In a Young's double slit experiment , the intensity of light at a point on the screen where the path difference is `lambda` is k units. Find the intensity at a point where the path difference is (a) `(lambda)/(4)` (b) `(lambda)/(3)` and (c) `(lambda)/(2)`

Options

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

Answer

C

Explanation

Intensity at any point on the screen is
`I=4I_(0)cos^(2).(phi)/(2)`
where `I_(0)` is the intensity of either wave and `phi` is the phase difference between two waves.
Phase difference, `phi=(2pi)/(lambda)xx` Path difference
When path difference is `lambda`, then
`phi=(2pi)/(lambda)xxlambda=2pi`
`:. I=4I_(0)cos^(2)((2pi)/(2))=4I_(0)cos^(2)(pi)=4I_(0)=K` ...(i)
When path difference is `(lambda)/(4)` , then
`phi=(2pi)/(lambda)xx(lambda)/(4)=(pi)/(2)`
`:.I=4I_(0)cos^(2)((pi)/(4))=2I_(0)=(K)/(2)` [Using (i)]