Wave Optics Test

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Wave Optics Test - 73

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Weekly Quiz Competition

Question 1
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The Young's double slit experiment is done in a medium of refractive index $$4/3$$. A light of $$600\ nm$$ wavelength is falling on the slits having $$0.45\ mm$$ separation. The lower slit $$S_{2}$$ is covered by a thin glass sheet of thickness $$10.4\ \mu m$$ and refractive index $$1.5$$. The interference pattern is observed on a screen placed $$1.5\ m$$ from the slits as shown in figure.
Now, if $$600\ nm$$ light is replaced by white light of range $$400$$ to $$700\ nm$$, find the wavelengths of the light that from maxima exactly at point $$O$$.
[All wavelengths in this problem are for the given medium of refractive index $$4/3$$. Ignore dispersion.]

A
$$1650\ nm, 433.33\ nm$$ have maxima at $$O$$.

B
$$2650\ nm, 433.33\ nm$$ have maxima at $$O$$.

C
$$650\ nm, 433.33\ nm$$ have maxima at $$O$$.

D
$$650\ nm, 433.33\ nm$$ have minima at $$O$$.
Question 2
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A whistle producing sound waves of frequencies 9500 Hz and above is approaching a stationary person with speed v $${ ms }^{ -1 }$$. the velocity of sound in air is 300 $${ ms }^{ -1 }$$. If the person hear frequencies up to a maximum of 10000 Hz, the maximum values of v up to which he can hear the whistle is then

A
30$${ ms }^{ -1 }$$

B
15$$\sqrt { 2 } $$$${ ms }^{ -1 }$$

C
15/$$\sqrt { 2 } $$$${ ms }^{ -1 }$$

D
15$${ ms }^{ -1 }$$

Solution

From Doppler's effect
$$1000=9500\left( \dfrac { 300 }{ 300-v } \right) \Rightarrow v=15m/s$$
Question 3
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In a modified Young's double slit experiment, a monochromatic uniform and parallel beam of light of wavelength $$6000$$ and intensity $$(10/\pi)Wm^{2}$$ is incident normally on two circular apertures $$A$$ and $$B$$ of radii $$0.001\ m$$ and $$0.002\ m$$ respectively. A perfectly transparent film of thickness $$2000A^{\circ}$$ and refractive index $$1.5$$ for the wavelength of $$6000A^{\circ}$$ is placed in front of aperture $$A$$, as shown in the figure. Calculate the power (in watts) received at the focal spot $$F$$ of the lens. The lens is symmetrically placed with respect to the apertures. Assume the $$10\%$$ of the power received by each aperture goes in the original direction and is brought to the focal spot.

A
$$9\times 10^{-6} watt$$.

B
$$17\times 10^{-6} watt$$.

C
$$8\times 10^{-6} watt$$.

D
$$7\times 10^{-6} watt$$.

Solution
Question 4
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Consider a two slit interference arrangements such that the distance of the screen from the slits is half the distance between the slits. Obtain the value of D in terms of $$\lambda$$ such that the first minima on the screen falls at a distance D from the centre O.

A
$$\dfrac{\lambda}{2.472}$$

B
$$\dfrac{\lambda}{2.236}$$

C
$$\dfrac{\lambda}{1.227}$$

D
$$\dfrac{\lambda}{3.412}$$

Solution

From diagram
$$T_1P = T_1O - OP = (D - x)$$
$$T_1P = T_1O + OP = (D + x)$$
Now $$S_1P = \sqrt{(S_1T_1)^2 +(T_1P)^2} = \sqrt{D^2 + (D-x)^2}$$

$$S_2P =\sqrt{(S_2T_2)^2+(T_2P)^2} = \sqrt{D^2 + (D+x)^2}$$
Path difference, $$S_2P - S_1P =\dfrac{\lambda}{2}$$; for first minimum to occur

$$\sqrt{D^2 +(D +x)^2} \,-\, \sqrt{D^2 +(D -x)^2} = \dfrac{\lambda}{2} $$

The first minimum falls at a distance D from the center,
i.e., $$x = D$$.
$$[D^2 + 4D^2]^{1/2} - D = \dfrac{\lambda}{2}$$

$$D(\sqrt{5} - 1) = \dfrac{\lambda}{2}$$

$$D(2.236 - 1) =\dfrac{\lambda}{2}$$ ; $$D = \dfrac{\lambda}{2.472}$$
Question 5
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In a YDSE with two identical slits, when the upper slit is covered with a thin, perfectly transparent sheet of mica, the intensity at the centre of screen reduces from its initial value. Second order minima is observed to be above this point and second order maxima below it. Which of the following can not be a possible value of phase difference caused by the mica sheet?

A
$$\dfrac{\pi }{3}$$

B
$$\dfrac{{7\pi }}{2}$$

C
$$\dfrac{{10\pi }}{3}$$

D
$$\dfrac{{11\pi }}{3}$$

Solution

When $$2nd$$ order minima replace central bright spot change in path differ =$$3\pi$$
as $$2nd$$ order minima just goes about the central phase difference =$$\dfrac{\pi}{2}$$
Net phase differ = $$3\pi+\dfrac{\pi}{2}$$
= $$\dfrac{7\pi}{2}$$
Hence the option (B) $$\dfrac{7\pi}{2}$$ is correct.
Question 6
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Initially interference is observed with the entire experimental set up inside a chamber filled with air, Now the chamber is evacuated. With the same source of light used, a careful observer will find that.

