In Young's double slit experiment, one of the slit is wider than other, so that amplitude of the light from one slit is double of that from other slit. If `I_m` be the maximum intensity, the resultant intensity I when they interfere at phase difference `phi` is given by:

Options

A. `(I_(m))/(9) (4 + 5 cos phi)`
B. `(I_(m))/(3) (1 + 2 cos^(2) (phi)/(2))`
C. `(I_(m))/(5) (1 + 4 cos^(2) (phi)/(2))`
D. `(I_(m))/(9) (1 + 8 cos^(2) (phi)/(2))`

Answer

D

Explanation

Let `A_(1) = a_(0) :. A_(2) = a_(0)`
At phase difference `phi`, the resultant intensity would be `I = A_(1)^(2) + A_(2)^(2) + 2 A_(1) A_(2) cos phi`, and
`I_(max) = I_(m) = A_(1)^(2) + A_(2)^(2) + 2 A_(1) A_(2)` `(because cos phi = 1)`
`:. (I)/(I_(m)) = (A_(1)^(2) + A_(2)^(2) + 2 A_(1)A_(2) cos phi)/(A_(1)^(2) + A_(2)^(2) + 2 A_(1) A_(2))`
`= (a_(0)^(2) + (2 a_(0))^(2) + 2(a_(0)) (2 a_(0)) cos phi)/(a_(0)^(2) + (2 a_(0))^(2) + 2(a_(0)) (2 a_(0))`
`= (5 a_(0)^(2) + 4 a_(0)^(2) cos phi)/(9 a_(0)^(2))`
`(I)/(I_(m)) = (5 + 4(2 cos^(2) phi//2 - 1))/(9)`
`I = (I_(m))/(9)(1 + 8 cos^(2) phi//2)`
Here, `d = 2 mm = 2 xx 10^(-3)m`,
`lambda_(1) = 12000 Å, lambda_(2) = 1000 Å`
`:. n_(1) lambda_(1) = n_(2) lambda_(2) or (n_(1))/(n_(2)) = (lambda_(2))/(lambda_(1)) = (10000)/(12000) = (5)/(6)`
For minimum distance `x, n_(1) = 5` and `n_(2) = 6`
`:. x_(min) = n_(1)(lambda_(1)D)/(d) = (5 xx (12000 xx 10^(-10)) xx 2)/(2 xx 10^(-3))`
`= 6 xx 10^(-3)m = 6 mm`