In a Young's double slit experiment the intensity at a point where tha path difference is `(lamda)/(6)` (`lamda` being the wavelength of light used) is I. If `I_0` denotes the maximum intensity, `(I)/(I_0)` is equal to

Options

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

Answer

A

Explanation

In Young.s double slit experiment intensity at a point is given by
`I=I_(0)cos^(2)((phi)/(2))`
or `(I)/(I_(0))=cos^(2)((phi)/(2))` ...(i)
Phase difference, `phi=(2pi)/(lambda)xx "path difference" `
`:.phi=(2pi)/(lambda)xx(lambda)/(6)` or `phi=(pi)/(3)` ...(ii) Substitute eqn. (ii) in eqn. (i), we get
`(I)/(I_(0))=cos^(2)((pi)/(6))` or `(I)/(I_(0))=(3)/(4)`