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

As fringe width is given by `beta = D lambda//d` and by presence of medium the wavelength becomes `lambda//mu`, so the fringe width in liquid will be
`beta' = (D lambda)/(mu_(m) d) = (1.33 xx 6300 xx 10^(-10))/(1.33 xx 1 xx 10^(-3)) = 0.63 mm`
Now, as the distance of a minima from adjacent maxima is `beta'//2,` so according to given problem, shift
`y_(0) = (D)/(d) (mu - 1)t = (beta')/(2)`
`implies (D(mu - 1)t)/(d) = (1)/(2) ((D lambda)/(mu_(M) d))`
`implies t = (lambda)/(2 mu_(M) (mu - 1)) = (6300)/(2 xx 1.33 xx (1.53 - 1)`
`4468.7 Å = 446.87 mu m`