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

When path difference is `lambda and I_("max")=4I=K`
When path difference is `(lambda)/(4)`, phase difference, `phi=pi//2`
`therefore I_(R )=I_(1)+I_(2)+2sqrt(I_(1)I_(2))cosphi=I+I=2I=(K)/(2)`