A double slit apparatus is immersed in a liquid of refractive index 1.33. It has slit and the screen 1 mm. The slits are illuminated by a parallel beam of light whose wavelength in air is `6300 Å`
a. calculate the fringe width.
b. One of the slits of the apparatus is covered by a thin glass sheet of refractive index 1.53. Find the smallest thickness of the sheet to bring athe 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`