Tag: wave optics

Questions Related to wave optics

Two coherent plane light waves of equal amplitude makes a small angle $\alpha (<<1)$ with each other. They fall almost normally on a screen. If $\gamma $ is the wavelength of light waves, the fringe width $\Delta x$ of interference patterns of the two sets of wave on the screen is  

physics wave optics interference

  1. $\dfrac { 2\lambda }{ \alpha } $

  2. $\dfrac { \lambda }{ \alpha } $

  3. $\dfrac { \lambda }{ (2\alpha ) } $

  4. $\dfrac { \lambda }{ \sqrt { \alpha } } $

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Correct Option: B

What is the amplitude of resultant wave, when two waves $y _1=A _1\sin (\omega t-B _1)$ and $y _2=A _2\sin (\omega t-B _2)$ superimpose ?

physics wave optics interference

  1. $A _1+A _2$

  2. $|A _1-A _2|$

  3. $\sqrt{A _1^2+A _2^2+2A _1A _2\cos (B _1-B _2)}$

  4. $\sqrt{A _1^2+A _2^2+2A _1A _2\cos B _1 B _2}$

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Correct Option: C

An isotropic point source emits light. A screen is situated at  a given distance. If the distance between sources and screen is decreased by $2\%$, illuminance will increase by:

physics wave optics interference

  1. $1\%$

  2. $2\%$

  3. $3\%$

  4. $4\%$

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Correct Option: D
Explanation:
For isotropic point source
$E\propto\dfrac{1}{r^{2}}$

For small change, $\dfrac{\Delta E}{\Delta r}=\dfrac{-2k }{r^{3}}$

$\dfrac{\Delta E}{\Delta r}=-2\dfrac{k}{r^{2}}\dfrac{1}{r}$ or $\dfrac{\Delta E}{\Delta r}=-2\dfrac{E}{r}$

or $\dfrac{\Delta E}{I}=2\left(-\dfrac{\Delta r}{r}\right)\therefore \% \Delta E=2\times 2\%=4\%$

Hence, (d) is correct.

The path difference between two wavefronts emitted by coherent sources of wavelength 5460 $\overset{o}{A}$ is 2.1 micron. The phase difference between the wavefronts at that point is

physics wave optics interference

  1. 7.962

  2. 7.962 $\pi$

  3. $\displaystyle\frac{7.962}{\pi}$

  4. $\displaystyle\frac{7.962}{3\pi}$

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Correct Option: B
Explanation:

Phase diff. = $\displaystyle\frac{2\pi x}{\lambda}$
Path difference = $\displaystyle\frac{2\pi \times 2.1 \times 10^{-6}}{5460 \times 10^{-10}}$ = 7.692 $\pi$ radian.

Two light rays having the same wavelength $\lambda$ in vacuum are in phase initially. Then the first ray travels a path ${L} _{1}$ through a medium of refractive index ${n} _{1}$ while the second ray travels a path of length ${L} _{2}$ through a medium of refractive index ${n} _{2}$. The two waves are then combined to produce interference. The phase difference between the two waves is:

physics wave optics interference

  1. $\dfrac { 2\pi }{ \lambda } \left( { L } _{ 2 }-{ L } _{ 1 } \right) $

  2. $\dfrac { 2\pi }{ \lambda } \left( { n } _{ 1 }{ L } _{ 1 }-{ n } _{ 2 }{ L } _{ 2 } \right) $

  3. $\dfrac { 2\pi }{ \lambda } \left( { n } _{ 2 }{ L } _{ 1 }-{ n } _{ 1 }{ L } _{ 2 } \right) $

  4. $\dfrac { 2\pi }{ \lambda } \left( \dfrac { { L } _{ 1 } }{ { n } _{ 1 } } -\dfrac { { L } _{ 2 } }{ { n } _{ 2 } } \right) $

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Correct Option: B
Explanation:

The optical path between any two points is proportional to the time of travel.
The distance traversed by light in a medium of refractive index $\mu $ in time $t$ is given by
$d=vt$            .....(i)


where $v$ is velocity of light in the medium. The distance traversed by light in a vacuum in this time,

$\Delta =ct$

  $=c\cdot \dfrac { d }{ v } $        [from equation (i)]

  $=d \dfrac { c }{ v } =\mu d$          .......(ii)                   (Since, $\mu =\dfrac { c }{ v } $)

This distance is the equivalent distance in vacuum and is called optical path.

Here, optical path for first ray $={ n } _{ 1 }{ L } _{ 1 }$

Optical path for second ray $={ n } _{ 2 }{ L } _{ 2 }$

Path difference $={ n } _{ 1 }{ L } _{ 1 }-{ n } _{ 2 }{ L } _{ 2 }$

Now, phase difference

    $=\dfrac { 2\pi  }{ \lambda  } \times $ path difference

    $=\dfrac { 2\pi  }{ \lambda  } \times \left( { n } _{ 1 }{ L } _{ 1 }-{ n } _{ 2 }{ L } _{ 1 } \right) $

Electrons accelerated from rest by an electrostatic potential are collimated and sent through a Young's double slit setup. The figure width is w. If the accelerating potential is doubled then the width is now close to.

physics wave optics interference

  1. $0.5$ w

  2. $0.7$ w

  3. $1.0$ w

  4. $2.0$ w

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Correct Option: B
Explanation:
$\beta=\dfrac{\lambda D}{d}$

$\lambda=\dfrac{h}{mV}=\dfrac{h}{\sqrt{2mq\Delta V}}$

$\beta \propto \lambda$

Therefore,

$\beta \propto \dfrac{1}{\sqrt{\Delta V}}$

$As $\Delta V$ is double,

$\beta$ is $\dfrac{1}{\sqrt 2}$ times of $\beta_{old}$

Therefore,

$\beta_{new}=0.7\beta=0.7\,w$

Which of the following statement is incorrect about the Raman effect?

physics wave optics refraction of water waves reflection and refraction at plane surfaces theories on light

  1. Raman effect can be seen by passing a monochromatic beam of light through benzene

  2. During scattering, the wavelength of the light used may change

  3. The wavelength of the monochromatic light used to show the Raman effect is approximately $4358$ $\overset{o}{A}$

  4. In Raman scattering the wavelength of the scattered light is the same as that of the incident light

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Correct Option: D

Which one is not produced by sound waves in air?

physics wave optics polarization of light polarisation of light polarisation

  1. Polarisation

  2. Diffraction

  3. Refraction

  4. Reflection

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Correct Option: A
Explanation:

As sound waves are longitudinal waves, they don't exhibit Polarization phenomenon, which is exhibited by the transverse wave only.

A point source of monochromatic light is situated at the centre of a circle, what is the phase difference between the light waves passing through the end points of any diameter

physics wave optics polarization of light polarisation of light polarisation

  1. $\dfrac{\pi}{2}$

  2. $\pi$

  3. $\dfrac{3\pi}{2}$

  4. $zero$

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Correct Option: D

An unpolarised light of intensity $32  \mathrm{W} / \mathrm{m}^{2}  $ passes through three polarisers, such that the transmission axis of last polarizer is perpendicular with the first. If the intensity of emergent light is $3  \mathrm{Wh}  $ Im $ ^{2} $ then the angle between the transmission axes of the first two polarisers is:

physics wave optics polarization of light polarisation of light polarisation

  1. $ 30^{\circ} $

  2. $ 19^{\circ} $

  3. $ 45^{\circ} $

  4. $ 90^{\circ} $

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Correct Option: A