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

Options

A. `0.866`
B. `0.5`
C. `0.707`
D. `0.75`

Answer

D

Explanation

Phase difference, `phi=(2pi)/(lamda)xx(lamda)/(6)=(pi)/(3)=60^(@)`
The intensity at a point is
`I=I_(max)"cos"^(2)(phi)/(2),I=I_(0)"cos"^(2)(phi/2)`
`I/I_(0)="cos"^(2)(phi)/(2),I=I_(0)"cos"^(2)((phi)/(2))`
`I/I_(0)=cos^(2)((phi)/(2))=cos^(2)((60)/(2))=cos^(2)30^(@)`
`=((sqrt(3))/(2))^(2)=(3)/(4)=0.75`