bright fringe.

From Eqs. (i) and (ii), we find that dark and bright fringes are of same width given by
β = \(\frac{λD}{d}\).

(c) In an interference experiment using monochromatic light of wavelength A, the intensity of light of point, where the path difference is X, on the screen is K units. Find out the Intensity of light at a point when path difference is λ/4. (CBSE 2019C)
Answer:
The intensity of light on the screen where the waves meet having phase difference Φ is:


Question 24.
(a) Two monochromatic waves emanating from two coherent sources have the displacements represented by y₁ = a cos ωt and y₂ = a cos (ωt + Φ), where Φ is the phase difference between the two displacements. Show that the resultant intensity at a point due to their superposition is given by l = 4l_o cos² Φ/2, where l_o = a².
Answer:
Let the displacements of the waves from the sources S₁ and S₂ at a point on the screen at any time t be given by y₁ = a cos ωt and y₂ = a cos (ωt + Φ), where Φ is the constant phase difference between the two waves. By the superposition principle, the resultant displacement at point P is given by

Thus the amplitude of the resultant displacement is A = 2a cos (Φ/2)

Therefore the intensity at that point is
l = A² = 4a² cos² \(\frac{Φ}{2}\)

(b) Hence obtain the conditions for constructive and destructive interference. (CBSE AI 2014C)
Answer:
Bright fringes: For bright fringes I = max, therefore Φ = 0° or cos Φ = +1
or
Φ = 2 n π
Dark fringes: For dark fringes l = 0,
therefore Φ = π or cos Φ = -1
or
Φ = (2n + 1) π

Question 25.
A parallel beam of monochromatic light falls normally on a narrow slit of width ‘a’ to produce a diffraction pattern on the screen placed parallel to the plane of the slit. Use Huygens’ principle to explain that
(a) the central bright maxima is twice as wide as the other maxima.
Answer:
The width of the central maxima is the distance between the first secondary minima on either side of the centre of the screen. The width of the central maxima is twice the angle 6 subtended by the first minima on either side of the central maxima.

Now sin θ = \(\frac{λ}{a}\). Since θ is small there, a fore it can be replaced by tan 0, hence
tan θ = \(\frac{λ}{a}\) = \(\frac{y}{L}\)
or
y = \(\frac{Lλ}{a}\). This gives the distance of the first secondary min¬ima on both sides of the centre of the screen. Therefore, the width of the central maxima is 2y, hence
2y = \(\frac{2Lλ}{a}\)

(b) the Intensity falls as we move to successive maxima away from the centre on either side. (CBSE Delhi 2014C)
Answer:
The maxima in diffraction pattern are formed at (n +1 /2) λ/a, with n = 2, 3, etc. These become weaker with increasing n, since only one third, one- fifth, one-seventh, etc., of the slit contributes in these cases.

Question 26.
Answer the following questions:
(a) In a double-slit experiment using the light of wavelength 600 nm, the angular width of the fringe formed on a distant screen is 0.1°. Find the spacing between the two slits.
Answer:

(b) Light of wavelength 5000 A propagating 1n air gets partly reflected from the surface of the water. How will the wavelengths and frequencies of the reflected and refracted light be affected? (CBSE Delhi 2015)
Answer:
No change in the wavelength and frequency of reflected light. In the case of refracted light, there is no change in frequency but the wavelength becomes 1/1.33 times the original wavelength.

Question 27.
(a) Good quality sun-glasses made of polaroids are preferred over ordinary coloured glasses. Justify your answer,
Answer:
A Polaroid sun-glass limits the light entering the eye, thus providing a soothing effect.

(b) Two polaroids and P₂ are placed In crossed positions. A third Polaroid P₃ is kept between P₁ and P₂ such that the pass axis of P₃ is parallel to that of P₁. How would the Intensity of light l₂ transmitted through P₂ vary as P₃ Is rotated? Draw a plot of Intensity l₂ Vs the angle ‘θ’, between pass axes of P₁ and P₃. (CBSE AI 2015C)
Answer:
Let l0 be the intensity of polarised light after passing through the first polarizer P₁ Then the intensity of light after passing through the second polarizer P₃ will be
l = l_o cos² θ
where θ is the angle between pass axes of P₁ and P₃. Since P₁ and P₂ are crossed the angle between the pass axes of P₂ and P₃ will be (π/2 – θ).

Hence the intensity of light emerging from P2 will be

Therefore, the transmitted intensity will be maximum when θ = π/4

For graph

Question 28.
In a single-slit diffraction pattern, how does the angular width of central maximum change, when
(a) slit width is decreased,
(b) distance between the slit and screen is increased, and
(c) light of smaller visible wavelength is used? Justify your answer in each case.
Answer:
We know that the angular width of the central maximum of the diffraction pattern of a single-slit is given by
w = \(\frac{2Dλ}{a}\).

(a) If slit width ‘a’ is decreased, the angular width will increase because
x ∝ \(\frac{1}{a}\)

(b) Increase in distance between the slit and the screen does not affect the angular width of diffraction maxima. However, linear width of the maxima
w = \(\frac{2Dλ}{a}\) will increase.

(c) If the light of a smaller visible wavelength is used, the angular width is decreased because x ∝ λ.

Question 29.
Light, from a sodium lamp, is passed through two polaroid sheets P₁ and P₂ kept one after the other. Keeping P₁ fixed, P₂ is rotated so that its ‘pass axis can be at different angles, θ, with respect to the pass-axis of P₁
An experimentalist records the following data for the Intensity of light coming out of P₂ as a function of the angle θ.

(a) l_o = Intensity of beam falling on P₁ One of these observations is not in agreement with the expected theoretical variation of l, Identify this observation and write the correct expression.
Answer:
The observation \(\frac{l_{0}}{2 \sqrt{2}}\) is not correct. It should be l_o/4.

(b) Define the Brewster angle and write the expression for it in terms of the refractive index of the medium.
Answer:
It is the angle of incidence at which the refracted and the reflected rays are perpendicular to each other. It is related to the refractive index as tan i_p = n

Question 30.
(a) Light, from a monochromatic source, is made to fall on a single-slit of variable width. An experimentalist records the following data for the linear width of the principal maxima ‘ on a screen kept at a distance of 1 m from the plane of the slit.

