Wave Optics Test

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Wave Optics Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

The wavefront of a light beam is given by the equation $$x+2y+3z=c$$ (where $$c$$ is arbitrary constant of light). What is the angle made by the light with the y-axis is

A
$$\cos ^{ -1 }{ \cfrac { 1 }{ \sqrt { 14 } } } $$

B
$$\sin ^{ -1 }{ \cfrac { 2 }{ \sqrt { 14 } } } $$

C
$$\cos ^{ -1 }{ \cfrac { 2 }{ \sqrt { 14 } } } $$

D
$$\sin ^{ -1 }{ \cfrac { 3 }{ \sqrt { 14 } } } $$

Solution

the wavefront equations must be of form $$ax+by+cz=1$$ where $$a^{2}+b^{2}+c^{2}=1$$ (a,b,c are cosines of angles made with x,y,z axes respectively)

applying this to the given equation ,
$$1^{2}+2^{2}+3^{2}=c^{2}$$
$$c^{2}=14$$
$$c=\sqrt{14}$$
angle made with y-axis is $$cos ^{-1}\dfrac{2}{\sqrt{14}}$$
option $$C$$ is correct.
Question 2
1 / -0

In Young's interference experiment, if the slits are of unequal width, then

A
no fringes will be formed

B
the positions of minimum intensity will not be completely dark

C
bright fringe as displaced from the original central position

D
distance between two consecutive dark fringes will not be equal to the distance between two consecutive right fringes.

Solution

Unequal width of slits will cause unequal intensity of lights entering from both slits.
As a result, during interference complete cancelling of light intensity will not take place at regions of otherwise dark fringe.
As the value of $$\beta$$ does not depend on intensity of light, there will be no shifting of fringes as well as no change in fringe width.
Answer. B) the positions of minimum intensity will not be completely dark.
Question 3
1 / -0

In a double-slit experiment, instead of taking slits of equal width, one slit is made twice as wide as the other. Then, in the interference pattern

A
the intensities of both maxima and the minima increases

B
the intensity of maximum increase and minima has zero intensity

C
the intensity of the maxima decreases and that of minima increases

D
the intensity of the maxima decreases and the minima has zero intensity

Solution

Net intensity at any point on the screen due to two slits with incident intensities $$I_1,I_2$$ is given by
$$I=I_1+I_2+2\sqrt{I_1I_2}\cos\phi$$
When the slits are equal, $$I_1=I_2$$
Thus $$I'=I_!+I_1+2\sqrt{I_1I_1}\cos\phi$$
Not let the intensity due to second slit be $$I_2>I_1$$
Then $$I''=I_1+I_2+2\sqrt{I_1I_2}\cos\phi>I'$$ for any $$\phi$$
Question 4
1 / -0

The resolving power of an electron microscope operated at 16 kV is R. The resolving power of the electron microscope when operated at 4 kV is

A
R/4

B
R/2

C
4R

D
2R

Solution

$$\sqrt{\dfrac{16}{4}}=\dfrac{R}{x}$$

$$x=\dfrac{R}{2}$$

option $$B$$ is correct
Question 5
1 / -0

One of the two slits in YDSE is painted over, so that it transmits only light waves having intensity half of the intensity of the light waves having half of the intensity of the light waves through the other slit. As a result of this

A
fringe pattern disappears

B
bright fringes become brighten and dark ones become darker

C
dark and bright fringes get fainter

D
dark fringes get brighter and bright fringes get darker

Solution

Case 1: When slit are of equal width, if $$ { I }_{ o }$$ is intensity of light then,
$$ { I }_{ max }={ I }_{ o }+{ I }_{ o }+2\sqrt { { I }_{ o }{ I }_{ o } } =4{ I }_{ o }$$
$$ { I }_{ min }={ I }_{ o }+{ I }_{ o }-2\sqrt { { I }_{ o }{ I }_{ o } } =0$$
Case 2: When one of the slit is painted over, its intensity would reduce to half then,
$$ { I }_{ newmax }={ I }_{ o }/2+{ I }_{ o }+2\sqrt { { I }_{ o }{ I }_{ o }/2 } =2.932{ I }_{ o }$$
$$ { I }_{ newmin }={ I }_{ o }/2+{ I }_{ o }-2\sqrt { { I }_{ o }{ I }_{ o }/2 } =0.068{ I }_{ o }$$
Therefore, dark fringes get brighter and bright fringes get darker
Question 6
1 / -0

A certain region of a soap bubble reflects red light of vacuum wavelength $$\lambda=650nm$$. What is the minimum thickness that this region of the soap bubble could have? Take the index of reflection of the soap film to be $$1.41$$

