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

Options

A. `(K)/(4)`
B. `(K)/(2)`
C. K
D. zero

Answer

B

Explanation

For path diff. `lambda` phase diff. `=2pi` and dintesnity of light `=I_(b)=4I=K " "....(i)`
For path diff` (lambda)/(4)`, phase diff. `(2pi)/(lambda)=(lambda)/(4)=(pi)/(2)` and intensity of light is,
`I_(d)=I_(1)+I_(2)+sqrt(I_(1)I_(1)) 2 cos 90 2I " "....(ii)`
`:.I_(1)I_(2)=I`
Now, `(I_(b))/(I_(d))=(4I)/(4I) :. I_(d)=(2I)/(4I)I_(b)=(I_(0))/(2)=(K)/(2)`