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

Answer

D

Explanation

`I = I_(1)+I_(2)+2sqrt(I_(1)I_(2))cos phi I_(1) = 4I_(2) , I_(max) = 9I_(2)`