A screen is at distance `D = 80` cm form a diagram 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)`

Options

A. `3/4`
B. `2/4`
C. `3/3`
D. `3/2`

Answer

A

Explanation

Optical path = (Refractive Index) x (Geometrical Path Length)
Path difference at point C on the screen
`Deltax = mu_(m)t_(1) + mu_(w) (S_(1)C - t_(1)) - [mu_(m)t_(2) + mu_(w) (S_(2)C-t_(2))]`
`Deltax = mu_(m) (t_(1)-t_(2)) - mu_(w)t_(1) + mu_(w)t_(2) (therefore s_(1)c = s_(2) c)`
`Deltax = (mu_(m) - mu_(w))(t_(1) - t_(2))`
`Deltax = (1.4 - 4/3) (2.5 xx 10^(-6) -1.25 xx 10^(-6))`
`Deltax = 0.2/3 xx 1.25 xx 10^(-6)`
`Deltax = 2500/3 A = 5000/6 A`
`Deltax = lambda/6`
Phase difference `phi = pi/3 = 60^(@)`
At any point on the screen `I_("net") = I_(1) + I_(2) + 2sqrt(I_(1)I_(2)) cos phi`
Resultant intensity `I_( c) = I + I + 2sqrt(I xx I) cos 60`
`I_( C) = 3I`
and `I_("max") = I + I + 2I` when `phi = 0^(@) rArr cos phi = +1`
`I_("max") = 4I`
`I_( C)/I_("max") = 3/4`