Use any two observations from this data to estimate the value of the wavelength of light used.
Answer:
The width of the central maxima is given by the expression β = \(\frac{2Lλ}{a}\)
or
λ₁ = \(\frac{βa}{2L}\)

Using the values of the first observation we have
λ₁ = \(\frac{6 \times 10^{-3} \times 0.1 \times 10^{-3}}{2 \times 1}\) = 0.3 × 10^-6

Using the values of the second observation we have
λ₁ = \frac{3 \times 10^{-3} \times 0.2 \times 10^{-3}}{2 \times 1} = 0.3 × 10^-6

Thus the wavelength of light used is λ = 0.3 × 10^-6 m

(b) Show that the Brewster angle iB for a given pair of transparent media is related to their critical angle ic through the relation i_c = sin^-1 (ωt i_B)
Answer:
We know that

Question 31.
For a single-slit of width, “a” the first minimum of the interference pattern of monochromatic light of wavelength λ occurs at an angle of λ/a. At the same angle λ/a, we get a maximum for two narrow slits separated by a distance ‘a’. Explain. (CBSE Delhi 2014)
Answer:
The path difference between two secondary wavelets is given by nλ = a sin θ. Since θ is a very small sin θ = 0. So, for the first-order diffraction n = 1, the angle is λ/a. Now we know that θ must be very small θ = 0 (nearly) because of which the diffraction pattern is minimum.

Now for interference case, for two interfering waves of intensity l1 and l2 we must have two slits separated by a distance. We have the resultant intensity,
l = l₁ + l₂ + 2\(\sqrt{l_{1} l_{2}}\) cos θ

Since θ = 0° (nearly) corresponding to angle λ/a so cos θ = 1 (nearly)
So,
l = l₁ + l₂ + 2\(\sqrt{l_{1} l_{2}}\) cos θ
l = l₁ + l₂ + 2\(\sqrt{l_{1} l_{2}}\)

We see the resultant intensity is the sum of the two intensities, so there is a maxima corresponding to the angle λ/a.

This is why, at the same angle of λ/a, we get a maximum for two narrow slits separated by a distance “a”.

Question 32.
Show using a proper diagram of how unpolarised light can be linearly polarized by reflection from a transparent glass surface. (CBSE AI 2018, Delhi 2018)
Answer:
An ordinary beam of light, on reflection from a transparent medium, becomes partially polarised. The degree of polarisation increases as the angle of incidence is increased. At a particular value of the angle of incidence, the reflected beam becomes completely polarised. This angle of incidence is called the polarising angle (i_p).

Question 33.
Answer the following questions:
(a) In a single-slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band?
Answer:
The size reduces by half according to the relation: size = λ/d. Intensity increases fourfold.

(b) In what way is diffraction from each slit related to the interference pattern in a double-slit experiment?
Answer:
The intensity of interference fringes in a double-slit arrangement is modulated by the diffraction pattern of each slit.

(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why?
Answer:
Waves diffracted from the edge of the circular obstacle interfere constructively at the centre of the shadow producing a bright spot.

(d) Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily? (NCERT)
Answer:
For diffraction or bending of waves by obstacles/apertures by a large angle, the size of the latter should be comparable to the wavelength. If the size of the obstacle/aperture is much too large compared to the wavelength, diffraction is by a small angle. Here the size is of the order of a few metres. The wavelength of light is about 5 × 10^-7 m, while sound waves of, say, 1 kHz frequency have a wavelength of about 0.3 m. Thus, sound waves can bend around the partition while light waves cannot.

Question 34.
(a) When an unpolarized light of intensity l0 is passed through a polaroid, what is the intensity of the linearly polarised light? Does it depend on the orientation of the polaroid? Explain your answer.
Answer:
(a) The intensity of the linearly polarised light would be l_o/2.
No, it does not depend on the orientation.
Explanation: The polaroid will let the component of the unpolarized light, parallel to its pass axis, to pass through it irrespective of its orientation.

(b) A plane polarised beam of light is passed through a polaroid. Show graphically the variation of the intensity of the transmitted light with an angle of rotation of the polaroid incomplete one rotation. (CBSE Delhi 2018C)
Answer:
We have l = l_o cos² θ
∴ The graph is as shown below

Question 35.
(a) If one of two identical slits producing interference in Young’s experiment is covered with glass so that the light intensity passing through it is reduced to 50%, find the ratio of the maximum and minimum intensity of the fringe in the Interference pattern.
Answer:
As intensity is directly proportional to the square of the amplitude.


(b) What kind of fringes do you expect to observe if white light is used Instead of monochromatic light?
Answer:
The central fringe will be white and the remaining will be coloured fringes of different width in the VIBGYOR sequence.

Question 36.
Find an expression for intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids. In which position of the polaroid sheet will the transmitted intensity be maximum? (CBSE Delhi 2015)
Answer:
Let l0 be the intensity of polarised light after passing through the first polarizer P₁ Then the intensity of light after passing through the second polarizer P₂ will be
l = l_o cos² θ
where 0 is the angle between pass axes of P₁ and P₂. Since P₁ and P₃ are crossed the angle between the pass axes of P₂ and P3 will be (π/2 – θ). Hence the intensity of light emerging from P₃ will be

The transmitted intensity will be maximum when 2θ = π/2 or θ = π/4

Question 37.
Derive the expression for the fringe width In Young’s double-slit experiment.
Answer:
Consider a point P on the viewing screen; the screen is Located at a perpendicular distance D from the screen containing the slits S₁ and S₂, which are separated by a distance d as shown in the figure. Let us assume that the source of tight is monochromatic having a waveLength X. Under these conditions, the waves emerging from S and S have the same frequency and amplitude and are In phase.

If δ is the path difference between the waves from slits S₂ and S₁ at point P then
δ = r₂ – r₁ = d sin θ …(1)

where it is assumed that r₁ and r₂ are parallel, which is approximately true because D is much greater than d.

