Wave Optics — Page 8 | JEE Physics

Q71

Two coherent sources of light interfere. The intensity ratio of two sources is

1: 4

. For this interference pattern if the value of

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

is equal to

\dfrac{2 \alpha+1}{\beta+3}

, then

\dfrac{\alpha}{\beta}

will be :

A 1.5

B 2

C 0.5

D 1

Correct Answer

Option B

Solution

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

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

$\therefore$

{{{I_{\max }} + {I_{\min }}} \over {{I_{\max }} - {I_{\min }}}} = {{2({I_1} + {I_2})} \over {4 \times \sqrt {{I_1}{I_2}} }}

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

= {1 \over 2} \times {{\left( {{1 \over 4} + 1} \right)} \over {\left( {{1 \over 2}} \right)}}

= {5 \over 4} = {{2 \times 2 + 1} \over {1 + 3}}

$\therefore$

{\alpha \over \beta } = {2 \over 1} = 2

Q72

In a Young's double slit experiment, the separation between the slits is 0.15 mm. in the experiment, a source of light of wavelengh 589 nm is used and the interference pattern is observed on a screen kept 1.5 m away. The separation between the successive bright fringes on the screen is :

A 4.9 mm

B 5.9 mm

C 6.9 mm

D 3.9 mm

Correct Answer

Option B

Solution

\beta = {{\lambda D} \over d} = {{589 \times {{10}^{ - 9}} \times 1.5} \over {0.15 \times {{10}^{ - 3}}}}

= 5.9 mm

Q73

A monochromatic light of wavelength

6000 ~\mathring{A}

is incident on the single slit of width

0.01 \mathrm{~mm}

. If the diffraction pattern is formed at the focus of the convex lens of focal length

20 \mathrm{~cm}

, the linear width of the central maximum is :

A

12 \mathrm{~mm}

B

24 \mathrm{~mm}

C

60 \mathrm{~mm}

D

120 \mathrm{~mm}

Correct Answer

Option B

Solution

To find the linear width of the central maximum in a single-slit diffraction pattern, we can use the formula that relates the position of the first minima on either side of the central maximum.

The angle, $\theta$ , at which the first minimum occurs is given by:

a \sin(\theta) = m\lambda

Where: $a$ is the width of the slit $\lambda$ is the wavelength of monochromatic light $m$ is the order number of the minimum (for the first minimum, $m = \pm 1$ ) In our case, we want to find the position of the first minima to determine the width of the central maximum on the screen.

So we will use $m = 1$ and $m = -1$ which correspond to the first minima on either side of the central peak.

Given the width of the slit $a = 0.01$ mm, which we need to convert to meters for consistency with the wavelength:

a = 0.01 \times 10^{-3} \text{m}

And the wavelength $\lambda = 6000 \mathring{A}$ , also converting to meters:

\lambda = 6000 \times 10^{-10} \text{m}

We're using a convex lens of focal length $f = 20$ cm to project the pattern onto a screen.

We'll need to convert the focal length to meters as well:

f = 20 \times 10^{-2} \text{m}

We can calculate the angle $\theta$ needed for the first minima using the approximation of small angles, where $\sin(\theta) \approx \theta$ :

a \theta = m\lambda

Now, solve for $\theta$ for the first minimum (m = 1):

\theta = \frac{\lambda}{a}

Plugging in the values:

\theta = \frac{6000 \times 10^{-10}}{0.01 \times 10^{-3}}

\theta = \frac{6000 \times 10^{-10}}{10^{-5}}

\theta = 6000 \times 10^{-5}

Now, linear width of the central maximum on the screen (from -m to m, or -1 to +1) will be twice the distance from the center to the first minimum, which can be found using the focal length of the lens $f$ and the angle $\theta$ :

y = 2f\theta

Plugging the focal length and calculated theta:

y = 2 \times 20 \times 10^{-2} \times 6000 \times 10^{-5}

y = 40 \times 10^{-2} \times 6000 \times 10^{-5}

y = 240 \times 10^{-3} \text{m}

y = 24 \times 10^{-2} \text{m}

y = 24 \text{mm}

This means the linear width of the central maximum is 24 mm. Hence, the correct answer is Option B.

Q74

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}

Q75

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

Q76

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)

Q77

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

Q78

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}

Q79

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

Q80

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.