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

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)

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

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 ?

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

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 $\pi$.

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). 

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

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:

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

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  

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 ?

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:

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

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:

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.