If this path difference is either zero or some Integral multiple of the wavelength, the waves are in phase at P and constructive Interference results. Therefore, the condition for bright fringes, or constructive interference, at P is given by
δ = d sin θ = m λ …(2)

Here m is called the order number. In order to find the position of maxima measured vertically from O to P, i.e. y we assume that D >> d, that is, the distance from the slits to the screen is much larger than the distance between the two slits. (In practice D = 1 m and d = a few millimetre) Under these conditions, θ is small, and so the approximation sin θ ≈ tan θ. Therefore, from triangle POQ in the figure, we find that
sin θ = tan θ = \(\frac{Y}{D}\) …(3)

Substituting this in equation (2) we have
d \(\frac{Y}{D}\) = m λ …(2)

rewriting the above equation we have
y = \(\frac{mDλ}{d}\) …(4)
This gives the distance of the mth maxima from the centre 0 of the screen. For m = 0, 1, 2, 3, 4, ……. the values of y are
y_o0 = 0 …….. position of the central bright fringe

y₁ = \(\frac{Dλ}{d}\) ……. position of the first bright fringe

y₂ = \(\frac{2Dλ}{d}\) ……. position of the second bright fringe

y_m = \(\frac{mDλ}{d}\) …….position of the mth bright fringe

The spacing between any two bright fringes gives the width of a dark fringe. Therefore,

This gives the width of the dark fringe. Similarly, when the path difference is an odd multiple of λ/2, the waves arriving at P will be 180° out of phase and will give rise to destructive interference. There¬fore, the condition for dark fringes, or de¬structive interference, at P is given by

This gives the distance of the mth minima from the centre O of the screen. For m is equal to 0, 1, 2, 3, ……. we have
y’_o = \(\frac{Dλ}{2d}\), ……. position of the first dark fringe,

y’₁ = \(\frac{3Dλ}{2d}\) ……. position of the second dark fringe

y’₂ = \(\frac{5Dλ}{2d}\) ……. position of the third dark fringe and,

y’_m = \(\frac{(2 m+1) D \lambda}{2 d}\) ……. position of the mth dark fringe

Now the spacing between two consecutive dark fringes gives the width of the bright fringe, therefore, a width of the bright fringe is

From equations 5 and 9 we find that the fringe width of the bright and the dark fringe is the same, therefore, the fringe width is given by
β = \(\frac{Dλ}{d}\) …(10)

Question 38.
(a) Using Huygens’ principle, draw a diagram to show the propagation of a wave-front originating from a monochromatic point source.
Answer:

(b) Describe diffraction of light due to a single-slit. Explain the formation of a pattern of fringes obtained on the screen and plot showing a variation of intensity with path difference in single-slit diffraction.
Answer:
Let us discuss the nature of the Fraunhofer diffraction pattern produced by a single-slit. Let us examine the waves coming from the various portions of the slit, as shown in the figure. According to Huygens principle, each portion of the slit acts as a source of waves. Hence, light from one portion of the slit can interfere with the light of another portion, and the resultant intensity on the screen will depend on the direction of θ. The secondary waves coming from the different parts of the slit interfere to produce either maxima or minima, thereby giving rise to a diffraction pattern.

For the diagram

2y = \(\frac{2Lλ}{a}\)

Question 39.
What is the effect on the interference fringes in Young’s double-slit experiment due to each of the following operations:
(a) the screen is moved away from the plane of the slits;
Answer:
The angular separation of the fringes remains constant (= λ/d). The actual separation of the fringes increases in proportion to the distance of the screen from the plane of the slits

(b) the (monochromatic) source is replaced by another (monochromatic) source of shorter wavelength;
Answer:
The separation of the fringes (and also angular separation) decreases.

(c) the separation between the two slits is Increased
Answer:
The separation of the fringes (and also angular separation) decreases.

(d) the source slit is moved closer to the double-slit plane;
Answer:
Let s be the size of the source and S its distance from the plane of the two slits. For interference, fringes to be seen the
condition \(\frac{s}{S}<\frac{\lambda}{d}\) should be satisfied otherwise interference patterns produced by different parts of the source overlap and no fringes are seen. Thus as S decreases (i.e. the source slit is brought closer), the interference pattern gets less and less sharp and when the source is brought too close for this condition to be valid the fringes – disappear. Till this happens, the fringe separation remains fixed.

(e) the width of the source slit is Increased;
Answer:
Same as in (d). As the source slit width increases fringe pattern gets less and less sharp. When the source slit is so wide that the condition \(\frac{s}{S}<\frac{\lambda}{d}\) is not satisfied the interference pattern disappears.

(f) the monochromatic source is replaced by a source of white light? (In each operation take all parameters other than the one specified to remain unchanged.)
Answer:
The interference patterns due to different component colours of white light overlap (incoherently). The central bright fringes for different colours are in the same position. Therefore, the central fringe is white. For a point P for which S₂P – S₁P= λ_b /2, where (λ_b = 400 nm) represents the wavelength for the blue colour, the blue component will be absent and the fringe will appear red in colour. Slightly farther away from where S₂Q – S₁Q = λ_b = λ_r/2 where λ_r (800 nm) is the wavelength for the red colour, the fringe will be predominantly blue. Thus, the fringe closest on either side of the central white fringe is red and the farthest will appear blue. After a few fringes, no clear fringe pattern is seen.

Question 40.
What is the diffraction of light? Draw a graph showing the variation of intensity with the angle in a single-slit diffraction experiment. Write one feature which distinguishes the observed pattern from the double-slit interference pattern. How would the diffraction pattern of a single-slit be affected when:
(i) the width of the slit is decreased?
(ii) the monochromatic source of light is replaced by a source of white light?
Answer:
Diffraction is the bending of light around obstacles or openings. It is a consequence of the wave nature of light. For diffraction to take place the obstacle should be of the order of the wavelength of light.

The intensity distribution for the single-slit diffraction pattern is as shown.

The intensity of all bright fringes is the same in Young’s interference pattern, but in diffraction, the intensity of bright fringes falls off on both sides of the central fringe.
We know that the angular width of the central maximum of the diffraction pattern of a single-slit is given by = \(\frac{2Dλ}{a}\)
(i) If slit width ‘a’ is decreased, the angular width will increase because
x ∝ \(\frac{1}{a}\)

(ii) When monochromatic light is replaced by white light, all the seven wavelengths form their own diffraction pattern, so coloured fringes are formed. The first few fringes are visible but due to overlapping the clarity of fringes decreases as the order increases.

