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

Given `mu_(1) = 1.33, d = 1 mm = 10^(-3) mm, D = 1.33 m, lambda = 6300 Å`
`= 6.3 xx 10^(-7) m`
When the experiment is performed in liquid, `lambda` change to
` lambda ' = (lambda)/(mu_(1))`
a. Fringe width,
`beta = (D lambda ')/mu_(l)`
`(1.33 xx 6.3 xx 10^(7))/(1.33 xx 10^(-3)) = 6.3 xx 10^(-4) m = 0.63 mm`
Displacement of fringes,
`Delta y = (beta)/(lambda ') (l mu_(g) - 1) t = (beta)/(lambda// mu_(l)) ((mu_(g) )/(mu_(l)) -1)t`
`= (beta)/(lambda) (mu_(g) - mu_(l))t = (Delta y lambda)/(beta.(mu_(g) - mu_(l)))`
`But `Delta y =` separation between bright and adjacent dark fringe `=(beta)/(2)`
`:. t = (((beta)/(2)) lambda)/(beta (mu_(g) - mu_(l))) = (lambda)/(2 (mu_(g) - mu_(l))`
Given `mu_(g) = 1.53, mu_(1) = 1.33`.
`:.t = (6.3 xx 10^(-7))/(2(1.53 - 1.33))`
`=(6.3 xx 10^(-7))/(2 xx 0.20) = 1.575 xx 10^(-6) m = 1.575 mu m`.