Two optical media of refractive indices `n_1` and `n_2` contain x and y waves of the same colour in the same thickness. Then their relative refractive index `n_1//n_2`

Two optical media of refractive indices n_1 and n_2 contain x and y waves of the same colour in the same Ihickness. Then their relative refractive index ^1n_2 is equal to

Obtain the equation for optical path of a medium of thickness d and refractive index n.

When light rays enter from one medium to another, refraction takes place. Consider refraction of light at a spherical surface separating two media of refractive indices n and n_2 (n_2 > n_1) .In the above case let the first medium be air and the second medium glass of refractive index 1.5, R of spherical surface = 20cm. An object is placed in air 100cm from the glass surface. At what position the image is formed?

What will be refractive index n_12 ? Refractive index, n_12 =

The optical fibres have in an inner core of refractive index n1 and a cladding of refractive index n2 such that

A light ray moving in medium-I (of refractive index n_(1) ) is incident on interface of two media and it is totally internally reflected at the interface. Now, refractive index n_(2) of medium-II is decreased, then

A light ray moving in medium-I (of refractive index n_(1) ) is incident on interface of two media and it is totally internally reflected at the interface. Now, refractive index n_(2) of medium-II is decreased, then

A and B are two parallel sided transparent slabs of refractive indices n_1 and n_2 respectively. A ray is incident at an angle theta on the surface of separation of A and B and after refraction from B into air grazes the surface of B. Then sin theta =1/(n_2) sin theta=1/n_1 sin theta =n_2/n_1 sin theta =n_1/n_2