Wave Optics — Page 8 | NEET Physics

Q71

In a Young's double slit experiment, the ratio of the slit's width is 4 : 1. The ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be :

A 25 : 9

B 4 : 1

C

{\left( {\sqrt 3 + 1} \right)^4}:16

D 9 : 1

Correct Answer

Option D

Solution

{I_1} = 4{I_0}

{I_2} = {I_0}

{I_{\max }} = {\left( {\sqrt {{I_0}} + \sqrt {{I_2}} } \right)^2}

= {\left( {2\sqrt {{I_0}} + \sqrt {{I_0}} } \right)^2} = 9{I_0}

{I_{\min }} = {\left( {\sqrt {{I_1}} - \sqrt {{I_2}} } \right)^2}

= {\left( {2\sqrt {{I_0}} - \sqrt {{I_0}} } \right)^2} = {I_0}

$\therefore$

{{{{\mathop{\rm I}\nolimits} _{\max }}} \over {{I_{\min }}}} = {{9{I_0}} \over {{I_0}}} = {9 \over 1}

Q72

A beam of plane polarised light of large cross-sectional area and uniform intensity of 3.3 Wm-2 falls normally on a polariser (cross sectional area 3

\times

10-4 m2) which rotates about its axis with an angular speed of 31.4 rad/s. The energy of light passing through the polariser per revolution, is close to :

A 1.0

\times

10-5 J

B 1.0

\times

10-4 J

C 1.5

\times

10-4 J

D 5.0

\times

10-4 J

Correct Answer

Option B

Solution

Intensity, I = 3.3 Wm–3 Area, A = 3 × 10–4 m2 Angular speed, $\omega$ = 31.4 rad/s

I = {I_0}{\cos ^2}(\omega t)

\Rightarrow {I_{av}} = {{{I_0}} \over 2}

$\because$

\left\langle {{{\cos }^2}\theta } \right\rangle

=

{1 \over 2}

, in one time period $\therefore$

E = {{{I_0}} \over 2} \times A \times (\Delta t)

and

\Delta t = {{2\pi } \over \omega } = {{2 \times 3.14} \over {31.4}} = {1 \over 5}s

$\therefore$

E = {{3.3} \over 2} \times 3 \times {10^{ - 4}} \times {1 \over 5} = 1 \times {10^{ - 4}}\,J

Q73

Two light waves having the same wavelength

\lambda

in vacuum are in phase initially. Then the first wave travels a path L1 through a medium of refractive index n1 while the second wave travels a path of length L2 through a medium of refractive index n2 . After this the phase difference between the two waves is :

A

{{2\pi } \over \lambda }\left( {{n_1}{L_1} - {n_2}{L_2}} \right)

B

{{2\pi } \over \lambda }\left( {{n_2}{L_1} - {n_1}{L_2}} \right)

C

{{2\pi } \over \lambda }\left( {{{{L_1}} \over {{n_1}}} - {{{L_2}} \over {{n_2}}}} \right)

D

{{2\pi } \over \lambda }\left( {{{{L_2}} \over {{n_1}}} - {{{L_1}} \over {{n_2}}}} \right)

Correct Answer

Option A

Solution

Phase difference =

{{2\pi } \over \lambda }

$\times$ optical path difference =

{{2\pi } \over \lambda }\left( {{n_1}{L_1} - {n_2}{L_2}} \right)

Q74

Interference fringes are observed on a screen by illuminating two thin slits 1 mm apart with a light source (

\lambda

= 632.8 nm). The distance between the screen and the slits is 100 cm. If a bright fringe is observed on a screen at a distance of 1.27 mm from the central bright fringe, then the path difference between the waves, which are reaching this point from the slits is close is

A 1.27

\mu

m

B 2.05

\mu

m

C 2.87 nm

D 2 nm

Correct Answer

Option A

Solution

y =

{{nD\lambda } \over d}

$\Rightarrow$ n =

{{yd} \over {D\lambda }}

=

{{1.27 \times {{10}^{ - 3}} \times {{10}^{ - 3}}} \over {1 \times 632.8 \times {{10}^{ - 9}}}}

= 2 Path difference

\Delta

x = n $\lambda$ = 2 $\times$ 632.8 nm = 1265.6 nm = 1.27 $\mu$ m

Q75

When a polaroid sheet is rotated between two crossed polaroids then the transmitted light intensity will be maximum for a rotation of :

A

90^\circ

B

30^\circ

C

45^\circ

D

60^\circ

Correct Answer

Option C

Solution

Let

\mathrm{I}_0

be intensity of unpolarised light incident on first polaroid.

\mathrm{I}_1=

Intensity of light transmitted from

1^{\text{st }}

polaroid

=\frac{\mathrm{I}_0}{2}

$\theta$ be the angle between

1^{\mathrm{st}}

and

2^{\text{nd }}

polaroid $\phi$ be the angle between

2^{\text{nd }}

and

3^{\text{rd }}

polaroid

\theta+\phi=90^{\circ}

(as

1^{\text{st }}

and

3^{\text{rd }}

polaroid are crossed)

