Tag: interference

Questions Related to 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  

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,

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

To demonstrate the phenomenon of interference we require two sources which emit radiation of

physics superposition of waves coherence young's double slit experiment interference

  1. nearly the same frequency

  2. the same frequency

  3. different wavelength

  4. the same frequency and having a definite phase relationship

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Correct Option: D

Which of the following is not an essential condition for interference?

physics superposition of waves coherence young's double slit experiment interference

  1. The two interfering waves must propagate in almost the same direction

  2. The waves must have the same period and wavelength

  3. The amplitudes of the two waves must be equal

  4. The two interfering beams of light must originate from the same source

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Correct Option: A
Explanation:

When two waves are propagate in same direction

If two sources have a randomly varying phase difference $\varphi ( t )$  the resultant intensity will be given by 

physics superposition of waves coherence young's double slit experiment interference

  1. $I _ { 0 }$

  2. $\dfrac { I _ { 0 } } { 2 }$

  3. $2 I _ { 0 }$

  4. $\dfrac { I _ { 0 } } { \sqrt { 2 } }$

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Correct Option: C
Explanation:

Phase difference is $\phi (t)$,

Since, the phase difference is varying, then the waves are said to be incoherent. So, the intensity of resultant wave is the sum of intensities of each wave i.e. $2\,{{I} _{0}}$ 

In a biprism experiment, the distance of 20 th bright bandfrom the center of the interference pattern is 8$\mathrm { mm }$ . The distance of 30th bright band from the center is

physics superposition of waves coherence young's double slit experiment interference

  1. $11.8\mathrm { mm }$

  2. 12$\mathrm { mm }$

  3. 14$\mathrm { mm }$

  4. 16$\mathrm { mm }$

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Correct Option: A
Explanation:

Given

$\begin{array}{l} 20\beta =8mm \ \beta =\dfrac { 8 }{ { 20 } }  \ Now, \ 30th\, \, \max  ima=30\beta =\dfrac { { 30\times \beta  } }{ { 20 } } =12mm \ 30th\, \min  ima=\dfrac { { \left( { 2\left( { 30 } \right) -1 } \right)  } }{ 2 } \beta  \ =\dfrac { { 59 } }{ 2 } \beta  \ =\dfrac { { 59 } }{ 2 } \times \dfrac { 8 }{ { 20 } }  \ =11.8mm \ Hence,\, option\, A\, is\, the\, correct\, answer. \end{array}$