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 thikness
`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. Amondichromatic 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)`

Options

A. `(3)/(4)`
B. `(2)/(3)`
C. `(8)/(9)`
D. `(5)/(7)`

Answer

A

Explanation

Path difference at 'C"
`Delta x = t_(1) (.omega_mu_(g)-1)-t_(2) (.omega_mu_(g)-1)= .omega_mu_(g)(t_(1)-t_(2))-(t_(1) - t_(2))`
`Delta x = (t_(1) - t_(2)) .omega_mu_(g) -1)`
`Delta x = (t_(1) - t_(2)) ((mu_(g))/(mu_(omega))-1) = (2.5-1.25) [(1.4)/(4)xx3-1]`
`= 1.25xx[(42)/(40) - 1] = 1.25xx (2)/(40) = (2.5)/(40)`
`Delta x = (25)/(400) = (1)/(16)mu m`.
Phase difference `(phi) = (2pi)/(lambda_(omega))xxDelta x`
`phi = (2pi)/(mu_(a)lambda_(a))xxmu_(omega)xxDelta x = (2pixx4)/(1xx5000xx10^(-10)xx3)xx(1)/(16)xx10^(-6)`
`phi = (pixx10^(-6))/(30xx10^(-7))=(pixx10^(-6))/(3xx10^(-6))=(pi)/(3)`
intensity at 'C'
`I_(C ) = 4I_(0) cos^(2) (phi//2) = 4I_(0)cos^(2) (pi//6) = 4I_(0)xx(3)/(4)`
`I_(C ) = 3I_(0)`
maximum intensity `I_(max) = 4I_(0)`
`(I_(C ))/(I_(max)) = (3I_(0))/(4I_(0)) implies (I_(C ))/(I_(max)) = (3)/(4)`