Wave Optics — Page 9 | JEE Physics

Q81

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

Q82

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)

Q83

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}

Q84

Unpolarized light of intensity I passes through an ideal polarizer A. Another identical polarizer B is placed behind A. The intensity of light beyond B is found to be I/2. Now another identical polarizer C is placed between A and B. The intensity beyond B is now found to be I/8. The angle between polarizer A and C is :

A 60o

B 30o

C 45o

D 0o

Correct Answer

Option C

Solution

As after B intensity of light does not drops, it means both A and B are alligned in single line means their plane of polarization is same.

Let C makes an angle $\theta$ with A then C will make $\theta$ with B also, as both A and B are alligned in a single line.

So, after C intensity is =

{{\rm I} \over 2}

cos2 $\theta$ , and , intensity after B =

{{\rm I} \over 2}

cos2 $\theta$ $\times$ cos2 $\theta$ According to question,

{{\rm I} \over 2}

cos4 $\theta$ =

{{\rm I} \over 8}

$\Rightarrow$ \,\,\, CO4 $\theta$ =

{{\rm I} \over 4}

$\Rightarrow$

\,\,\,

cos $\theta$ =

= {1 \over {\sqrt 2 }}

= cos45o

\therefore\,\,\,

$\theta$ = 45o

Q85

In a Young’s double slit experiment, light of 500 nm is used to produce an interference pattern. When the distance between the slits is 0.05 mm, the angular width (in degree) of the fringes formed on the distance screen is close to

A 0.17o

B 1.7o

C 0.57o

D 0.07o

Correct Answer

Option C

Solution

$\beta$ =

{{\lambda D} \over d}

and $\theta$ =

{\beta \over D}

$\Rightarrow$ $\theta$ =

{\lambda \over d}

=

{{500 \times {{10}^{ - 9}}} \over {0.05 \times {{10}^{ - 3}}}}

= 0.01 rad = 0.57o

Q86

A single slit of width b is illuminated by a coherent monochromatic light of wavelength

\lambda

. If the second and fourthminima in the diffraction pattern at a distance 1 m from the slit are at 3 cm and 6 cm respectively from the central maximum, what is the width of the central maximum ? (i.e. distance between first minimum on either side of the central maximum)

A 1.5 cm

B 3.0 cm

C 4.5 cm

D 6.0 cm

Correct Answer

Option B

Solution

For single slit diffraction, sin $\theta$ =

{{n\lambda } \over b}

From central maxima the position of nth minima =

{{n\lambda D} \over b}

Now when, n = 2, then x2 =

{{2\lambda D} \over b}

= 0.03 . . . .(1) n = 4, then x4 =

{{4\lambda D} \over b}

= 0.06 . . . .(2) Performing (2) $-$ (1) we get, x4 $-$ x2 =

{{2\lambda D} \over b}

= 0.03

\therefore\,\,\,

{{\lambda D} \over b}

=

{{0.03} \over 2}

As, width of crntral maximum =

{{2\lambda D} \over b}

= 2 $\times$

{{0.03} \over 2}

= 0.03 m = 3 cm

Q87

Two coherent sources produce waves of different intensities which interfere. After interference, the ratio of the maximum intensity to the minimum intensity is 16. The intensity of the waves are in the ratio :

A 16 : 9

B 25 : 9

C 4 : 1

D 5 : 3

Correct Answer

Option B

Solution

Given that,

{{{{\rm I}_{\max }}} \over {{{\mathop{\rm I}\nolimits} _{min}}}} = {{16} \over 1}

We know, Imax $=$

{\left( {\sqrt {{{\rm I}_1}} + \sqrt {{{\rm I}_2}} } \right)^2}

and Imin

= {\left( {\sqrt {{{\rm I}_1}} - \sqrt {{{\rm I}_2}} } \right)^2}

$\therefore$

{{{{\left( {\sqrt {{{\rm I}_1}} + \sqrt {{{\rm I}_2}} } \right)}^2}} \over {{{\left( {\sqrt {{{\rm I}_1}} - \sqrt {{{\rm I}_2}} } \right)}^2}}} = {{16} \over 1}

