Question

In a young's double slit experiment ,let Aand B be the two slits.A thin film film of thickness t and refractive indexμ is pklaced in front of A.Let β =fringe width.The central maximum will shift:

A. towards A
B. towards B
C. by t(μ−1)β/λ
D. by μt β/λ

Answer 1

Correct option: C. by t(μ−1)β/λ .

The central maximum in Young's double slit experiment will shift by t(μ - 1)β/λ when a thin film with refractive index μ and thickness t is placed in front of one of the slits.

Step-by-step explanation:

In the context of Young's double slit experiment, when a thin film with a refractive index μ and thickness t is placed in front of one of the slits (say slit A), it causes an additional optical path length for the light passing through it.

This optical path length is given by μt, but since the film is only μ times more optically dense than the air (or vacuum), the effective change in the path length that causes a shift in the interference pattern is t(μ - 1).

Now, considering the fringe width β and the wavelength of light λ, we can relate the shift in the central maximum to these quantities.

The shift Δy of the central maximum can be calculated by dividing the extra path length by the wavelength and then multiplying by the fringe width, resulting in the formula Δy = t(μ - 1)β/λ.

This shows how the position of the central maximum in an interference pattern is affected when one of the slits is covered by a film.