Wave Optics Test

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Wave Optics Test - 69

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Question 1
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In the arrangement shown in figure, $$D>>d$$. For what minimum value of $$d$$ is there a dark band at point $$O$$ on the screen?

A
$$\sqrt{\cfrac{D\lambda}{4}}$$

B
$$\sqrt{\cfrac{3D\lambda}{4}}$$

C
$$\sqrt{\cfrac{D\lambda}{8}}$$

D
$$\sqrt{\cfrac{2D\lambda}{3}}$$

Solution

Path difference between A and B is given by $$\Delta x =\dfrac{dy}{D}$$ where $$y=d$$
net path difference $$= 2 \Delta x =\dfrac{2d^2}{D}$$
For minima at O it should be multiple of $$\dfrac{\lambda}{2}$$
$$\dfrac{2d^2}{D}=\dfrac{\lambda}{2}$$
$$d=\sqrt{\dfrac{\lambda D}{4}}$$
Question 2
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Directions For Questions

In a modified YDSE source $$S$$ is kept in front of slit $${S}_{1}$$. Find the phase difference at point $$O$$ that is equidistant from slits $${S}_{1}$$ and $${S}_{2}$$, and point $$P$$ that is in front of slit $${S}_{1}$$ in the following situations.

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A liquid of refractive index $$\mu$$ is filled between the screen and slits.

A
$$\cfrac { 2\pi }{ \lambda } \left[ \left[ \sqrt { { d }^{ 2 }+{ x }_{ 0 }^{ 2 } } +{ x }_{ 0 } \right] +\cfrac { \mu { d }^{ 2 } }{ 2D } \right] $$

B
$$\cfrac { 2\pi }{ \lambda } \left[ \left[ \sqrt { { d }^{ 2 }+{ x }_{ 0 }^{ 2 } } -{ x }_{ 0 } \right] +\cfrac { \mu { d }^{ 2 } }{ 2D } \right] $$

C
$$\cfrac { 2\pi }{ \lambda } \left[ \left[ \sqrt { { d }^{ 2 }-{ x }_{ 0 }^{ 2 } } +{ x }_{ 0 } \right] +\cfrac { \mu { d }^{ 2 } }{ 2D } \right] $$

D
$$\cfrac { 2\pi }{ \lambda } \left[ \left[ \sqrt { { d }^{ 2 }-{ x }_{ 0 }^{ 2 } } -{ x }_{ 0 } \right] +\cfrac { \mu { d }^{ 2 } }{ 2D } \right] $$

Solution

Path Difference due to Position of source:
$$\delta _{source} = \sqrt{d^{2}+x_{o}^{2}} - x _{o}$$

Path Difference at O is Zero
Path Difference at P:
$$\delta _{P} = \dfrac{d}{2}\times\dfrac{d}{D}$$

For Phase Difference: Multiply $$\delta _{source} $$ by $$\dfrac{2\pi}{\lambda}$$ and $$\delta _{P} $$ by $$\dfrac{2\pi\mu}{\lambda}$$

Therefore, Answer is $$\phi = \dfrac{2\pi}{\lambda}(\delta _{source} + \mu\times\delta _{P})$$
$$\Rightarrow \phi =\cfrac { 2\pi }{ \lambda } \left[ \left[ \sqrt { { d }^{ 2 }+{ x }_{ 0 }^{ 2 } } -{ x }_{ 0 } \right] +\cfrac { \mu { d }^{ 2 } }{ 2D } \right] $$
Question 3
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Directions For Questions

Consider the situation shown in figure. The two slits $${S}_{1}$$ and $${S}_{2}$$ placed symmetrically around the central line are illuminated by monochromatic light of wavelength $$'\lambda\ '$$. The separation between the slits is $$'d'$$. The light transmitted by the slits falls on a screen $${S}_{0}$$ placed at a distance $$'D'$$ from the slits. The slit $${S}_{3}$$ is at the central line and the slit $${S}_{4}$$ is at a distance z from $${S}_{3}$$. Another screen $${S}_{c}$$ is placed a further distance $$'D\ '$$ away from $${S}_{c}$$. Find the ration of the maximum and minimum intensity observed on $${S}_{c}$$.

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If $$z=\cfrac{\lambda D}{2d}$$

A
$$1$$

B
$$1/2$$

C
$$3/2$$

D
$$2$$

Solution

Intensity at:
$$S _{3} = 4 I _{o}$$
$$S _{4} = 0$$

$$I _{max} = 4I _{o}$$
$$I _{min} = 4I _{o}$$

$$\therefore \dfrac{I _{max}}{I _{min}} = 1$$
Question 4
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Directions For Questions

Consider the situation shown in figure. The two slits $${S}_{1}$$ and $${S}_{2}$$ placed symmetrically around the central line are illuminated by monochromatic light of wavelength $$'\lambda\ '$$. The separation between the slits is $$'d'$$. The light transmitted by the slits falls on a screen $${S}_{0}$$ placed at a distance $$'D'$$ from the slits. The slit $${S}_{3}$$ is at the central line and the slit $${S}_{4}$$ is at a distance z from $${S}_{3}$$. Another screen $${S}_{c}$$ is placed a further distance $$'D\ '$$ away from $${S}_{c}$$. Find the ration of the maximum and minimum intensity observed on $${S}_{c}$$.

