Wave Optics — Page 4 | JEE Physics

Q31

In Young's double slit experiment, if the source of light changes from orange to blue then :

A the central bright fringe will become a dark fringe.

B the distance between consecutive fringes will decrease.

C the distance between consecutive fringes will increases.

D the intensity of the minima will increase.

Correct Answer

Option B

Solution

The answer is Option B : the distance between consecutive fringes will decrease.

Young's double-slit experiment depends on the principle of interference of light waves, and the resulting pattern of light and dark bands (fringes) depends on the wavelength of the light used.

The formula for the distance between fringes (y) in the double slit experiment is : y =

{{\lambda D} \over d}

where : λ is the wavelength of the light D is the distance from the slits to the screen d is the distance between the slits The wavelength of orange light is approximately 600 nm while the wavelength of blue light is around 475 nm.

Therefore, if you change the light from orange to blue, the wavelength λ decreases.

As λ is directly proportional to y, if λ decreases, y will also decrease.

So, the distance between consecutive fringes will decrease.

Q32

Using Young's double slit experiment, a monochromatic light of wavelength 5000

\mathop A\limits^o

produces fringes of fringe width 0.5 mm. If another monochromatic light of wavelength 6000

\mathop A\limits^o

is used and the separation between the slits is doubled, then the new fringe width will be :

A 0.5 mm

B 1.0 mm

C 0.6 mm

D 0.3 mm

Correct Answer

Option D

Solution

Fringe width =

{{\lambda D} \over d}

$\Rightarrow$ Fringe width $\propto$

{\lambda \over d}

$\Rightarrow$ New fringe width = 0.5 mm

\times {{1.2} \over 2} = 0.3

mm

Q33

In Young's double slit experiment performed using a monochromatic light of wavelength

\lambda

, when a glass plate (

\mu

= 1.5) of thickness x

\lambda

is introduced in the path of the one of the interfering beams, the intensity at the position where the central maximum occurred previously remains unchanged. The value of x will be :

A 3

B 2

C 1.5

D 0.5

Correct Answer

Option B

Solution

For the intensity to remain same the position must be of a maxima so path difference must be n $\lambda$ so (1.5 $-$ 1) x $\lambda$ = n $\lambda$ x = 2n (n = 0, 1, 2 ....)

So, value of x will be x = 0, 2, 4, 6 ....

Q34

The interference pattern is obtained with two coherent light sources of intensity ratio 4 : 1. And the ratio

{{{I_{\max }} + {I_{\min }}} \over {{I_{\max }} - {I_{\min }}}}

is

{5 \over x}

. Then, the value of x will be equal to :

A 3

B 4

C 2

D 1

Correct Answer

Option B

Solution

{{{I_{\max }} + {I_{\min }}} \over {{I_{\max }} - {I_{\min }}}} = {{{I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} + {I_1} + {I_2} - 2\sqrt {{I_1}{I_2}} } \over {{I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} - {I_1} - {I_2} + 2\sqrt {{I_1}{I_2}} }}

= {{2({I_1} + {I_2})} \over {4\sqrt {{I_1}{I_2}} }}

= {{\left( {{{{I_1}} \over {{I_2}}} + 1} \right)} \over {2\sqrt {{{{I_1}} \over {{I_2}}}} }} = {{4 + 1} \over {2 \times 2}} = {5 \over 4}

So

x = 4

Q35

Two light beams of intensities in the ratio of 9 : 4 are allowed to interfere. The ratio of the intensity of maxima and minima will be :

A 2 : 3

B 16 : 81

C 25 : 169

D 25 : 1

Correct Answer

Option D

Solution

{{{I_{\max }}} \over {{I_{\min }}}} = {\left( {{{\sqrt {{I_1}} + \sqrt {{I_2}} } \over {\sqrt {{I_1}} - \sqrt {{I_2}} }}} \right)^2} = {\left( {{5 \over 1}} \right)^2}

= {{25} \over 1}

Q36

Find the ratio of maximum intensity to the minimum intensity in the interference pattern if the widths of the two slits in Young's experiment are in the ratio of 9 : 16. (Assuming intensity of light is directly proportional to the width of slits)

A 3 : 4

B 4 : 3

C 7 : 1

D 49 : 1

Correct Answer

Option D

Solution

Let, Width of first slit = w1 and width of second slit = w2 Given,

{{{w_1}} \over {{w_2}}} = {9 \over {16}}

Given, Intensity of light

(I) \propto w

$\therefore$

{{{I_1}} \over {{I_2}}} = {{{w_1}} \over {{w_2}}} = {9 \over {16}}

We know,

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

= {\left( {{{\sqrt {{{{I_1}} \over {{I_2}}}} + 1} \over {\sqrt {{{{I_1}} \over {{I_2}}}} - 1}}} \right)^2}