\phi=90^{\circ}-\theta

\mathrm{I}_2=

Intensity from

2^{\text{nd }}

polaroid

\mathrm{I}_2=\mathrm{I}_1 \cos ^2 \theta=\frac{\mathrm{I}_0}{2} \cos ^2 \theta

\mathrm{I}_3=

Intensity from

3^{\text{rd }}

polaroid

\mathrm{I}_3=\mathrm{I}_2 \cos ^2 \phi

\mathrm{I}_3=\mathrm{I}_1 \cos ^2 \theta \cos ^2 \phi

\mathrm{I}_3=\frac{\mathrm{I}_0}{2} \cos ^2 \theta \cos ^2 \phi

\phi=90-\theta

\begin{aligned} & I_3=\frac{I_0}{2} \cos ^2 \theta \sin ^2 \theta \\ & I_3=\frac{I_0}{2}\left[\frac{2 \sin \theta \cos \theta}{2}\right]^2 \\ & I_3=\frac{I_0}{8} \sin ^2 2 \theta \end{aligned}

\mathrm{I}_3

will be maximum when

\sin 2 \theta=1

\begin{aligned} & 2 \theta=90^{\circ} \\ & \theta=45^{\circ} \end{aligned}

Q76

A polarizer - analyser set is adjusted such that the intensity of light coming out of the analyser is just 10% of the original intensity. Assuming that the polarizer - analyser set does not absorb any light, the angle by which the analyser need to be rotated further to reduce the output intensity to be zero, is :

A 71.6o

B 90o

C 18.4o

D 45o

Correct Answer

Option C

Solution

I = I0 cos2 $\theta$ $\Rightarrow$

{{{I_0}} \over {10}}

= I0 cos2 $\theta$ $\Rightarrow$ cos $\theta$ =

{1 \over {\sqrt {10} }}

$\Rightarrow$ $\theta$ = 71.6o $\therefore$ $\phi$ = 90 - 71.6 = 18.4o

Q77

The Young's double slit interference experiment is performed using light consisting of 480 nm and 600 nm wavelengths to form interference patterns. The least number of the bright fringes of 480 nm light that are required for the first coincidence with the bright fringes formed by 600 nm light is

A 8

B 4

C 5

D 6

Correct Answer

Option C

Solution

d \sin \theta = m \lambda

For two wavelengths,

\lambda_1 = 600\,\text{nm}

and

\lambda_2 = 480\,\text{nm}

, the bright fringes coincide when the conditions

m \lambda_2 = n \lambda_1

are simultaneously satisfied for integers

m

and

n

, representing the bright fringe order for each wavelength. Rearranging the equation gives:

\frac{m}{n} = \frac{\lambda_1}{\lambda_2} = \frac{600}{480} = \frac{5}{4}

Thus, the smallest positive integers that satisfy this ratio are:

m = 5 \quad \text{and} \quad n = 4.

This means the first coincidence of the bright fringes occurs at the 5th bright fringe for the 480 nm light.

Q78

A system of three polarizers P1, P2, P3 is set up such that the pass axis of P3 is crossed with respect to that of P1. The pass axis of P2 is inclined at 60o to the pass axis of P3. When a beam of unpolarized light of intensity I0 is incident on P1, the intensity of light transmitted by the three polarizers is I. The ratio (

{{{I_0}} \over I}

) equals (nearly) :

A 10.67

B 5.33

C 16.00

D 1.80

Correct Answer

Option A

Solution

When unpolarized light of intensity I0 passes through P1, then intensity I1 =

{{{I_0}} \over 2}

as we know, I = I0 cos2 $\theta$ Given that angle between P2 & P3 = 60o and P1 and P3 are crossed that means angle between P1 and P3 is 90o so angle between P1 and P2 = 90° – 60° = 30° I2 =

{{{I_0}} \over 2}{\cos ^2}30^\circ

=

{{3{I_0}} \over 8}

I3 =

{{3{I_0}} \over 8}{\cos ^2}60^\circ

=

{{3{I_0}} \over {32}}

= I $\therefore$

{{{I_0}} \over I}

=

{{{I_0}} \over {{{3{I_0}} \over {32}}}}

=

{{32} \over 3}

= 10.67

Q79

In a Young's double slit experiment, the path difference, at a certain point on the screen, between two interfering waves is

{1 \over 8}

th of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to :

A 0.94

B 0.85

C 0.74

D 0.80

Correct Answer

Option B

Solution

\Delta

x $=$

{\lambda \over 8}

\Delta

$\phi$ $=$

{{\left( {2\pi } \right)} \over \lambda }{\lambda \over 8} = {\pi \over 4}

I $=$ I0cos2

\left( {{\pi \over 8}} \right)

{{\rm I} \over {{{\rm I}_0}}} =

cos2

\left( {{\pi \over 8}} \right)

Q80

The box of a pin hole camera, of length

L,

has a hole of radius a. It is assumed that when the hole is illuminated by a parallel beam of light of wavelength

\lambda

the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say

{b_{\min }}

) when :

A

a = \sqrt {\lambda L} \,

and

{b_{\min }} = \sqrt {4\lambda L}

B

a = {{{\lambda ^2}} \over L}

and

{b_{\min }} = \sqrt {4\lambda L}

C

a = {{{\lambda ^2}} \over L}

and

{b_{\min }} = \left( {{{2{\lambda ^2}} \over L}} \right)

D

a = \sqrt {\lambda L}

and

{b_{\min }} = \left( {{{2{\lambda ^2}} \over L}} \right)

Correct Answer

Option A

Solution

Given geometrical spread

=a

Diffraction spread

= {\lambda \over a} \times L = {{\lambda L} \over a}

The sum

b = a + {{\lambda L} \over a}

For

b

to be minimum

{{db} \over {da}} = 0

{d \over {da}}\left( {a + {{\lambda L} \over a}} \right) = 0

a = \sqrt {\lambda L}

b_{min} = \sqrt {\lambda L} + \sqrt {\lambda L} = 2\sqrt {\lambda L} = \sqrt {4\lambda L}