Question 41.
(a) Define a wavefront. Using Huygens’ geometrical construction, explain with the help of a diagram how the plane wavefront travels from the instant t₁ to t₂ in the air.
Answer:
(a) Wavefront: It is the locus of the medium or points of a medium that is in the same phase of disturbance
First, consider a plane wave moving through free space as shown in the figure. At t = 0, the wavefront is indicated by the plane labelled AA’. In Huygens’s construction, each point on this wavefront is considered a point source. For clarity, only a few points on AA’ are shown. With these points as sources for the wavelets, we draw circles each of radius cΔt, where c is the speed of light in free space and Δt is the time of propagation from one wavefront to the next. The surface drawn tangent to these wavelets is the plane BB’, (Wavefront at a later time t). Here A₁ A₂ = B₁ B₂ = C₁C₂ = C_τ

(b) A plane wavefront is incident on a convex lens. Explain, with the help of the diagram, the shape of the refracted wavefront formed. (CBSE AI 2019)
Answer:
Spherical wavefront

Question 42.
(a) In Young’s double-slit experiment, derive the condition for
(i) constructive Interference and
(ii) destructive Interference at a point on the screen.
Answer:
Consider a point P on the viewing screen; the screen is Located at a perpendicular distance D from the screen containing the slits S₁ and S₂, which are separated by a distance d as shown in the figure. Let us assume that the source of tight is monochromatic having a wavelength X. Under these conditions, the waves emerging from S and S have the same frequency and amplitude and are In phase.

If δ is the path difference between the waves from slits S₂ and S₁ at point P then
δ = r₂ – r₁ = d sin θ …(1)

where it is assumed that r₁ and r₂ are parallel, which is approximately true because D is much greater than d.

If this path difference is either zero or some Integral multiple of the wavelength, the waves are in phase at P and constructive Interference results. Therefore, the condition for bright fringes, or constructive interference, at P is given by
δ = d sin θ = m λ …(2)

Here m is called the order number. In order to find the position of maxima measured vertically from O to P, i.e. y we assume that D >> d, that is, the distance from the slits to the screen is much larger than the distance between the two slits. (In practice D = 1 m and d = a few millimetre) Under these conditions, θ is small, and so the approximation sin θ ≈ tan θ. Therefore, from triangle POQ in the figure, we find that
sin θ = tan θ = \(\frac{Y}{D}\) …(3)

Substituting this in equation (2) we have
d \(\frac{Y}{D}\) = m λ …(2)

rewriting the above equation we have
y = \(\frac{mDλ}{d}\) …(4)
This gives the distance of the mth maxima from the centre 0 of the screen. For m = 0, 1, 2, 3, 4, ……. the values of y are
y_o = 0 …….. position of the central bright fringe

y₁ = \(\frac{Dλ}{d}\) ……. position of the first bright fringe

y₂ = \(\frac{2Dλ}{d}\) ……. position of the second bright fringe

y_m = \(\frac{mDλ}{d}\) …….position of the mth bright fringe

The spacing between any two bright fringes gives the width of a dark fringe. Therefore,

This gives the width of the dark fringe. Similarly, when the path difference is an odd multiple of λ/2, the waves arriving at P will be 180° out of phase and will give rise to destructive interference. There¬fore, the condition for dark fringes, or de¬structive interference, at P is given by

This gives the distance of the mth minima from the centre O of the screen. For m is equal to 0, 1, 2, 3, ……. we have
y’_o = \(\frac{Dλ}{2d}\), ……. position of the first dark fringe,

y’₁ = \(\frac{3Dλ}{2d}\) ……. position of the second dark fringe

y’₂ = \(\frac{5Dλ}{2d}\) ……. position of the third dark fringe and,

y’_m = \(\frac{(2 m+1) D \lambda}{2 d}\) ……. position of the mth dark fringe

Now the spacing between two consecutive dark fringes gives the width of the bright fringe, therefore, the width of the bright fringe is

From equations 5 and 9 we find that the fringe width of the bright and the dark fringe is the same, therefore, the fringe width is given by
β = \(\frac{Dλ}{d}\) …(10)

(b) A beam of light consisting of two wavelengths, 800 nm and 600 nm, is used to obtain the interference fringes in Young’s double-slit experiment on a screen placed 1.4 m away. If the two slits are separated by 0.28 mm, calculate the least distance from the central bright maximum where the bright fringes of the two wavelengths coincide. (CBSE Al 2012)
Answer:
Let at a distance y from central maxima the bright fringes due to wavelengths l1 and l2 coincide first time. For this to happen, nl₁ = (n + 1)l₂, where n is an integer.
or
\(\frac{n+1}{n}=\frac{\lambda_{1}}{\lambda_{2}}=\frac{800 \times 10^{-9}}{600 \times 10^{-9}}=\frac{4}{3}\) solving for n we have n = 3

It means that at distance nth (3^th) maxima for wavelength λ₁ is just coinciding with (n + 1)^th (4^th) maxima for wavelength l₂

Question 43.
(a) How does an unpolarized light incident on a Polaroid get polarised? Describe briefly, with the help of a necessary diagram, the polarisation of light by reflection from a transparent medium.
(b) Two polaroids ‘A’ and ‘B’ are kept in a crossed position. How should a third Polaroid ‘C’ be placed between them so that the intensity of polarised light transmitted by Polaroid B reduces to 1/8th of the intensity of unpolarized light incident on A? (CBSE AI 2012)
Answer:
(a) The polariser allows only those vibrations of light to pass which are parallel to the pass axis of the polariser and blocks the remaining vibrations. Thus the light vibrations are restricted to only one plane.

Suppose an unpolarised light beam is an incident on a surface as shown in (figure a). The beam can be described by two electric field components, one parallel to the surface (the dots) and the other perpendicular to the first and to the direction of propagation (the arrows). It is found that the parallel components reflect more strongly than the other component, this results in a partially polarised beam. Furthermore, the refracted ray is also partially. polarised. Now suppose the angle of incidence 6, is varied until the angle between the reflected and the refracted beam is 90° (fig b). At this particular angle of incidence, the reflected beam is completely polarised with its electric vector parallel to the surface, while the refracted beam is partially polarised.