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If $$z=\cfrac{\lambda D}{4d}$$

A
$${ \left[ 3-2\sqrt { 2 } \right] }^{ 2 }$$

B
$${ \left[ 3+\sqrt { 2 } \right] }^{ 2 }$$

C
$${ \left[ 3-\sqrt { 2 } \right] }^{ 2 }$$

D
$${ \left[ 3+2\sqrt { 2 } \right] }^{ 2 }$$

Solution

$$S _{3} = 4 I _{o}$$
$$S _{4} = 2 I _{o}$$

$$I _{max} = 4I _{o} + 2I _{o} + 2\sqrt{4I _{o}2I _{o}} = (6 + 4\sqrt{2})I _{o}$$
$$I _{min} = 4I _{o} + 2I _{o} - 2\sqrt{4I _{o}2I _{o}} = (6 - 4\sqrt{2})I _{o}$$

Hence $$\dfrac{I _{max}}{I _{min}} = [3 + 2\sqrt{2}]^2$$
Question 5
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Directions For Questions

A coherent parallel beam of microwaves of wavelength $$\lambda=0.5mm$$ falls on Young's double-slit apparatus. The separation between the slits is $$1.0mm$$. The intensity of microwaves is measured on a screen placed parallel to the plane of the slits at a distance of $$1.0m$$ from it as shown in figure.

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If the incident beam makes an angle of $${30}^{o}$$ with the x-axis (as in the dotted arrow shown in the figure), find the y-coordinates of the first minima on either side of the central maximum..

A
$$\cfrac { 3 }{ \sqrt { 7 } } $$ and $$\cfrac { 1 }{ \sqrt { 15 } }m $$

B
$$\cfrac { 3 }{ \sqrt { 7 } } $$ and $$\cfrac { 2 }{ \sqrt { 15 } }m $$

C
$$\cfrac { 3 }{2 \sqrt { 7 } } $$ and $$\cfrac { 1 }{ \sqrt { 15 } }m $$

D
$$\cfrac { 6 }{ \sqrt { 7 } } $$ and $$\cfrac { 3 }{ \sqrt { 15 } }m $$

Solution

Path Difference due to Angle:
$$\delta _{source} = 0.5mm $$

For Central Maxima, Path Difference should be Zero:
$$\delta _{source} = x\times\dfrac{d}{D}$$
which gives us $$ x = 0.5m$$
Question 6
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In the arrangement shown in figure, $$D>>d$$. Find the fringe width.

A
$$d$$

B
$$2d$$

C
$$4d$$

D
$$3d$$
Question 7
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In the arrangement shown in figure, $$D>>d$$. Find the distance $$x$$ at which the next bright fringe is formed.

A
$$\cfrac{3\lambda}{2}$$

B
$$\cfrac{\lambda}{4}$$

C
$$\cfrac{\lambda}{2}$$

D
$$\cfrac{5\lambda}{2}$$
Question 8
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Directions For Questions

