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.

Fringe width on the screen is

Options

A. `1mu m`
B. `2mu m`
C. `0.5mu m`
D. `1.5mu m`

Answer

A

Explanation

Area `(A) = mu_(3) (dx) = ((1+3)t)/(2) = 2t`
Path difference
`Delta x = mu_(1)SS_(2)+mu_(2)S_(2)P-[mu_(1)SS_(1)+mu_(2)S_(1)P-overset(t)underset(0)int (mu_(3)-mu_(2))dx]`
`= mu_(1)[SS_(2)-SS_(1)]+SS_(1)+mu_(2)[S_(2)P-S_(1)P] - overset(t)underset(0)int mu_(3)dx + overset(t)underset(0)int mu_(2)dx`
In order to get central maxima at the centre 'O' of screen is `Delta x = 0`
`O = (2xx(1xx10^(-3))^(2))/(2xx1) + 0 - 2t+(3)/(2)t`
`1xx10^(-6)= 2t - (3t)/(2) = (t)/(2) t = 2xx10^(-6)m implies t = 2mu m`