$\Rightarrow$

{{\sqrt {{{\rm I}_1}} + \sqrt {{{\rm I}_2}} } \over {\sqrt {{{\rm I}_1}} - \sqrt {{{\rm I}_2}} }} = {4 \over 1}

$\Rightarrow$

4\sqrt {{{\rm I}_1}} - 4\sqrt {{{\rm I}_2}} = \sqrt {{{\rm I}_1}} + \sqrt {{{\rm I}_2}}

$\Rightarrow$

3\sqrt {{{\rm I}_1}} = 5\sqrt {{{\rm I}_2}}

$\Rightarrow$

{{\sqrt {{{\rm I}_1}} } \over {\sqrt {{{\rm I}_2}} }} = {5 \over 3}

$\Rightarrow$

{{{{\rm I}_1}} \over {{{\rm I}_2}}} = {{25} \over 9}

Q88

To demonstrate the phenomenon of interference, we require two sources which emit radiation

A of nearly the same frequency

B of the same frequency

C of different wavelengths

D of the same frequency and having a definite phase relationship

Correct Answer

Option D

Solution

To observe the phenomenon of interference, you need two sources that emit radiation of the same frequency and have a definite phase relationship.

This is because interference is a result of the superposition of waves, which requires the waves to be coherent.

Coherence is achieved when the waves have a constant phase difference, which implies that they are of the same frequency and have a definite phase relationship (a phase relationship that does not change with time).

So, the correct answer is : Option D : of the same frequency and having a definite phase relationship.

Q89

Given below are two statements : Statement I : If the Brewster's angle for the light propagating from air to glass is

\mathrm{\theta_B}

, then the Brewster's angle for the light propagating from glass to air is

\dfrac{\pi}{2}-\theta_B

Statement II : The Brewster's angle for the light propagating from glass to air is

{\tan ^{ - 1}}({\mu _\mathrm{g}})

where

\mathrm{\mu_g}

is the refractive index of glass. In the light of the above statements, choose the correct answer from the options given below :

A Both Statement I and Statement II are false

B Both Statement I and Statement II are true

C Statement I is false but Statement II is true

D Statement I is true but Statement II is false

Correct Answer

Option D

Solution

Case I : Transmitted is $\perp$ to reflected.

$i+r=90^{\circ}$ Snell's law $\mu_{a} \sin i=\mu_{g} \sin r$ $\tan i=\dfrac{\mu_{g}}{\mu_{a}}$ $i=\tan ^{-1}\left(\dfrac{\mu_{g}}{\mu_{a}}\right)=\theta_{B}$ Case II : $i+r=90^{\circ}$ as transmitted is $\perp$ to reflected.

$\tan i=\dfrac{\mu_{a}}{\mu_{g}} \Rightarrow i=\tan ^{-1} \dfrac{\mu_{a}}{\mu_{g}}=\dfrac{\pi}{2}-\theta_{B}$

Q90

A single slit of width 0.1 mm is illuminated by a parallel beam of light of wavelength 6000

\mathop A\limits^ \circ

and diffraction bands are observed on a screen 0.5 m from the slit. The distance of the third dark band from the central bright band is :

A 3 mm

B 9 mm

C 4.5 mm

D 1.5 mm

Correct Answer

Option B

Solution

The slit width is a = 0.1 mm = 10 $-$ 4 m.

The wavelength of the light is $\lambda$ = 6000 $\times$ 10 $-$ 10 = 6 $\times$ 10 $-$ 7.

The distance from the slit to diffraction bands is D = 0.5 m.

We calculate the third dark band from the central band as follows:

A\sin \theta = 3\lambda \Rightarrow \sin \theta = {{3\lambda } \over a} = {x \over D}

\Rightarrow x = {{3\lambda D} \over a} = {{3 \times 6 \times {{10}^{ - 7}} \times 0.5} \over {{{10}^{ - 4}}}} = 9

mm