Wave Optics Test

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Wave Optics Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

The ratio of slit widths in Young's double slit experiment is $$4:9$$. The ratio of maximum and minimum intensities will be

A
$$169:25$$

B
$$81:16$$

C
$$13:5$$

D
$$25:1$$

Solution

The intensity ratio will be same as slit width ratio
$$I_{1} : I_{2} = 4:9$$

$$\dfrac {I_{max}} {I_{min}} =( \dfrac {\sqrt {I_{2}} + \sqrt {I_{1}}} {\sqrt{I_2} - \sqrt{I_{1}}})^{2}$$ = $$(\dfrac {3+2} {3-2})^2$$ = 25:1

Answer. D) 25:1
Question 2
1 / -0

A narrow monochromatic beam of light of intensity $$I$$ is incidents on a glass plate as shown, in the figure. Another identical glass plate is kept close to the first-one and parallel to it. Each glass plate reflects $$25{\%}$$ of the light incident on it and transmits the remaining. Then, the ratio of the maximum and minimum intensities in the interference pattern formed by the two beams obtained after one reflection at each plate is

A
$$7:1$$

B
$$49:1$$

C
$$4:1$$

D
$$16:9$$

Solution

As given that at each reflection 25% of incident energy is reflected, therefore at each refraction 75% of incident energy will be refracted.
Intensity of incident light =I
Hence, Intensity of reflected ray AB
$$I_{AB}=25I/100=I/4$$....................eq1

Intensity of refracted ray AC
$$I_{AC}=75I/100=3I/4$$

Intensity of reflected ray CA′
$$I_{CA'}=25I_{AC}/100=1/4\times3I/4=3I/16$$

Intensity of refracted ray A′B′
$$I_{A'B'}=75I_{CA'}/100=3/4\times 3I/16=9I/64$$ .......................eq2

Now, rays AB and A'B' interfere, therefore
$$\dfrac{I_{max}}{I_{min}}=\dfrac{I_{AB}+I_{A'B'}+2\sqrt{I_{AB}I_{A'B'}}}{I_{AB}+I_{A'B'}-2\sqrt{I_{AB}I_{A'B'}}}$$
$$\dfrac{I_{max}}{I_{min}}=\dfrac{I/4+9I/64+2\sqrt{I/4\times9I/64}}{I/4+9I/64-2\sqrt{I/4\times9I/64}}=\dfrac{49}{1}$$
Question 3
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Two nearby objects are just resolved, if the principle maximum in the diffraction pattern of one coincides with

A
principal maxima of other

B
first minimum of the other

C
half-way between principal maximum and first minimum.

D
second maximum of the other

Solution

The Rayleigh criterion is the generally accepted criterion for the minimum resolvable detail - the imaging process is said to be diffraction-limited when the first diffraction minimum of the image of one source point coincides with the maximum of another.
Question 4
1 / -0

An un-publicized beam of intensity $$2a^2$$ passes through a thin Polaroid. Assuming zero absorption in the Polaroid the intensity of emergent planes polarized light is

A
$$2a^2$$

B
$$a^2$$

C
$$\sqrt2a^2$$

D
$$\dfrac {\ a^2}{2}$$

Solution

When an upolarized beam passes through a polaroid, the intensity of the beam halves. When this polarized beam passes through a polaroid again, the beam's intensity varies by Malus' Law.
Thus the intensity becomes $$\dfrac{2a^2}{2}=a^2$$
Question 5
1 / -0

White light is normally incidents on a soap film. The thickness of the film is $$0.5\mu m$$ and its refractive index is $$1.33$$. Which wave length will be reflected maximum in the visible region?

A
$$26600A^o$$

B
$$8860A^o$$

C
$$5320A^o$$

D
$$3800A^o$$

Solution

Given : $$n =1.33$$, $$d = 5\times 10^{-7}$$ m
For maximum reflection, $$2nd =\bigg(m+\dfrac{1}{2}\bigg) \lambda_m$$
$$\therefore$$ $$2(1.33)(5\times 10^{-7}) =\bigg(m+\dfrac{1}{2}\bigg) \lambda_m$$ $$\implies \lambda_m = \dfrac{26600}{2m+1}$$ $$A^o$$
Thus $$\lambda_o = 26600$$ $$A^o$$, $$\lambda_1 = 8866.67$$ $$A^o$$, $$\lambda_2 = 5320$$ $$A^o$$
Hence, $$5320$$ $$A^o$$ will be reflected maximum in the visible region.
Question 6
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Light of wavelength $$5880A^o$$ is incidents on a thin glass plate $$(\mu=1.5)$$ such that the angle of refraction in the plate is $$60^0$$. The minimum thickness of the plate, so that it appears dark in the reflected light will be

A
$$3920A^o$$

B
$$4372A^o$$

C
$$5840A^o$$

D
$$6312A^o$$

Solution

From equation of destructive interference,
$$2tcos\theta=\dfrac{\lambda}{n}$$
where, $$n$$ is refractive index.
Using other values from the problem statement,
$$t=\dfrac{5880A^o}{1.5}=3920A^o$$
Question 7
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Directions For Questions

Figure illustrates the interference experiment with Fresnel mirror. The angle between the mirrors $$a=12'$$. The distances from the mirrors intersection line to the narrow slit $$S$$ and the screen $$E$$ is equal to $$r=10cm$$ and $$b=130cm$$ respectively. The monochromatic wavelength of light $$\lambda=550nm$$.