(b) Let l0 be the intensity of unpolarized light incident on Polaroid A. Then l_o/2 will be the intensity of polarised light after passing through the first polariser A. Then the intensity of light after passing through second polariser B will be l = (l_o/2)cos²θ

where θ is the angle between pass axes of A and B. Since A and B are crossed the angle between the pass axes of B and C will be (π/2 – θ). Hence the intensity of light emerging from C will be

Therefore, we have
\(\frac{l_{0}}{8}=\frac{l_{0}}{8}\) sin² 2θ or sin 2θ = 1 or 2θ = 90°
i.e. θ = π/4

Therefore, the transmitted intensity will be maximum when θ = π/4

Question 44.
(a) Describe any two characteristic features which distinguish between interference and diffraction phenomena. Derive the expression for the intensity at a point of the interference pattern in Young’s double-slit experiment.
(b) In the diffraction due to a single slit experiment, the aperture of the slit is 3 mm. If monochromatic light of wavelength 620 nm is incident normally on the slit, calculate the separation between the first order minima and the 3^rd order maxima on one side of the screen. The distance between the slit and the screen is 1.5 m. (CBSE Delhi 2019)
Answer:
(a) (i) Interference pattern has a number of equally spaced bright and dark bands, while the diffraction pattern has a central bright maximum which is twice as wide as the other maxima.

(ii) In interference pattern, the intensity of all bright fringes is the same, while in diffraction pattern intensity of bright fringes goes on decreasing with the increasing order of the maxima.

Let the displacements of the waves from the sources S₁ and S₂ at a point P on the screen at any time t be given by y₁ = a cos cot and y₂ = a cos (ωt + Φ), where Φ is the constant phase difference between the two waves.

By the superposition principle, the resultant displacement at point P is given by

Thus the amplitude of the resultant displacement is
A = 2a cos (Φ/2) …(4)

Therefore the intensity at that point is
l = A² = 4a² cos²\(\frac{Φ}{2}\) …(5)

(b) Given a = 3mm = 3 × 10^-3m, λ = 620nm = 620 × 10^-9 m, D = 1.5 m Distance of first-order minima from centre of the central maxima = X_D1 = λ_D/a

Distance of third order maxima from centre of the central maxima X_D3 = 7λ_D/2a

Distance between first order minima and third order maxima = X_D3 – X_D1

Question 45.
(a) In Young’s double-slit experiment, deduce the conditions for obtaining constructive and destructive interference fringes. Hence deduce the expression for the fringe width.
Answer:
Consider a point P on the viewing screen; the screen is Located at a perpendicular distance D from the screen containing the slits S_D1 and S₂, which are separated by a distance d as shown in the figure. Let us assume that the source of tight is monochromatic having a waveLength X. Under these conditions, the waves emerging from S and S have the same frequency and amplitude and are In phase.

If δ is the path difference between the waves from slits S2 and S1 at point P then
δ = r₂ – r₁ = d sin θ …(1)

where it is assumed that r1 and r2 are parallel, which is approximately true because D is much greater than d.

If this path difference is either zero or some Integral multiple of the wavelength, the waves are in phase at P and constructive Interference results. Therefore, the condition for bright fringes, or constructive interference, at P is given by
δ = d sin θ = m λ …(2)

Here m is called the order number. In order to find the position of maxima measured vertically from O to P, i.e. y we assume that D >> d, that is, the distance from the slits to the screen is much larger than the distance between the two slits. (In practice D = 1 m and d = a few millimetre) Under these conditions, θ is small, and so the approximation sin θ ≈ tan θ. Therefore, from triangle POQ in the figure, we find that
sin θ = tan θ = \(\frac{Y}{D}\) …(3)

Substituting this in equation (2) we have
d \(\frac{Y}{D}\) = m λ …(2)

rewriting the above equation we have
y = \(\frac{mDλ}{d}\) …(4)
This gives the distance of the mth maxima from the centre 0 of the screen. For m = 0, 1, 2, 3, 4, ……. the values of y are
y_o = 0 …….. position of the central bright fringe

y₁ = \(\frac{Dλ}{d}\) ……. position of the first bright fringe

y₂ = \(\frac{2Dλ}{d}\) ……. position of the second bright fringe

y_m = \(\frac{mDλ}{d}\) …….position of the mth bright fringe

The spacing between any two bright fringes gives the width of a dark fringe. Therefore,

This gives the width of the dark fringe. Similarly, when the path difference is an odd multiple of λ/2, the waves arriving at P will be 180° out of phase and will give rise to destructive interference. Therefore, the condition for dark fringes, or destructive interference, at P is given by

This gives the distance of the mth minima from the centre O of the screen. For m is equal to 0, 1, 2, 3, ……. we have
y’_o = \(\frac{Dλ}{2d}\), ……. position of the first dark fringe,

y’₁ = \(\frac{3Dλ}{2d}\) ……. position of the second dark fringe

y’₂ = \(\frac{5Dλ}{2d}\) ……. position of the third dark fringe and,

y’_m = \(\frac{(2 m+1) D \lambda}{2 d}\) ……. position of the mth dark fringe

Now the spacing between two consecutive dark fringes gives the width of the bright fringe, therefore, the width of the bright fringe is

From equations 5 and 9 we find that the fringe width of the bright and the dark fringe is the same, therefore, the fringe width is given by
β = \(\frac{Dλ}{d}\) …(10)

(b) Show that the fringe pattern on the screen is actually a superposition of single-slit diffraction from each slit.
Answer:
It is a broader diffraction peak in which there appear several fringes of smaller width due to the double-slit interference pattern. This is shown below:

(c) What should be the width of each slit to obtain 10 maxima of the double-slit pattern within the central maximum of the single-slit pattern, for the green light of wavelength 500 nm, if the separation between two slits is 1 mm? (CBSE AI 2015)
Answer:
Given 10 b = width of central maxima in diffraction pattern λ = 500 nm, d= 1 mm,
Now 10\(\frac{Dλ}{d}\) = \(\frac{Dλ}{a}\)
or
a = \(\frac{d}{5}=\frac{1}{5}\) = 0.2 mm

Question 46.
(a) Define a wavefront. How is it different from a ray?
Answer:
A wavefront is defined as the locus of all adjacent points at which the phase of vibration of a physical quantity associated with the wave is the same.

It is in two dimensions while a ray is in one dimension.