A
The interference pattern is almost absent as it is very much diffused

B
There is no change in the interference pattern

C
The fringe width is slightly decreased

D
The fringe width is slightly increased

Solution

The refractive index of vacuum is less than that of air. The fringe width will vary as follows

$${{\beta }_{air}}=\dfrac{{{\beta }_{vacuum}}}{{{\mu }_{air}}}$$
And $${{\mu }_{air}}$$is greater than 1. So, fringe width will slightly increase.
Question 7
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A glass plate of refractive index, $$1.5$$ is coated with a thin layer of thickness t and refractive index $$1.8$$. Light of wavelength $$648$$ nm travelling in air is incident normally on the layer. It is partly reflected at upper and lower surfaces of the layer and the rays interfere constructively is?

A
$$30$$ nm

B
$$60$$ nm

C
$$90$$ nm

D
$$120$$ nm

Solution

Incident ray $$AB$$ is partly reflected as ray $$1$$ from the upper surface and partly reflected as ray $$2$$ from the lower surface of the layer of thickness $$t$$ and refractive index $${\mu _1} = 1.8$$ as shown in figure path difference between the two rays would be $$\Delta x = 2{\mu _1}t = 2\left( {1.8} \right)t = 3.6t$$
Ray $$1$$ is reflected from a denser medium, therefore, it undergoes a phase change $$\pi $$ whereas the ray $$2$$ gets reflected from a rarer medium, therefore, there is no change in phase of ray $$2$$.

For constructive interference
$$\Delta x = \left( {n - \frac{1}{2}} \right)\lambda ,$$, where $$n$$=1,2,3,4...................

or $$3.6t = \left( {n - \frac{1}{2}} \right)\lambda $$
lest value of $$t$$ is corresponding to $$n=1$$ or $$t\min = \dfrac{2}{{2 \times 3.6}}$$ or $$t\min = \dfrac{{648}}{{7.2}}nm$$
or
$$t\min = 90nm$$.
Question 8
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In the adjacent, $$CP$$ represents a wavefront and $$AO$$ & $$BP$$, the corresponding two rays. Find the condition on $$\theta$$ for constructive interference at $$P$$ between the ray $$BP$$ and reflected ray $$OP$$.

A
$$\cos { \theta } =3\lambda /2d$$

B
$$\cos { \theta } =\lambda /4d$$

C
$$\sec { \theta } -\cos { \theta } =\lambda d$$

D
$$\sec { \theta } -\cos { \theta } =4\lambda d$$

Solution

For constructive interference
$$\Delta \phi =\left( d\sec { \theta } +d\sec { \theta } \cos { 2\theta } \right) \dfrac { 2\pi }{ \lambda } +\pi =2n\pi $$
$$\Rightarrow \cos { \theta } =\dfrac { \left( 2n-1 \right) \lambda }{ 4d }$$
Hence $$(B)$$ is correct.
Question 9
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In Youngs double slit experiment, the fringes are displaced by a distance $$x$$ when a glass plate of one refractive index $$1.5$$ is introduced in the path of one of the beams. When this plate in replaced by another plate of the same thickness, the shift of fringes is $$(3/2)x$$. The refractive index of the second plate is

A
$$1.75$$

B
$$1. 50$$

C
$$1.25$$

D
$$1.00$$

Solution

The path difference introduced by a plate of thickness $$t$$ and refractive index $$\mu$$ is given by $$\Delta=(\mu-1)t$$
A path difference of $$\lambda$$ introduced a phase shift of $$\beta$$
Where $$\beta=\lambda D/(2d)$$. $$2d=$$ separation between slits
So, a path difference of $$(\mu-1)t$$ introduced a shift $$x$$ on the screen given by
$$x=\dfrac { \left( \mu -1 \right) t\beta }{ \lambda }$$
For furst plate, $$x=\dfrac { \left( { \mu }_{ 1 }-1 \right) t\beta }{ \lambda }$$
For second plate
$$\dfrac { 3 }{ 2 } x=\dfrac { \left( { \mu }_{ 2 }-1 \right) t\beta }{ \lambda }$$
$$\therefore \dfrac { \left( { \mu }_{ 2 }-1 \right) }{ \left( { \mu }_{ 1 }-1 \right) } =\dfrac { 3 }{ 2 }$$
or $$\dfrac { \left( { \mu }_{ 2 }-1 \right) }{ \left( 1.5-1 \right) } =\dfrac { 3 }{ 2 }$$
$${ \mu }_{ 2 }=1.75$$
Hence $$(A)$$ is correct.
Question 10
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In Youngs double slit experiment, the slits are $$2mm$$ apart and are illuminated with a mixture of two wavelengths $$\lambda=12000\ {A}^{o}$$ and $$\lambda=10000\ {A}^{o}$$. At what minimum distance from the common central bright fringe on a screen $$2m$$ from the slits will a bright fringe from one interference coincide with a bright fringe from the other?

A
$$3.2\ mm$$

B
$$6.0\ mm$$

C
$$7.2\ mm$$

D
$$9.2\ mm$$

Solution

$$n\lambda \dfrac { D }{ d } =\left( x-1 \right) \lambda '\dfrac { D }{ d }$$
$$n=5$$; $$x=6.0\ mm$$
Hence $$(B)$$ is correct.