Wave Optics Test

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Wave Optics Test - 12

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Question 1
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The question has a paragraph followed by two statements, Statement 1 and Statement 2. Of the given four alternatives after the statements, choose the one that best describes the statements.
A thin air film is formed by putting the convex surface of a plane convex lens over a plane glass plate. With monochromatic light, this film gives an interference pattern due to light reflected from the top (convex) surface and the bottom (glass plate) surface of the film.
Statement 1 : When light reflects from the air-glass plate interface, the reflected wave suffers a phase change of $$\pi$$.
Statement 2 : The center of the interference pattern is dark.

A
Statement 1 is True, Statement 2 is False.

B
Statement 1 is True, Statement 2 is True; Statement 2 is a correct explanation for Statement 1.

C
Statement 1 is True, Statement 2 is True; Statement 2 is not the correct explanation for Statement 1.

D
Statement 1 is False, Statement 2 is True.

Solution

As the centre of the pattern is dark, we know that the phase difference is $$ (2n+1)\pi $$
Furthermore, we also know that for the center of the interference pattern, there is no path difference. So the difference can only be due to reflection which is $$\pi$$
Therefore, both the assertion and the reason are correct and the reason is explanatory
Question 2
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The box of a pinhole 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
$$a=\dfrac { { \lambda }^{ 2 } }{ L } $$ and $${ b }_{ min }=\dfrac { { 2\lambda }^{ 2 } }{ L } $$

B
$$a=\sqrt { \lambda L } $$ and $${ b }_{ min }=\dfrac { { 2\lambda }^{ 2 } }{ L } $$

C
$$a=\sqrt { \lambda L } $$ and $${ b }_{ min }=\sqrt { 4\lambda L } $$

D
$$a=\dfrac { { \lambda }^{ 2 } }{ L } $$ and $${ b }_{ min }=\sqrt { 4\lambda L } $$

Solution

We know that spot size $$b=2(\dfrac { \lambda L }{ 2a } )+2a$$.
$${ a }^{ 2 }+\lambda L-ab=0$$.
For roots to be real D >= 0,
Hence, $${ b }_{ min }=\sqrt { 4\lambda L } $$ in which case $$a=\sqrt { \lambda L } $$
Question 3
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An initially parallel cylindrical beam travels in a medium of refractive index $$\mu(I) =\mu_0 + \mu_2 I$$, where $$\mu_0$$ and $$\mu_2$$ are positive constants and $$I$$ is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius. The initial shape of the wave front of the beam is

A
Convex

B
Concave

C
Convex near the axis and concave near the periphery

D
Planar

Solution

For a parallel cylinderical beam, wavefront will be planar.
Question 4
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To demonstrate the phenomenon of interference we require two soruces which emit radiation of

A
nearly the same frequency

B
the same frequency

C
different wavelength

D
the same frequency and having a definite phase relationship.

Solution

To demonstrate interference, to coherent sources are required. Sources are called coherent when they emit waves of nearly equal or equal frequency and a constant phase difference throughout.
Question 5
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An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency $$10\,GHz$$. What is the frequency of the microwave measured by the observer?
(speed of light $$=3\times 10^8 ms^{-1}$$)

A
$$15.3\, GHz$$

B
$$10.1 \,GHz$$

C
$$12.1\, GHz$$

D
$$17.3\, GHz$$

Solution

Speed of observer $$V=\dfrac{C}{2}$$

where speed of Microwave is $$C$$

Original frequency of source is given as $$f_0=10 GH_2$$

Using relativistic Doppler effect in case observer moves towards the stationary source.

Apparent frequency measured by observer $$f' = f_0\sqrt{\dfrac{C+v}{C-v}}$$

$$f' = 10\sqrt{\dfrac{C+(C/2)}{C-(C/2)}}$$

$$f' = 10\sqrt{\dfrac{(3/2)C}{(1/2)C}}$$

$$F'=10\times 1.73 GH_z$$

$$F' = 17.3 GH_z$$
Question 6
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In a double-slit experiment, at a certain point on the screen the path difference between the two interfering waves is $$\cfrac { 1 }{ 8 } th$$ of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe is:

A
$$0.672$$

B
$$0.853$$

C
$$0.760$$

D
$$0.568$$

Solution

Given path difference $$\Delta x = \dfrac{\lambda}{8}$$

Phare difference $$\Delta \phi = \dfrac{2\pi}{\lambda} . \Delta x$$

$$\Delta \phi = \dfrac{2\pi }{\lambda} . \left(\dfrac{\lambda}{8}\right)$$

$$\Delta \phi = \dfrac{\pi}{4}$$

In $$YDSE$$ we know that
at any point, intensity $$I = I_{max} \cos^2\left(\dfrac{\Delta \phi}{2}\right)$$

$$I = I_{max} \cos^2\left(\dfrac{\pi/4}{2}\right)$$

$$\dfrac{I}{I_{max}} = \left[ \cos\left(\dfrac{\pi}{8}\right)\right]^2$$

$$\dfrac{I}{I_{max}} = (0.9238)^2$$

$$\dfrac{I}{I_{max}} = 0.853$$
Question 7
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In Young's double slit experiment, one of the slit is wider than other, so that amplitude of the light from one slit double of that from other slit. If $$\mathrm{I}_{\mathrm{m}}$$ be the maximum intensity. The resultant intensity $$I$$ when they interfere at phase difference $$\phi$$ is given by