= {\left( {{{{3 \over 4} + 1} \over {{3 \over 4} - 1}}} \right)^2}

= {\left( {{7 \over 1}} \right)^2}

= {{49} \over 1}

Q37

In Young's double slit experiment, the fringe width is

12 \mathrm{~mm}

. If the entire arrangement is placed in water of refractive index

\dfrac{4}{3}

, then the fringe width becomes (in mm):

A 16

B 9

C 48

D 12

Correct Answer

Option B

Solution

B = 12 \times {10^{ - 3}}

\beta ' = {\beta \over \mu } = {{12 \times {{10}^{ - 3}}} \over {{4 \over 3}}}

= 9 \times {10^{ - 3}}

m = 9 mm

Q38

An unpolarised light beam of intensity

2 I_{0}

is passed through a polaroid P and then through another polaroid Q which is oriented in such a way that its passing axis makes an angle of

30^{\circ}

relative to that of P. The intensity of the emergent light is

A

\dfrac{\mathrm{I}_{0}}{4}

B

\dfrac{\mathrm{I}_{0}}{2}

C

\dfrac{3 I_{0}}{4}

D

\dfrac{3 \mathrm{I}_{0}}{2}

Correct Answer

Option C

Solution

$\mathrm{I}_1=\dfrac{1}{2}\left(2 \mathrm{I}_0\right)=\mathrm{I}_0$ $\mathrm{I}_2=\mathrm{I}_1 \cos ^2 30^{\circ}$ $=\mathrm{I}_0 \cdot \dfrac{3}{4}=\dfrac{3 \mathrm{I}_0}{4}$

Q39

Two polaroide

\mathrm{A}

and

\mathrm{B}

are placed in such a way that the pass-axis of polaroids are perpendicular to each other. Now, another polaroid

\mathrm{C}

is placed between

\mathrm{A}

and

\mathrm{B}

bisecting angle between them. If intensity of unpolarized light is

\mathrm{I}_{0}

then intensity of transmitted light after passing through polaroid

\mathrm{B}

will be:

A

\dfrac{I_{0}}{4}

B

\dfrac{I_{0}}{8}

C Zero

D

\dfrac{I_{0}}{2}

Correct Answer

Option B

Solution

$\mathrm{I}_{\mathrm{A}}=\dfrac{\mathrm{I}_{\mathrm{o}}}{2}$ $\mathrm{I_C}=\dfrac{\mathrm{I}_{\mathrm{o}}}{2} \cos ^{2} 45=\dfrac{\mathrm{I}_{\mathrm{o}}}{4}$ $\mathrm{I}_{\mathrm{B}}=\mathrm{I}_{\mathrm{C}} \cos ^{2} 45=\dfrac{\mathrm{I}_{\mathrm{o}}}{8}$

Q40

A single slit of width

a

is illuminated by a monochromatic light of wavelength

600 \mathrm{~nm}

. The value of '

a

' for which first minimum appears at

\theta=30^{\circ}

on the screen will be :

A

{3} \mu \mathrm{m}

B

0.6 \mu \mathrm{m}

C

1.8 \mu \mathrm{m}

D

1.2 \mu \mathrm{m}

Correct Answer

Option D

Solution

When light passes through a narrow slit, it diffracts and produces a diffraction pattern on a screen.

The pattern consists of a central bright maximum flanked by a series of alternating bright and dark fringes.

The condition for the first minimum in the diffraction pattern is given by :

a\sin\theta = \lambda

where $a$ is the width of the slit, $\theta$ is the angle of diffraction, and $\lambda$ is the wavelength of the incident light.

In this problem, we are given that the wavelength of the incident light is $\lambda = 600\,\mathrm{nm}$ , and the angle of diffraction for the first minimum is $\theta = 30^\circ$ .

We want to find the width of the slit, $a$ , for which this occurs.

Substituting the given values into the condition for the first minimum, we get :

a\sin 30^\circ = 600\,\mathrm{nm}

Simplifying, we get :

a\cdot \frac{1}{2} = 600\,\mathrm{nm}

Multiplying both sides by 2, we get :

a = 1200\,\mathrm{nm} = \boxed{1.2\,\mu\mathrm{m}}

Therefore, the width of the slit for which the first minimum appears at $\theta = 30^\circ$ is $\boxed{1.2\,\mu\mathrm{m}}$ .