(b) Depict the shape of a wavefront in each of the following cases.
(i) Light diverging from a point source,
Answer:

(ii) Light emerging out of a convex lens when a point source is placed at its focus.
Answer:

(c) Using Huygen’s construction of secondary wavelets, draw a diagram showing the passage of a plane wavefront from a denser into a rarer medium. (CBSE AI 2015C)
Answer:
The diagram is as shown

Question 47.
(a) Why does unpolarised light from a source show a variation in intensity when viewed through a Polaroid which is rotated? Show with the help of a diagram how unpolarised light from sun got linearly polarised by scattering.
(b) Three identical Polaroid sheets P1, P2 and P3 are oriented so that the pass axis of P2 and P3 are inclined at angles of 60° and 90° respectively with the pass axis of P1. A monochromatic source S of unpolarised light of intensity l0 is kept in front of the polaroid sheet P! as shown in the figure. Determine the intensities of light as observed by the observer at O when Polaroid P3 is rotated with respect to P2 at angles θ = 30° and 60°. (CBSE AI 2016)

Answer:
(a) When light passes through a Polaroid it absorbs all vibrations which are not parallel to its pass axis.
It is found that when unpolarised sunlight is incident on a small dust particle or air molecules it is scattered in all directions. The light scattered in a direction perpendicular to the incident light is found to be completely polarised.

The electric vector of incident light has components both in the plane of the paper as well as perpendicular to the plane of the paper. The electrons under the influence of the electric vector acquire motion in both directions. However, radiation scattered by the molecules in perpendicular direction contains only components represented by dots (.), i.e. polarised perpendicular to the plane of the paper.

(b) Let l0 be the intensity of light before passing P₁ Intensity of light coming out of P₁ will be l_o/2.

Then the intensity of light after passing through second poloroid (analyser) is
l₂ = \(\frac{l_{0}}{2}\)cos²60° = \(\frac{l_{0}}{8}\)

When P₃ is rotated with respect to P₂ at an angle of 30°, then angle be¬tween the pass axis of P₂ and P₃ will be θ = 30°+ 30° = 60°.
Hence l3 = l2 cos²60° = \(\frac{l_{0}}{8} \times \frac{1}{4}=\frac{l_{0}}{32}\)
or
θ = 30° – 30° = 0°

Therefore, l₃ = l₂ cos² 0° = l₂ = l_o/8
When P₃ is rotated with respect to P₂ at an angle of 60°, then the angle between the pass axis of P₂ and P₃ will be
θ = 30° + 60° = 90°

Question 48.
(a) Derive an expression for path difference in Young’s double-slit experiment and obtain the conditions for constructive and destructive interference at a point on the screen.
Answer:
Consider a point P on the viewing screen; the screen is Located at a perpendicular distance D from the screen containing the slits S₁ and S₂, which are separated by a distance d as shown in the figure. Let us assume that the source of tight is monochromatic having a waveLength X. Under these conditions, the waves emerging from S and S have the same frequency and amplitude and are In phase.

If δ is the path difference between the waves from slits S₂ and S₁ at point P then
δ = r₂ – r₁ = d sin θ …(1)

where it is assumed that r₁ and r₂ are parallel, which is approximately true because D is much greater than d.

If this path difference is either zero or some Integral multiple of the wavelength, the waves are in phase at P and constructive Interference results. Therefore, the condition for bright fringes, or constructive interference, at P is given by
δ = d sin θ = m λ …(2)

Here m is called the order number. In order to find the position of maxima measured vertically from O to P, i.e. y we assume that D >> d, that is, the distance from the slits to the screen is much larger than the distance between the two slits. (In practice D = 1 m and d = a few millimetre) Under these conditions, θ is small, and so the approximation sin θ ≈ tan θ. Therefore, from triangle POQ in the figure, we find that
sin θ = tan θ = \(\frac{Y}{D}\) …(3)

Substituting this in equation (2) we have
d \(\frac{Y}{D}\) = m λ …(2)

rewriting the above equation we have
y = \(\frac{mDλ}{d}\) …(4)
This gives the distance of the mth maxima from the centre 0 of the screen. For m = 0, 1, 2, 3, 4, ……. the values of y are
y_o = 0 …….. position of the central bright fringe

y₁ = \(\frac{Dλ}{d}\) ……. position of the first bright fringe

y₂ = \(\frac{2Dλ}{d}\) ……. position of the second bright fringe

y_m = \(\frac{mDλ}{d}\) …….position of the mth bright fringe

The spacing between any two bright fringes gives the width of a dark fringe. Therefore,

This gives the width of the dark fringe. Similarly, when the path difference is an odd multiple of λ/2, the waves arriving at P will be 180° out of phase and will give rise to destructive interference. There¬fore, the condition for dark fringes, or de¬structive interference, at P is given by

This gives the distance of the mth minima from the centre O of the screen. For m is equal to 0, 1, 2, 3, ……. we have
y’_o = \(\frac{Dλ}{2d}\), ……. position of the first dark fringe,
y’₁ = \(\frac{3Dλ}{2d}\) ……. position of the second dark fringe
y’₂ = \(\frac{5Dλ}{2d}\) ……. position of the third dark fringe and,
y’_m = \(\frac{(2 m+1) D \lambda}{2 d}\) ……. position of the mth dark fringe

Now the spacing between two consecutive dark fringes gives the width of the bright fringe, therefore, the width of the bright fringe is

From equations 5 and 9 we find that the fringe width of the bright and the dark fringe is the same, therefore, the fringe width is given by
β = \(\frac{Dλ}{d}\) …(10)

(b) The Intensity at the central maxima in Young’s double-slit experiment is l0. Find out the intensity at a point where the path difference is λ/6, λ/4 and λ/3. (CBSE AI 2016)
Answer:
Let l be the intensity of light coming out of each slit. Since waves at the central maxima are in phase, therefore, we have

Question 49.
(a) Define a wavefront. Using Huygens’ Principle, verify the laws of reflection at a plane surface.
Answer:
The wavefront is a locus of points that oscillate in the same phase.

Consider a plane wavefront AB incident obliquely on a plane reflecting surface MM^–. Let us consider the situation when one end A of wavefront strikes the mirror at an angle i but the other end B has still to cover distance BC. The time required for this will be t = BC/c.

According to Huygen’s principle, point A starts emitting secondary wavelets and in time t, these will cover a distance c t = BC and spread. Hence, with point A as centre and BC as radius, draw a circular arc. Draw tangent CD on this arc from point C. Obviously, the CD is the reflected wavefront inclined at an angle ‘r’. As incident wavefront and reflected wavefront, both are in the plane of the paper, the 1^st law of reflection is proved.

To prove the second law of reflection, consider ΔABC and ΔADC. BC = AD (by construction),
∠ABC = ∠ADC = 90° and AC is common.

