In YDSE arrangement as shown in figure, fringes are seen on screen using monochromatic source S having wavelength 3000 Å (in air). `S_1 `and `S_2` are two slits seperated by d = 1 mm and D = 1m. Left of slits `S_1` and `S_2` medium of refractive index `n_1 = 2` is present and to the right of `S_1` and `S_2` medium of `n_2 = 3/2`, is present. A thin slab of thickness 't' is placed in front of `S_1` . The refractive index of `n_3` of the slab varies with distance from it's starting face as shown in figure.

In order to get central maxima at the centre of screen, the thickness of slab required is :

Options

A. `1 mum`
B. ` 2 mum`
C. `0.5 mum`
D. `1.5 mum`

Explanation

Path difference
`Deltax=n_(1)S S_(2)+n_(2)S_(2)P-[(n_(1)S S_(1)+n_(2)S_(1)P)-int_(0)^(t)(n_(3)-n_(2)) dx]`
`= n_(1)(S S_(2)-S S_(1))+n_(2)(S_(2)P-S_(1)P)-int_(0)^(t) n_(3) dx+_ n_(2)t`
In order to get central maxima at centre of screen

`O=(2xx(1xx10^(-3))^(2))/(2xx1)+0-2t+(3t)/2`
`0.5 t =1 mu m`
`t= 2 mu m`