Tag: interference

Questions Related to interference

Two points P and Q are situated at the same distance from a source of light but on opposite sides.The phase difference between the light waves passing through P and Q will be

  1. $\pi$

  2. 2$\pi$

  3. $\pi$/2

  4. 0


Correct Option: D

Light waves of wave length $\lambda$ propagate in a medium. If $M$ and $N$ are two points on the wave front and they are separated by a distance $\lambda /4$, the phase difference between them will be (in radian)

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

  2. $\dfrac{\pi}{8}$

  3. $\dfrac{\pi}{45}$

  4. zero


Correct Option: D

In an interference experiment, distance between the lists is $2\ mm$ and screen is placed at distance $1\ m$ from the slits. Fourth dark fringe is formed exactly opposite to one of the slits. Wavelength of light used in nm is

  1. 480

  2. 600

  3. 570

  4. 500


Correct Option: A

A glass wedge of angle $0.01$ radian and $\mu=1.5$ is illuminated by monochromatic light of wavelength $6000\ A$ falling normally on it. At what distance from wedge will $10^{th}$ dark fringe be observed by reflected light ?

  1. $0.1\ mm$

  2. $0.2\ mm$

  3. $0.3\ mm$

  4. $0.4\ mm$


Correct Option: A

In YDSE $S _1$ and $S _2$ has intersity $I$ and $9I$. Find difference in intensity b/w point which has phase difference of  $\pi$

  1. $10 I$

  2. $6 I$

  3. $8I$

  4. $4I$


Correct Option: D

For interference between waves from two sources of intensities $I$ and $4I$, find the intensity at the point in the pattern where the phase difference is $\dfrac{\pi}{2}$ and $\pi$.

  1. $10I$ and $I$

  2. $5I$ and $5I$

  3. $5I$ and $I$

  4. $5I$ and $10I$


Correct Option: C

The path difference between two interfering waves at a point on the screen  is $ \lambda /6 $. The ratio of intensity at the point and that the central bright fringe will be (Assume that internally due to each slit in same). 

  1. 0.853

  2. 8.53

  3. 0.75

  4. 7.5


Correct Option: C
Explanation:

Given that,

Path difference $x=\dfrac{\lambda }{6}$

We know that,

  $ I={{I} _{0}}{{\cos }^{2}}\left( \dfrac{\phi }{2} \right) $

 $ \dfrac{I}{{{I} _{0}}}={{\cos }^{2}}\left( \dfrac{\phi }{2} \right)....(I) $

Now, the phase difference is

  $ \phi =\dfrac{2\pi }{\lambda }\times x $

 $ \phi =\dfrac{2\pi }{\lambda }\times \dfrac{\lambda }{6} $

 $ \phi =\dfrac{\pi }{3} $

Now, put the value of $\phi $ in equation (I)

  $ \dfrac{I}{{{I} _{0}}}={{\cos }^{2}}{{30}^{0}} $

 $ \dfrac{I}{{{I} _{0}}}=\dfrac{3}{4} $

 $ \dfrac{I}{{{I} _{0}}}=0.75 $

Hence, the value of ratio of the intensity at the point is $0.75$ 

The distance between the two slits in a Young's double slit experiment is $d$ and the distance of the screen from the plane of the slits is $b$,$P$ is a point on the screen directly in front of one of the slits. The path difference between the waves arriving at $P$ from the two slits is

  1. $\dfrac{d^{2}}{b}$

  2. $\dfrac{d^{2}}{2b}$

  3. $\dfrac{2d^{2}}{b}$

  4. $\dfrac{d^{2}}{4b}$


Correct Option: B

The phase difference between two waves, represented by
${ y } _{ 1 }={ 10 }^{ -6 }sin{ 100t+(x/50)+0.5} m$
${ y } _{ 2 }={ 10 }^{ -6 }cos{ 100t+\left( \frac { x }{ 50 }  \right) } m$
where x is expressed in meters and is expressed in seconds, is approximately:

  1. 2.07 Radians

  2. 0.5 Radians

  3. 1.5 Radians

  4. 1.07 Radians


Correct Option: D

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.


Correct Option: A