Therefore, the two triangles are congruent and, hence, ∠BAC = ∠DCA or ∠i = ∠r, i.e.the angle of reflection is equal to the angle of incidence, which is the second law of reflection.
Or
The refractive index of medium 2, w.r.t. medium 1 equals the ratio of the sine of the angle of incidence (in medium 1) to the sine of the angle of refraction (in medium 2), The diagram is as shown.

From the diagram

(b) In a single-slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band? Explain.
Answer:
Size of central maxima reduces to half, (Size of central maxima = 2λD/d) Intensity increases.
This is because the amount of light, entering the slit, has increased and the area, over which it falls, decreases.

(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the obstacle. Explain why. (CBSE AI 2018, Delhi 2018)
Answer:
This is because of the diffraction of light. Light gets diffracted by the tiny circular obstacle and reaches the centre of the shadow of the obstacle.
β = \(\frac{Dλ}{d}\).

(i) When D is decreased, the fringe width decreases.
(ii) When d is Increased, the fringe width decreases.

Question 50.
The following figure shows an experiment set up similar to Young’s double-slit experiment to observe interference of light.

Here SS₂ – SS₁ = λ/4
Write the condition of (i) constructive,
(ii) destructive Interference at any point P in terms of path difference Δ = S₂P – S₁P.

Does the central fringe observed in the above set up tie above or below O? Give reason in support of your answer.
Yellow light of wavelength 6000°A produces fringes of width 08 mm in Young’s double-slit experiment. What will be the fringe width if the light source is replaced by another monochromatic source of wavelength 7500° A and separation between the slits is doubled?
Answer:
Given SS₂ – SS₁ = λ/4
Now path difference between the two waves from slits S₁ and S₂ on reaching point P on screen is

(a) For constructive interference at point P both difference Δx = nλ.
Therefore

where n = 0, 1, 2, 3, …

(b) For destructive interference at point P path difference

For central bright fringe, putting n = 0 in equation (1), we get
\(\frac{y d}{D}=-\frac{\lambda}{4}\)
or
y = – \(\frac{\lambda D}{4 d}\)

The -ve sign indicates that the central bright fringe will be observed below centre O of the screen, at distance below it.
Given λ₁ = 6000°A, β₁ = 0.8 mm,
λ₂ = 7500°A, β₂ = ?, d₂ = 2d₁

Using the expression
β = \(\frac{Dλ}{d}\) we have
\(\frac{\beta_{2}}{\beta_{1}}=\frac{\lambda_{2} d_{1}}{\lambda_{1} d_{2}}=\frac{7500}{6000} \times \frac{d_{1}}{2 d_{1}}\)

Solving we have β₂ = 0.5 mm

Question 51.
(a) There are two sets of apparatus of Young’s double-slit experiment. Inset A, the phase difference between the two waves emanating from the slits does not change with time, whereas in set B, the phase difference between the two waves from the slits changes rapidly with time. What difference will be observed in the pattern obtained on the screen in the two setups?
(b) Deduce the expression for the resultant intensity in both the above- mentioned setups (A and B), assuming that the waves emanating from the two slits have the same amplitude A and same wavelength λ.
Or
(a) The two polaroids, in a given setup, are kept ‘crossed’ with respect to each other. A third polaroid, now put in between these two polaroids, can be rotated. Find an expression for the dependence of the intensity of light I, transmitted by the system, on the angle between the pass axis of the first and the third polaroid. Draw a graph showing the dependence of l on θ.
(b) When an unpolarized light is an incident on a plane glass surface, find the expression for the angle of incidence so that the reflected and refracted light rays are perpendicular to each other. What is the state of polarization, of reflected and refracted light, under this condition?
Answer:
(a) Set A: Stable interference pattern, the positions of maxima and minima, does not change with time.
Set B: Positions of maxima and minima will change rapidly with time and an average uniform intensity distribution will be observed on the screen.

(b) Let the displacements of the waves from the sources S1 and S2 at a point on the screen at any time t be given by y₁ = a cos ωt and y₂ = a cos (ωt + Φ), where Φ is the constant phase difference between the two waves. By the superposition principle, the resultant displacement at point P is given by

Thus the amplitude of the resultant displacement is A = 2a cos (Φ/2)

Therefore, the intensity at that point is l = A² = 4a² cos²\(\frac{Φ}{2}\) – 4l_o cos²\(\frac{Φ}{2}\)
Since Φ = 0, l.e. there is no phase difference, hence l = 4 l_o

Inset B, the intensity will be given by the average intensity
l = 4l_o\(\left(\cos ^{2} \frac{\phi}{2}\right)\) = 2l_o
Or
(a) Let P₁ and P₂ be the two crossed Polaroids and P₃ be the polaroid kept between the two. Let l_o, be the intensity of polarised light after passing through the first Polaroid P₁. Then the Intensity of light after passing through the second Polaroid P₂ will be,
l = l_o cos² θ
where θ is the angle between pass axes of P₁ and P₂. Since P₁ and P₂ are crossed the angle between the pass axes of P₂ and P₃ will be (π/2 – 0). Hence the intensity of light emerging from P2₂ will be

The transmitted intensity will be maximum when π = π/4. The graph is as shown

(b) The ray diagram is as shown.

When light Is incident at a certain angle called Brewster angle, the reflected beam is completely polarised with its electric vector parallel to the surface, while the refracted beam is partially polarised. The angle of incidence at which this occurs is called the polarising angle i_B.

From figure we see that at the polaris-ing angle
i_B + 90° + r = 180°
or
r = 90° – i_B.

Using Snell’s law we have
μ = \(\frac{\sin i_{\mathrm{B}}}{\sin r}\)

Now sin r = (90° – i_B), therefore the above expression becomes

Nature of polarisation: Reflected light and linearly polarised

Numerical Problems:
Formulae for solving numerical problems

• Fringe width is given by β = \(\frac{Dλ}{d}\)
• Brewster’s Law: μ = \(\frac{\sin \theta_{p}}{\cos \theta_{p}}\) = tan θ_p
• Intensity of light coming out of a polariser l = l_o cos² θ
• If l₁ and l₂ are the intensities of light emitted by the two sources, w₁ and w₂ be the widths of the two slits, a₁ and a₂ be the amplitudes of the waves from the two slits then,
  \(\frac{l_{1}}{l_{2}}=\frac{w_{1}}{w_{2}}=\frac{a_{1}^{2}}{a_{2}^{2}}\)
• lf l_max and lmin be the intensities of light at the maxima and the minima the
  \(\frac{l_{\max }}{l_{\min }}=\frac{\left(a_{1}+a_{2}\right)^{2}}{\left(a_{1}-a_{2}\right)^{2}}\)

Question 1.
Laser light of wavelength 630 nm incident on a pair of slits produces an interference pattern in which the bright fringes are separated by 7.2 mm. Calculate the wavelength of another source of laser light that produces interference fringes separated by 8.1 mm using the same pair of slits. (CBSE Al 2011C)
Answer:
Given λ₁ = 630 nm, β₁ = 7.2 mm, β₂ = 8.1 mm, λ₂ =?

