Question

In YDSE, the slits have different widths. As a result, amplitude of waves from slits are A and 2A respectively. If `I_(0)` be the maximum intensity of the intensity of the interference pattern then the intensity of the pattern at a point where the phase difference between waves is `phi` is given by `(I_(0))/(P)(5+4cos phi)` . Where P is in in integer. Find the value of P ?

Answer 1

Correct Answer - 9
As the amplitude are A and 2A then the ratio of intensities is `1:4`
`I_(max) = I_(0) = I_(1)+I_(2) +2sqrt(I_(1)+I_(2)) = I+4I+2xx2I`
`I_(0) = 9I implies I = (I_(0))/(9)`
Intensity at any point :
`I^(1) = I_(1)+I_(2)+2sqrt(I_(1)+I_(2)) cos phi`
`I^(1) = I + 4I + 2sqrt(Ixx4I) cos phi`
`I^(1) = 5I + 4I cos phi , I^(1) = I [5+4cos phi]`
`I^(1) = (I_(0))/(9) [5+4 cos phi] = (I_(0))/(P) [5+4cos phi]`
`P = 9`