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

Answer

a.,c.,d

Explanation

`Delta x` (at P) after insertion of slab
`(S_(1) P - t)_("air") + t_("medium") - S_(2) P_(air) = (mu -1)t`
Earlier, `Delta x` (at P) `= S_(1) P - S_(2) P = 0`
So, change in optical path due to insertion of slab is `(mu - 1) t`.
For intensity to be zero at P, we have
`Delta x = ((2n - 1)lambda)/(2) (n = 1, 2,...)`
`(mu - 1) t = (lambda)/(2) , (3 lambda)/(2) , (5 lambda)/(2) ,...`