A screen is at distance `D = 80` cm form a diaphragm having two narrow slits `S_(1)` and `S_(2)` which are `d = 2` mm apart.
Slit `S_(1)` is covered by a transparent sheet of thickness
`t_(1) = 2.5 mu m` slit `S_(2)` is covered by another sheet of thickness
`t_(2) = 1.25 mu m` as shown if Fig. 2.52.
Both sheets are made of same material having refractive index `mu = 1.40`
Water is filled in the space between diaphragm and screen. A monochromatic light beam of wavelength `lambda = 5000 Å` is incident normally on the diaphragm.
Assuming intensity of beam to be uniform, calculate ratio of intensity of C to maximum intensity of interference pattern obtained on the screen `(mu_(w) = 4//3)`

Explanation

Path difference at C.
`Delta x = t_(1) (mu - 1) - t_(2) (mu - 1)`
`mu (t_(1) - t_(2)) - (t_(1) - t_(2))`
`(t_(1 ) - t_(2) (t_(1) - (t_(2))`
`= (2.5 - 1.25) ((1.4 xx 3)/(4 xx 10 )1) = 1.25 xx (2)/(40) = (2.5)/(400)`
`implies Delta x = (1)/(16) mu m`
`phi = (2 pi)/(lambda) Delta x = (2 pi xx 4)/(5000 xx 3 xx 10^(-10)) xx (1)/(16) xx 10^(-6) = (pi)/(3)`
`I_(max) = 4 I_(0)`
Intensity at C, `I_(c) = 2 I_(0) (1 + cos (pi)/(3)) = 3 I_(0)`
`:.` Required raito `= (I_(C))/(I_(max)) = (3)/(4)`