A double - slit apparatus is immersed in a liquid of refractive index `1.33` it has slit separation of 1mm and distance between the plane of slits and screen `1.33` m the slits are illuminated by a parallel beam of light whose wavelength in air is 800 nm.
(i) Calculate the fringe width.
(ii) One of the slits of apparatus is covered by a thin glass sheet of refractive index `1.53` Find the smallest thickness of the sheet to bring the adjacent minima on the axis.

Explanation

(a) As fringe width `beta = (D lamda // d)` and by presence of medium the wavelength becomes `(lamda//mu)` , so the fringe-width in liquid will be
`beta. = (D lamda)/(mu d) = (1.33 xx 6300 xx 10^(-10))/(1.33 xx 1 xx 10^(-3))`
` =0.63 mm`
(b) Now as the distance of a minima from adjacent maxima is `beta. //2 ` so according to given problem, shift
`y_0 = D/d (mu_r - 1)t = (beta.)/(2)` with `mu_r = (mu_P)/(mu_M)`
`D/d [ (mu_P)/(mu_M) - 1] t = (D lamda)/(2mu_M d) i.e., t = (lamda)/(2(mu_P - mu_M))`
so `t = (6300 xx 10^(-10))/(2(1.53 - 1.33)) = 1.575 mu m`