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

Options

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

Answer

B

Explanation

Phase difference `(phi)=(2pi)/(lamda)` (path difference)
`=(2pi)/(lamda)xx(lamda)/(6)=(pi)/(3)` radians
In Young's double slit experiment, the intensity at any point
`I=I_(0)"cos"^(2)((phi)/(2))`
`therefore I/I_(0)=cos^(2)((pi//3)/(2))=cos^(2)((pi)/(6))`
`=((sqrt(3))/(2))^(2)=(3)/(4)`