Tag: physics

Questions Related to physics

In a YDSE, the central bright fringe can be identified :

physics wave optics interference

  1. as it has greater intensity than the other bright fringe.

  2. as it is wider than the other bright fringes.

  3. as it is narrower than the other bright fringes.

  4. by using white light instead of single wavelength light.

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Correct Option: A

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

Show answer
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,

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

The higher the frequency of light?

physics oscillations and waves huygen's wave theory refraction of water waves reflection and refraction at plane surfaces theories on light

  1. The longer (larger) its wavelength.

  2. The shorter (smaller) its wavelength.

  3. The greater its velocity in a vacuum.

  4. The redder it will be

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Correct Option: B

A point source of light is placed at origin, in air. The equation of wave front of the wave at time $t$, emitted by source at $t=0$, is $(Take\ refractive\ index\ of\ air\ ms\ 1)$

physics oscillations and waves huygen's wave theory refraction of water waves reflection and refraction at plane surfaces theories on light

  1. $x+y+z=et$

  2. $x^{2}+y^{2}+z^{2}=t^{2}$

  3. $xy+yz+zx=e^{2}t^{2}$

  4. $x^{2}+y^{2}+z^{2}=e^{2}t^{2}$

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Correct Option: A

When the beam of light travels in a medium with lesser velocity than that of in vaccum then the value of wavelength and frequency will respectively

physics oscillations and waves huygen's wave theory refraction of water waves reflection and refraction at plane surfaces theories on light

  1. Increase, decrease

  2. Increase, unchanged

  3. Decrease, unchanged

  4. Decrease, decrease

Show answer
Correct Option: C
Explanation:

Energy $=\dfrac{4c}{\lambda }=4\nu $     which is constant.
Here, velocity is decreasing so, to make constant energy $\lambda$ should also decrease and frequency remains constant.