In Young's double slit experiment, the maximum intensity is `I_(0)`. What is the intensity at a point on the screen where the path difference between the interfering waves is `lamda/4` ?

Options

A. `I_(0)`
B. `I_(0)/3`
C. `I_(0)/4`
D. `I_(0)/2`

Answer

D

Explanation

Phase diff`=(2pi)/(lamda)` (path diff.)
`therefore delta=(2pi)/(lamda)xx(lamda)/(4)=(pi)/(2)" "therefore (delta)/(2)=(pi)/4`
`therefore I=I_(0)cos^(2)((delta)/(2)) =I_(0)cos^(2)(45^(@))=I_(0)xx(1)/(2)=(I_(0))/(2)`