A transparent slab of thickness t and refractive index `mu` is inserted in front of upper of YDSE apparauts. The wavelength of ligth used. is `lambda`. Assume that there is no absorption of light by the slab. Mark the correct statement(s).

Options

A. The intensity of dark fringes will be zero, if slits are identical.
B. The change in optical path due to insertion o9f palte is `mu t` .
C. The change in optical path due to insertion of plate is `(mu - 1)t` .
D. For making intensity zero at centre of screen the thickness can be `(5lambda)/(2(mu-1))`.

Answer

A, C, D

Explanation

Solution: Path difference at 'P'
`Delta x = (S_(1)P - t)_(air) + t_(med) - S_(1)P_(air)`
`= S_(1)P-S_(2)P-t+mu t Delta x = (S_(1)P-S_(2)P)+(mu-1)t`

When there is no absorption of light by slab then path difference at P earlier is `S_(1)P-S_(2)P=0` . so change in optical path due to inserion of slab is `(mu-1)t` .
For Intensity to be zero at 'P' we have `Delta x = (2n-1)(lambda)/(2) = (mu - 1)t implies t = ((2n-1)lambda)/(2(mu-1))`
Then `t= (lambda)/(2(mu-1)) , (3lambda)/(2(mu-1)) , (5lambda)/(2(mu-1))` , - - - - -