In YDSE, when a glass plate of refractive index 1.5 and thickness t is placed in the path of one of the intefering beams (wavelength`lambda`), intensity at the position where central maximum occurred previously remains unchanged. The minimum thickness of the glass plate is

Options

A. `lambda//2`
B. `lambda`
C. `2lambda`
D. `4lambda`

Answer

C

Explanation

Optical path difference introduced at the centre due to thin film can be written as follows:
`Delta x = (mu -1)t`
We know that there is maximum at the centre of screen in standard set-up, hence for same intensity there should be maximum again at the centre. For maximum intensity, path difference must be integral multiple of wavelength.
`(mu - 1)t = n lambda`.
For minimum thickness `n = 1`
`t_("min") = lambda//(mu - 1) = lambda//(1.5 - 1) = 2 lambda`.