In Young's double slit experiment, the intensity on the screen at a pont where path difference `lamda` is K. what will be the intensity at the point where the path difference is `lamda/4`?

Options

A. `K/4`
B. `K/2`
C. `K`
D. Zero

Answer

B

Explanation

Phase difference `(phi)=(2pi)/(lamda)` (path difference)
`therefore phi_(1)=(2pi)/(lamda)(lamda)=2piand (phi_(1))/(2)=pi`
and `phi_(2)=(2pi)/(lamda)((lamda)/(4))=(pi)/(2) and phi_(3)=(pi)/(4)`
`therefore` The intensity at any point is `I=4I_(0)"cos"^(2)((phi)/(2))`
`therefore I_(1)/I_(2)=(4I_(0)cos^(2)(pi))/(4I_(0)"cos"^(2)((pi)/(4)))=(1)/((1)/(2))=2`
But `I_(1)=K" "therefore I_(2)=K/2`