In a Young's double slit experiment, the intensity at a point where the path difference is `(lambda)/(6) (lambda` = wavelength of the light) is I. If `I_(0)` denotes the maximum intensity, then `(I)/(I_(0))` is equal to

Options

A. `(1)/(2)`
B. `(sqrt(3))/(2)`
C. `(1)/(sqrt(2))`
D. `(3)/(4)`

Answer

D

Explanation

`phi = (lambda)/(6) = (360^(@))/(6) = 60^(@)`
`I = I_(0)cos^(2)theta = I_(0) cos^(2) 60^(@) = (3)/(4) xx I_(0)`
`(I)/(I_(0)) = (3)/(4)`