A
$$\displaystyle \frac{I_{m}}{9}(4+5\cos\phi)$$

B
$$\displaystyle \frac{I_{m}}{3}(1+2\cos^{2}\frac{\phi}{2})$$

C
$$\displaystyle \frac{I_{m}}{5}(1+4\cos^{2}\frac{\phi}{2})$$

D
$$\displaystyle \frac{I_{m}}{9}(1+8\cos^{2}\frac{\phi}{2})$$

Solution

$$I= I_0+ 4I_0+ 2\sqrt{I_0\times 4I_0 }cos\phi$$
$$I= I_0+ 4I_0+ 4I_0 cos\phi$$
$$I_m$$ will be at $$cos\phi=1$$,
So, $$I_m= 9I_0$$
$$\displaystyle \frac{I}{I_m}= \frac{5I_0+ 4I_0 cos\phi}{9}$$
$$\displaystyle \frac{I}{I_m}= \frac{1+ 8cos^2\frac{\phi}{2}}{9}$$
Question 8
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A parallel beam of light strikes a piece of transparent glass having cross section as shown in the figure below. Correct shape of the emergent wavefront will be (figures are schematic and not drawn to scale)

A

B

C

D

Solution

Refer Fig. 1
Glass can be divided into three parts from top to below characters of glass from top to bottom are converging, diverging and converging
Refer Fig. 2
We know, For converging a beam of light converges in a point while for diverging a beam of light diverges from a point.
So, the situation diagram of wavefront is shown below..
Refer Fig. 3
Hence, option A is correct
Question 9
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In a Young's double slit experiment, the slit separation d is $$0.3$$ mm and the screen distance D is $$1 m$$. A parallel beam of light of wavelength $$600 \, nm$$ is incident on the slits at angle $$\alpha$$ as shown in figure. On the screen, the point O is equidistant from the slits and distance PO is $$11.0 $$mm. Which of the following statement (s) is/are correct ?

A
For $$\alpha = 0$$, there will be constructive interference at point P.

B
For $$\alpha = \dfrac{0.36}{\pi} $$ degree, there will be destructive interference at point P.

C
Fro $$\alpha = \dfrac{0.36}{\pi}$$ degree, there will be destructive interference at point O.

D
Fringe spacing depends on $$\alpha$$.

Solution

$$\Delta x = d \sin \alpha + d \sin \theta$$
$$\theta , \alpha$$ : small angle $$ \sin \theta \simeq \tan \theta = \dfrac{y}{b}$$
$$\Delta x = d \alpha + \dfrac{dy}{D}$$
$$(1) \, \alpha = 0 $$ $$\therefore \Delta x = dy/ D = \dfrac{0.3 \times 11}{1000} = 33 \times 10^{-4} mm$$
$$\Delta x$$ in terms of $$ \lambda = \dfrac{33 \times 10^{-4}}{600 \times 10^{-6}} \lambda = \dfrac{11 \lambda}{2}$$
as $$\Delta x = (2n - 1) \dfrac{\lambda}{2}$$
There will be destructive interference
$$(2) \, \Delta x = 0.3 mm \times \dfrac{0.36}{\pi} \times \dfrac{\pi}{180} + \dfrac{0.3 mm \times 11 mm}{1000} = 39 \times 10^{-4} mm$$
$$39 \times 10^{-4} = (2n - 1) \times \dfrac{600 \times 10^{-9} \times 10^3}{2}$$
$$n = 7$$
There will be destructive interference
$$(3) \, \Delta x = 3 mm \times \dfrac{0.36}{\pi} \times \dfrac{\pi}{180} + 0 = 600 nm$$
$$600 nm = n \lambda$$
$$n = 1$$
constructive interference
(4) Fringe width does not depend on $$\alpha $$.
Question 10
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The diameter of the objective lens of microscope makes an angle $$'\beta\ '$$ at the focus of the microscope. Further, the medium between the object and the lens is an oil of refractive index 'n'. Then the resolving power of the microscope.

A
Increases with decreasing value of $$n$$

B
Increases with decreasing value of $$\beta$$

C
Increases with increasing value of $$n sin 2\beta$$

D
Increases with increasing value of $$\dfrac {1}{n sin 2\beta}$$

Solution

Resolving Power of the Microscope is given by $$\displaystyle \dfrac{1}{d} = \dfrac{2nsin\theta}{\lambda}$$

where $$n$$ is refractive index , $$\theta$$ is the angle of the cone and d is the distance between the object .

Here , The diameter of the objective lens of microscope makes an angle $$\beta$$ at the focus of the microscope.

Hence , $$\beta = \theta /2 $$

Hence , resolving power is $$\displaystyle \dfrac{2n \ sin2\beta }{\lambda }$$