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

At point P (central maxima), originally it has maximum intensity. If the intensity coming out of each slit is `I_(0),` the intensity at point P should be `4 I_(0).`
`I =2 I_(0) (1 + cos phi) implies (3)/(4) (4 I_(0)) = 2 I_(0) (1 + cos phi)`
`cos phi = (1)/(2) implies phi = 2n pi +- (pi)/(3)` (n is integer)
Path difference at P,
`4 lambda le Delta x le ((2 xx 5 - 1)/(2)) lambda`
`implies 4 lambda le Delta x le (9)/(2) lambda implies 4 lambda le Delta x le 4.5 lambda`
From Eq. (i),
`theta = (2 pi)/(lambda) Delta x implies Delta x = n lambda +- (lambda)/(6)`
`Delta x = lambda (4 +- (1)/(6))`
From Eqs. (ii) and (iii), `n = 4.` Hence,
`Delta x lambda (4 + (1)/(6)) = (25)/(6) lambda implies Delta x = t(mu - 1)`
`t = (25 lambda)/(6(mu_(2) - mu_(1))) = (25 xx 5400 xx 10)/(6 xx 0.3 xx 10^(-1)) = 7500 xx 10^(-9) m`
`= 7.5 mu m`