...view full instructions

The shift of the interference pattern on the screen when the slit is displayed by $$Sl=1mm$$ along the arc of radius $$r$$ with centre at $$0$$.

A
$$4mm$$

B
$$6mm$$

C
$$10mm$$

D
$$13mm$$

Solution

The mirror will rotate by $$\dfrac{\bigtriangleup l}{r}$$ or rotation of reflected ray $$=\dfrac{\bigtriangleup l}{r}$$ shift in fringe magnitude
$$=\displaystyle\frac{b\bigtriangleup l}{r}$$ $$=1.3\times (10)^2=13mm$$
Question 8
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Two identical lights sources $$S_1$$ and $$S_2$$ emit the light of same wavelength $$\lambda$$. These light rays will exhibit interference if

A
their phase difference remain constant.

B
their phase difference is distributed randomly.

C
their light intensities remain constant.

D
their light intensities change continuously.

Solution

For interference to take place the light sources need to be either in phase or have a constant phase difference. In case the phase difference keeps changing the interference pattern will keep on changing, as a result of interference pattern will be observed.

Answer. A) Thier phase difference remain constant.
Question 9
1 / -0

Directions For Questions

Figure illustrates the interference experiment with Fresnel mirror. The angle between the mirrors $$a=12'$$. The distances from the mirrors intersection line to the narrow slit $$S$$ and the screen $$E$$ is equal to $$r=10cm$$ and $$b=130cm$$ respectively. The monochromatic wavelength of light $$\lambda=550nm$$.

...view full instructions

At what maximum width $$\delta_{max}$$, of the slit are the interference fringes on the screen observed still sharp?

A
$$42\mu m$$

B
$$36\mu m$$

C
$$64\mu m$$

D
none of these

Solution

$$\displaystyle\frac{b\delta_{max}}{r}=\frac{\beta}{2}$$

$$\Rightarrow \delta_{max}=\displaystyle\frac{\beta r}{2b}=\displaystyle{10^{-3}\times 1.1\times (.1)}{2\times 1.3}=\displaystyle 42\mu m$$
Question 10
1 / -0

In a YDSE with identical slits, the intensity of the central bright fringe is $${I}_{0}$$.If one of the slits is covered, the intensity at the same point is

A
$$2{I}_{0}$$

B
$$ { I }_{ 0 }$$

C
$${ I }_{ 0 }/2$$

D
$${ { I }_{ 0 } }/{ 4 } $$

Solution

Let intensity due to each slit on the central bright fringe be $$I_1$$.
Thus intensity at the point of bright fringe due to interference of the light from two slits is $$I_0=I_1+I_1+2\sqrt{I_1.I_1}\cos\phi$$
For central bright fringe, $$\phi=0$$
$$\implies I_0=4I_1$$
$$\implies I_1=\dfrac{I_0}{4}$$

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Re-Attempt Weekly Quiz Competition

Wave Optics Test - 30

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Wave Optics Test - 30

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SHARING IS CARING

Question 1

$$169:25$$

$$81:16$$

$$13:5$$

$$25:1$$

Solution

Question 2

$$7:1$$

$$49:1$$

$$4:1$$

$$16:9$$

Solution

Question 3

principal maxima of other

first minimum of the other

half-way between principal maximum and first minimum.

second maximum of the other

Solution

Question 4

$$2a^2$$

$$a^2$$

$$\sqrt2a^2$$

$$\dfrac {\ a^2}{2}$$

Solution

Question 5

$$26600A^o$$

$$8860A^o$$

$$5320A^o$$

$$3800A^o$$

Solution

Question 6

$$3920A^o$$

$$4372A^o$$

$$5840A^o$$

$$6312A^o$$

Solution

Question 7

$$4mm$$

$$6mm$$

$$10mm$$

$$13mm$$

Solution

Question 8

their phase difference remain constant.

their phase difference is distributed randomly.

their light intensities remain constant.

their light intensities change continuously.

Solution

Question 9

$$42\mu m$$

$$36\mu m$$

$$64\mu m$$

none of these

Solution

Question 10

$$2{I}_{0}$$

$$ { I }_{ 0 }$$

$${ I }_{ 0 }/2$$

$${ { I }_{ 0 } }/{ 4 } $$

Solution

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