We know that β = \(\frac{Dλ}{d}\), for the same value of D and d we have β₂ ∝ λ,
Therefore, we have

Question 2.
What is the speed of light in a denser medium of polarising angle 30°? (CBSE Delhi 2019)
Answer:

Question 3.
Laser light of wavelength 640 nm incident on a pair of slits produces an interference pattern in which the bright fringes are separated by 7.2 mm. Calculate the wavelength of another source of light that produces interference fringes separated by 8.1 mm using the same arrangement. Also, find the minimum value of the order (n) of the bright fringe of the shorter wavelength which coincides with that of the longer wavelength. (CBSE AI 2012C)
Answer:
Given λ₁ = 640 nm, λ₂ = ?, b₁ = 7.2 mm and b₂ = 8.1 mm

Using the expression β = \(\frac{Dλ}{a}\) we have

Now n fringes of shorter wavelength will coincide with (n – 1) fringes of Longer wavelength
n₁λ₁ = n₂λ₂
or
n × 640 = (n – 1) × 720
Solving for n we have n =9

Question 4.
Yellow light (λ = 6000 Å) illuminates a single-slit of width 1 × 10 m. Calculate
(a) the distance between the two dark tines on either side of the central maximum, when the diffraction pattern is viewed on a screen kept 1.5 m away from the slit;
(b) the angular spread of the first diffraction minimum. (CBSE AI 2012C)
Answer:

Question 5.
Find the ratio of Intensities of two points P and Q on the screen in Young’s double-slit experiment when the waves from sources S₁ and S₂ have a phase difference of (i) π/3 and (11) π/2.
Answer:
Intensity at the screen when the phase difference is Φ = π/3 is

Now intensity at the screen when the phase difference is Φ = 90° is

Therefore, ratio of intensities is
\(\frac{l_{p}}{l_{Q}}=\frac{3l}{2l}=\frac{3}{2}\)

Question 6.
In young’s double-slit experiment, two sifts are separated by 3 mm distance and illuminated by the light of wavelength 480 nm. The screen is at 2 m from the plane of the slits. Calculate the separation between the 8th bright fringe and the 3N dark fringes observed with respect to the central bright fringe.
Answer:
Using the formuLa

Question 7.
Two coherent sources have Intensities In the ratio 25: 16. Find the ratio of the Intensities of maxima to minima, after the interference of light occurs.
Answer:
In the given problem

Question 8.
The ratio of intensities of maxima and minima in an interference pattern is found to be 25: 9. Calculate the ratio of light intensities of the sources producing this pattern.
Answer:
Given

Question 9.
In Young’s double-slit experiment using the light of wavelength 400 nm, Interference fringes of width X are obtained. Th. the wavelength of light is Increased to 600 nm and the separation between the slits Is halved. If on. wants the observed fringe width on the screen to b. the same in the two cases, find the ratio of the distance between the screen and the plan. of the Interfering sources In the two arrangements.
Answer:
Let D₁ be the distance between the screen and the sources when a tight of wavelength 400 nm Is used.

β = \(\frac{Dλ}{a}\)
or
X = \(\frac{D_{1} \times 400 \times 10^{-9}}{d}\) …(i)

Let D₂ be the distance between the screen and the sources to obtain the same fringe width, when light of wavelength 600 nm is used. Then,
X = \(\frac{D_{2} \times 600 \times 10^{-9}}{d}\) ….(ii)

From the equations (i) and (ii), we have
\(\frac{D_{1}}{D_{2}}=\frac{600 \times 10^{-9}}{400 \times 10^{-9}}\) = 1.5

Question 10.
In Young’s slit experiment, Interference fringes are observed on a screen, kept at D from the slits. If the screen Is moved towards the slits by 5 × 10^-2 m, the change in fringe width Is found to be 3 × 10^-5 m. If the separation between the slits is 10-3 m, calculate the wavelength of the light used.
Answer:
Given ΔD = 5 × 10^-2-2 m, Δβ = 3 × 10^-5-5 m, d = 10^-3 m,

Using the relation β = \(\frac{Dλ}{a}\) we have for the two cases
β₁ = \(\frac{D_{1} \lambda}{d}\) and
β₂ = \(\frac{D_{2} \lambda}{d}\) subtracting we have
β₁ – β₂ = \(\frac{λ}{d}\)(D1 – D2)
or

Question 11.
A slit of width ‘a’ is illuminated by monochromatic light of wavelength 700 nm at Normal Incidence. Calculate the value of ‘a’ for the position of
(a) the first minimum at an angle of diffraction of 30°.
(b) first maximum at an angle of diffraction of 30°.
Answer:
Given X = 700 nm = 7 × 10^-7 m

Question 12.
Estimate the angular separation between . first order maximum and third order minimum of the diffraction pattern due to a single-slit of width 1 mm, when light of wavelength 600 nm is incident normal on it. (CBSE AI 2015C)
Answer:
Given d = 1 mm = 10^-3 m,
λ = 600 nm = 6 × 10^-7 m,
For first order maxima
d sin θ = n λ
or
sin θ = \(\frac{n \lambda}{d}=\frac{6 \times 10^{-7}}{10^{-3}}\) = 6 × 10^-4
or
θ_max = 1.047°
Now for minima we have
d sin θ = (2n+1 )\(\frac{λ}{2}\)

For third-order minima we have n = 3
Therefore we have

sin θ = (2 × 3 + 1)\(\frac{λ}{2d}\) = \(\frac{7λ}{2d}\)
= \(\frac{7 \times 6 \times 10^{-7}}{2 \times 10^{-3}}\)
= 21 × 10^-4
or
θ_min = 6.397°

Therefore, angular separation is
θ_min – θ_max = 6.397 – 1.047 = 5.35°