The maximum intensity in Young's double slit experiment is `I_(0)` . What will be the intensity of light in front of one the slits on a screen where path difference is `(lambda)/(4)` ?

Options

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

Answer

A

Explanation

Corresponding phase difference will be
`phi = ((2pi)/(lambda)) (Delta x) = ((2pi)/(lambda))((lambda)/(4)) = (pi)/(2)`
or `(phi)/(2) = (pi)/(4) l :. I = I_(0) cos^(2)((phi)/(2))`
`= I_(0) cos^(2)((pi)/(4)) = (I_(0))/(2)`