Tag: wave optics

Quizzes (12) Questions (312)

Questions Related to wave optics

Multiple choice physics wave optics interference

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  

  1. $\dfrac { 2\lambda }{ \alpha } $
  2. $\dfrac { \lambda }{ \alpha } $
  3. $\dfrac { \lambda }{ (2\alpha ) } $
  4. $\dfrac { \lambda }{ \sqrt { \alpha } } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When two plane waves make a small angle alpha, the fringe width is given by beta = lambda / alpha. This is derived from the geometry of the interference pattern formed by the two waves.

Multiple choice physics wave optics interference

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 ?

  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. $\sqrt{A _1^2+A _2^2+2A _1A _2\cos B _1 B _2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When two waves y1 = A1 sin(omega*t - B1) and y2 = A2 sin(omega*t - B2) superimpose, the resultant amplitude A is given by the vector sum of the amplitudes: A = sqrt(A1^2 + A2^2 + 2*A1*A2*cos(B1 - B2)).

Multiple choice physics wave optics interference

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:

  1. $1\%$
  2. $2\%$
  3. $3\%$
  4. $4\%$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

Multiple choice physics wave optics interference

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

  1. 7.962
  2. 7.962 $\pi$
  3. $\displaystyle\frac{7.962}{\pi}$
  4. $\displaystyle\frac{7.962}{3\pi}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

Multiple choice physics wave optics interference

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:

  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) $
Reveal answer Fill a bubble to check yourself
B Correct answer
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) $

Multiple choice physics wave optics interference

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.

  1. $0.5$ w
  2. $0.7$ w
  3. $1.0$ w
  4. $2.0$ w
Reveal answer Fill a bubble to check yourself
B Correct answer
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,

$\beta_{new}=0.7\beta=0.7\,w$
Multiple choice physics wave optics refraction of water waves reflection and refraction at plane surfaces theories on light

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

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The Raman effect involves inelastic scattering of light, meaning the wavelength of the scattered light is different from the incident light (Stokes and anti-Stokes lines).

Multiple choice physics wave optics polarization of light polarisation of light polarisation

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

  1. $\dfrac{\pi}{2}$
  2. $\pi$
  3. $\dfrac{3\pi}{2}$
  4. $zero$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A point source emits spherical waves. Any points at the same distance from the source, such as the endpoints of a diameter, lie on the same wavefront and thus have zero phase difference.

Multiple choice physics wave optics polarization of light polarisation of light polarisation

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:

  1. $ 30^{\circ} $
  2. $ 19^{\circ} $
  3. $ 45^{\circ} $
  4. $ 90^{\circ} $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Malus's Law, the intensity after the first polarizer is I1 = I0/2 = 16. After the second, I2 = 16*cos^2(theta). After the third (perpendicular to the first), I3 = I2*cos^2(90-theta) = 16*cos^2(theta)sin^2(theta) = 16(sin(2*theta)/2)^2 = 4*sin^2(2*theta). Setting 4*sin^2(2*theta) = 3 gives sin^2(2*theta) = 3/4, so sin(2*theta) = sqrt(3)/2, meaning 2*theta = 60 degrees, or theta = 30 degrees.