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_(0))/2[1-8"cos"^(2)(pi)/(lamda) d sin alpha]`
B. `(I_(0))/6[1-4"cos"^(2)(2pi)/(lamda)2d sin alpha]`
C. `(I_(0))/9[1+8"cos"^(2)(pi)/(lamda)dsin alpha]`
D. `(I_(0))/9[1+4"cos"(pi)/(lamda)d sin alpha]`

Answer

C

Explanation

NA