In Young's experiment the upper slit is covered by a thin glass plate of refractive index `1.4` while the lower slit is covered by another glass plate, having the same thickness as the first one but having refractive index `1.7` interference pattern is observed using light of wavelength `5400 Å`
It is found that point P on the screen where the central maximum `(n = 0)` fell before the glass plates were inserted now has `3//4` the original intensity. It is further observed that what used to be the fourth maximum earlier, lies below point P while the fifth minimum lies above P.
Calculate the thickness of glass plate. (Absorption of light by glass plate may be neglected.
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Explanation

For sources of equal intensities, we can write resultant intensity as follows :
`I = 4I_0 cos^2 (delta)/2`
without films, intensity at the centre is `4I_0`, but after the introduction of films, resultant intensity reduces to `3//4` of the value , hence we can write the following equation:
`3/4 (4I_0) = 4I_0 cos^2 (delta)/2`
`implies cos (delta)/(2) = (sqrt3)/(2)`
`implies delta/2 = pi/6`
`implies delta = pi/3`
We know that refractive index of lower slit is more and hence fringe pattern will be downwards. It is given that fifth maximum earlier has crossed centre and is there below the centre. And hence total phase difference introdcued at the centre due to introduction of film can be written as follow :
`phi = 10pi + pi//3`
`implies phi = 31pi//3 " " .....(i)`
Path difference introduced at the centre due to films can be written as follows:
`Deltax = (mu_2 - mu_1) t`
Corresponding phase difference can be written as follows :
`phi = (2pi)/lambda xx (mu_2 - mu_1) t`
We can substitute from equation (i) to write the following:
`phi = (2pi)/lambda xx (mu_2 - mu_1)t = (31pi)/3`
`implies t = 31/6 xx (lambda)/(mu_2 - mu_1)`
`implies t = 31/6 xx (5400 xx 10^(-10))/(1.7 - 1.4)`
`implies t = 31/6 xx (5400 xx 10^(-10))/(0.3)`
`implies t = 9.3 xx 10^(-0)m`.