Consider an YDSE that has different slit width. As a result, amplitude of waves from two slits are A and 2A, respectively. If `I_(0)` be the maximum intensity of the interference pattern, then intensity of the pattern at a point where phase difference between waves is `phi` is

Options

A. `(I_(0))/(9) cos^(2) phi`
B. `(I_(0))/(3) sin^(2).(phi)/(2)`
C. `(I_(0))/(9) [5 + 4 cos phi]`
D. `(I_(0))/(9) [ 5 + 8 cos phi]`

Answer

c

Explanation

As amplitudes are A and 2A, So intensities would be in the ratio 1:4, let us say I and 4 I.
`I_(max) = I_(0) = I + I_(0) + 2 sqrt(4 I^(2)) = 9 I`
`implies I =(I_(0))/(9)`
Intensity at any point,
`I' = I + 4I + 2 sqrt(4 I^(2)) cos phi`
`implies I' = 5 I + 4 I cos phi = (I_(0))/(9) (5 + 4 cos phi)`