A
$$1.2\times {10}^{-7}m$$

B
$$650\times {10}^{-9}m$$

C
$$120\times {10}^{7}m$$

D
$$650\times {10}^{5}m$$

Solution

By equation of destructive interference,
thickness, $$t = \dfrac{1}{2}\dfrac{(m+0.5)\lambda}{1.41}$$
where $$m$$ is any non-negative integer,
$$\lambda=650*10^{-9}$$.
For minimum thickness, take $$m=0$$
So, $$t=1.152*10^{-7}\approx1.2*10^{-7}$$
Question 7
1 / -0

Light from a source travel out with a velocity c. If the source moves away from the observer with a velocity v, then relative velocity of light w.r.t observer is

A
$$ c $$

B
$$ c+v $$

C
$$ c-v $$

D
$$\sqrt{c^2+v^2} $$

Solution

Consider two particles with velocity u and w. The expression for the relative velocity can be given by

$$ v = \dfrac{w-u}{1-\dfrac{wu}{c^2}} $$

If one of the particles is light, $$w=c$$.

This gives relative velocity of light $$v=c$$.
Question 8
1 / -0

Directions For Questions

When light from two sources (say slits $${S}_{1}$$ and $${S}_{2}$$) interfere, they form alternate dark and bright fringes. Bright fringe is formed at all points where the path difference is an odd multiple of half wavelength. At the condition of equal amplitudes, $${A}_{1}={A}_{2}=a$$, the maximum intensity can also be indicated with phase factor as $$I=2{a}^{2}\cos ^{ 2 }{ \left( \phi /2 \right) } $$.

...view full instructions

If the path difference between the slits $${S}_{1}$$ and $${S}_{2}$$ is $$\lambda$$, the central fringe will have an intensity of

A
$$0$$

B
$${a}^{2}$$

C
$$2{a}^{2}$$

D
$$4{a}^{2}$$

Solution

Path difference at the central fringe will be the same as the path difference at the slits i.e. $$\lambda$$.
Hence, the waves reaching the central fringe will be $$180^o$$ out of phase and will result in destructive interference. Hence, the intensity of the central fringe will be 0.
Question 9
1 / -0

A thin slice is cut out of a glass cylinder along a plane parallel to its axis. The slice is placed on a flat glass plae as shown in figure. The observed interference fringes from this combination shall be

A
straight

B
circular

C
equally spaced

D
having fringe spacing which increases as we go outward

Solution

The locus of the equal path difference consists in lines going parallel to the axis of cylinder.
Therefore, interference fringes will be straight.
Question 10
1 / -0

Resolving power of a telescope increases with

A
increase in focal length of eye-piece

B
increase in focal length of objective

C
increase in aperture of eye piece

D
increase in aperture of objective

Solution

Resolving power $$= \displaystyle \dfrac{\lambda}{d \lambda}$$ plane transmission granting

Resolving power for telescope

$$= \displaystyle \frac{1}{\text{limit of resolution}} = \dfrac{d}{1.22 \lambda} = \dfrac{d_0}{d_1}$$

by increasing the aperture of objective resolving power can be increased.

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Wave Optics Test - 32

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Wave Optics Test - 32

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Question 1

$$\cos ^{ -1 }{ \cfrac { 1 }{ \sqrt { 14 } } } $$

$$\sin ^{ -1 }{ \cfrac { 2 }{ \sqrt { 14 } } } $$

$$\cos ^{ -1 }{ \cfrac { 2 }{ \sqrt { 14 } } } $$

$$\sin ^{ -1 }{ \cfrac { 3 }{ \sqrt { 14 } } } $$

Solution

Question 2

no fringes will be formed

the positions of minimum intensity will not be completely dark

bright fringe as displaced from the original central position

distance between two consecutive dark fringes will not be equal to the distance between two consecutive right fringes.

Solution

Question 3

the intensities of both maxima and the minima increases

the intensity of maximum increase and minima has zero intensity

the intensity of the maxima decreases and that of minima increases

the intensity of the maxima decreases and the minima has zero intensity

Solution

Question 4

R/4

R/2

4R

2R

Solution

Question 5

fringe pattern disappears

bright fringes become brighten and dark ones become darker

dark and bright fringes get fainter

dark fringes get brighter and bright fringes get darker

Solution

Question 6

$$1.2\times {10}^{-7}m$$

$$650\times {10}^{-9}m$$

$$120\times {10}^{7}m$$

$$650\times {10}^{5}m$$

Solution

Question 7

$$ c $$

$$ c+v $$

$$ c-v $$

$$\sqrt{c^2+v^2} $$

Solution

Question 8

$$0$$

$${a}^{2}$$

$$2{a}^{2}$$

$$4{a}^{2}$$

Solution

Question 9

straight

circular

equally spaced

having fringe spacing which increases as we go outward

Solution

Question 10

increase in focal length of eye-piece

increase in focal length of objective

increase in aperture of eye piece

increase in aperture of objective

Solution

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