Interference of Light and Necessary Conditions for Interference

Two light rays having the same wavelength in vacuum are in phase initially. Then, the first ray travels a path ${\mathrm{L}}_{\mathrm{l}}$ through a medium of refractive index ${\mathrm{\mu }}_1$ while the second ray travels a path ${\mathrm{L}}_2$ through a medium of refractive index ${\mathrm{\mu }}_2$. The two waves are then combined to observe interference. The phase difference between the two waves is

Two waves originating from source ${\mathrm{S}}_1$ and ${\mathrm{S}}_2$ having zero phase difference and common wavelength $\lambda$ will show completely destructive interference at a point $\mathrm{P}$ if $\left({\mathrm{S}}_1\mathrm{P}-{\mathrm{S}}_2\mathrm{P}\right)$ is

Two waves $y_1=A_1\sin \left(\omega t-{\beta }_1\right)$ and $y_2=A_2\sin \left(\omega t-{\beta }_2\right)$ superimpose to form a resultant wave whose amplitude is 

One important similarity between light waves and sound waves is that both -

If the ratio of the intensity of two coherent sources is $4$ then the visibility $\frac{{\mathrm{I}}_{\max }-{\mathrm{I}}_{\min }}{{\mathrm{I}}_{\max }+{\mathrm{I}}_{\min }}$ of the fringes is -

Two light waves are given by, $E_1=2\sin \left(100\pi t-kx+30°\right)$ and $E_2=3\cos \left(200\pi t-kx+60°\right)\text{.}$ The ratio of intensity of first wave to that of second wave is -

The equation of two light waves are $y_1=6\cos \left(\mathrm{\omega t}\right)$ and $y_2=8\cos (\mathrm{\omega t}+\mathrm{\varphi })\text{.}$ The ratio of maximum to minimum intensities produced by the superposition of these waves will be-

The intensity ratio of two waves is $9:1$. These waves produce the event of interference. The ratio of maximum to minimum intensity will be-

Consider interference between waves from two sources of intensities $I \& 4I$. Find intensities at points where the phase difference is $\pi$.

The intensities of two sources are $I$ and $9I$ respectively. If the phase difference between the waves emitted by them is $\pi$ then the resultant intensity at the point of observation will be-

In a YDSE, a small detector measures an intensity of illumination of $I$ units at the centre of the fringe pattern. If one of the two (identical) slits is now covered, the measured intensity will be :

Hologram is based on phenomenon of

Two beams of light having intensities $I$ and $4I$ interfere to produce a fringe pattern on a screen. The phase difference between the beams is $\frac{\pi }{2}$ at point $\mathrm{A}$ and $\pi$ at point $\mathrm{B}$. Then the difference between resultant intensities at $\mathrm{A}$ and $\mathrm{B}$ is

The two coherent sources of intensity that ratio $2:8$ produce an interference pattern. The values of maximum and minimum intensities will be respectively(intensity of first light is $I_1$)

In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is $\lambda \text{is} I$. The intensity of light at a point where the path difference becomes $\frac{\lambda }{3}$ is

In a double-slit experiment, instead of taking slits of equal widths, one slit is made twice as wide as the other. Then in the interference pattern:

In Young double-slit experiment, the two slits act as coherent sources of equal amplitude $A$ and wavelength $\lambda .$ In another experiment with the same set up the two slits are sources of equal amplitude $A$ and wavelength $\lambda$ but are incoherent. The ratio of the intensity of light at the midpoint of the screen in the first case to that in the second case is :

By a monochromatic wave, we mean-

Assertion: Speed of light in glass is independent of the colour of light. Reason: The violet colour travels faster than the red light in a glass prism.

A cylindrical wavefront spreads from a line source which is comparable to a long and narrow slit. The wavefront is at a distance d from source then the amplitude of wave is proportional to