Tag: physics

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Questions Related to physics

Multiple choice physics wave optics interference

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

  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.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In a YDSE, the central bright fringe occurs where the path difference is zero for all wavelengths, resulting in maximum constructive interference and thus the highest intensity.

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

The higher the frequency of 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
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The relationship between frequency (f) and wavelength (lambda) is given by c = f * lambda, where c is the speed of light. Therefore, frequency and wavelength are inversely proportional. As frequency increases, wavelength must decrease.

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

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

  1. Increase, decrease

  2. Increase, unchanged

  3. Decrease, unchanged

  4. Decrease, decrease
Reveal answer Fill a bubble to check yourself
C Correct answer
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.