Young's double slit experiment is first performed in air and then in a medium other than air. It is found that the $8^{th}$ bright fringe in the medium lies where the $5^{th}$ dark fringe lies in air. The refractive index of the medium is nearly:

In a Young's double slit experiment,if there is no initial phase difference between the light from the two slits,what is the path difference at a point on the screen corresponding to the fifth minimum?

State two conditions for obtaining sustained interference of light. In Young's double-slit experiment,using light of wavelength $400 \, nm$,interference fringes of width $'X'$ are obtained. If the wavelength of light is increased to $600 \, nm$ and the separation between the slits is halved,find the ratio of the distances between the screen and the slits in the two arrangements if the fringe width remains the same.

In Young's double-slit experiment, one slit is wider than the other, such that the amplitude of the light wave from one slit is twice that of the other. If the maximum intensity is $I_m$, then the resultant intensity $I$ when they interfere with a phase difference of $\phi$ is:

In an interference experiment,the spacing between successive maxima or minima is (where the symbols have their usual meanings):

$A$ beam of light consisting of two wavelengths $6500 \mathring{A}$ and $5200 \mathring{A}$ is used to obtain interference fringes in Young's double-slit experiment. The distance between slits is $2 \text{ mm}$ and the distance of the screen from the slits is $120 \text{ cm}$. What is the least distance from the central maximum where the bright fringes due to both wavelengths coincide (in $\text{ cm}$)?