The wavelength of the light used in Young's double slit experiment is `lambda`. The intensity at a point on the screen is `I`, where the path difference is `(lambda)/(6)`. If `I_(0)` denotes the maximum intensity, then the ratio of `I` and `I_(0)` is

Options

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

Answer

D

Explanation

Given, path difference
`therefore phi =(lambda)/(3)=(360^(@))/(3)=120`
Intensity `I=I_(0)cos^(2)phi//2=I_(0) cos^(2) 60^(@)=(I_(0))/(4), (I)/(I_(0))=(1)/(4)`