Thin films, including soap bubbles and oil show patterns of alternative dark and bright regions resulting from interference among the reflected light waves. If two waves are in phase, their crests and troughs will coincide. The interference will be constructive and the amplitude of resultant wave will be greater than either of constituent waves. If the two waves are out of phase by half a wavelength $$({180}^{o})$$, the crests of one wave will coincide with the troughs of the other wave. The interference will be destructive and the amplitude of the resultant wave will be less than that of either constituent wave.
At the interface between two transparent media, some light is reflected and some light is refracted.
(1) When incident light $$I$$, reaches the surface at point $$a$$, some of the light is reflected as ray $${R}_{a}$$ and some is refracted following the path $$ab$$ to the back of the film.
(2) At point $$b$$, some of the light is refracted out of the film and part is reflected back through the film along the path $$bc$$. At point $$c$$, some of the light is reflected back into thin film and part is reflected out of the film as ray $${R}_{c}$$.
$${R}_{a}$$ and $${R}_{c}$$ are parallel. however, $${R}_{c}$$ has travelled the extra distance within the film of $$abc$$. If the angle of incidence is small, then $$abc$$ is approximately twice the film's thickness.
If $${R}_{a}$$ and $${R}_{c}$$ are in phase, they will undergo constructive interference and the region $$ac$$ will be bright. If $${R}_{a}$$ and $${R}_{c}$$ are out of phase, they will undergo destructive interference and the region $$ac$$ will be dark.
The thickness of the film and the refractive indices of teh media at each interface, determine the final phase relationship between $${R}_{a}$$ and $${R}_{c}$$.
$$I$$. Refraction at an interface never changes the phase of the wave.
$$II$$. For reflection at the interface between two media 1 and 2, if $${n}_{1}>{n}_{2}$$, the reflected wave will change phase. If $${n}_{1}<{n}_{2}$$, the reflected wave will not undergo a phase change.
For reference, $${n}_{air}=1.00$$
$$III$$.If the waves are in phase after reflection at all interfaces, then the effects of path length in the film are:
Constuctive interference occurs when $$2t=m\lambda /n,m=0,1,2,3,......$$
Destructive interference occurs when $$2t=\left( m+\cfrac { 1 }{ 2 } \right) \cfrac { \lambda }{ n } , m=0,1,2,3,....$$
If the waves are $${180}^{o}$$ out if the phase after reflection at all interfaces, then the effects of path length in the film are:
Constructive interference occurs when
$$2t=\left( m+\cfrac { 1 }{ 2 } \right) \cfrac { \lambda }{ n } $$
Destructive interference occurs when
$$2t=\cfrac{m\lambda}{n}, m=0,1,2,3,....$$

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A thin film with index of refraction $$1.50$$ coats a glass lens with index of refraction $$1.80$$. What is the minimum thickness of the film that will strongly reflect light with wavelength $$600nm$$?

A
$$150nm$$

B
$$200nm$$

C
$$300nm$$

D
$$450nm$$

Solution

Both waves $$R_a$$ and $$R_b$$ are out of phase ($$180^o$$) with the incident wave, so $$R_a$$ and $$R_b$$ are in phase with each other.
thus using $$2t= m\lambda/n$$ for constructive interference (i-e. complete reflection)
let t be the minimum thickness of the film( i-e. $$m = 1$$)
Given ; $$n = 1.5 \lambda=600 nm$$
$$ 2t = \dfrac{1 \times 600 nm}{1.5}$$
On solving $$t = 200 nm$$
Question 9
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Directions For Questions

The interference film is used to measure the thickness of slides, paper, etc. The arrangement is as shown in figure.
For the sake of clarity, the two strips are shown thick. Consider the wedge formed in between strips 1 and 2. If the interference pattern because of the two waves reflected from wedge surface is observed, then from the observed data we can compute thickness of paper, refractive index of the medium filled in wedge, number of bonds formed, etc.
Consider the strips to be thick as compared to wavelength of light and light is incident normally.
Neglect the effect due to reflection from top surface of strip 1 and bottom surface of strip 2. Take $$L=5cm$$ and $${\lambda}_{air}=40nm$$.

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Consider an air wedge formed by two glass plates having refractive index $$1.5$$ by placing a piece of paper of thickness $$20mm$$. Determine the number of dark bands formed.

A
$$1000$$

B
$$500$$

C
$$5000$$

D
$$100$$
Question 10
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Directions For Questions

A block of plastic having a thin air cavity (whose thickness is comparable to wavelength of light waves) is shown in figure. The thickness of air cavity (which can be considered as air wedge for interference pattern) is varying linearly from one end to other as shown.
A broad beam of monochromatic light is incident normally from the top and some from the bottom of cavity. The plastic layers above and below the cavity are having thickness much larger than wavelength of incident light. An observer when looking down from top sees an interference pattern consisting of eight dark fringes and seven bright fringes along the wedge. Take wavelength of incident light in air as $${\lambda}_{0}$$ and refractive index of plastic as $$\mu$$.
Assume that the thickness of the ends of air cavity are such that formation of fringes takes place there.

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Determine the difference $${L}_{1}-{L}_{2} (=\Delta L)$$ in terms of $${\lambda}_{0}$$.

A
$$\cfrac { 4{ \lambda }_{ 0 } }{ \mu } $$

B
$$\cfrac { 7{ \lambda }_{ 0 } }{2 \mu }$$

C
$$\cfrac { 3{ \lambda }_{ 0 } }{ \mu }$$

D
none of the above

Solution

According to the question, at A and B dark fringe is formed.Thus there are 7 pairs of fringes ( bright + dark).
Hence, $$\beta= \dfrac{L}{7}$$
$$\alpha= \dfrac{L_1-L_2}{L}$$
$$\beta= \dfrac{\lambda_o}{2\mu \alpha}$$
Thus $$\beta=\dfrac{\lambda_o L}{2\mu (L_1-L_2)} $$
$$\implies L_1-L_2= \dfrac{7\lambda_o}{2\mu}$$