In a Young's double slit experiment with light of wavelength `lamda` the separation of slits is d and distance of screen is D such that `D gt gt d gt gt lamda`. If the fringe width is `bea`, the distance from point of maximum intensity to the point where intensity falls to half of maximum intensity on either side is

Options

A. `(beta)/(3)`
B. `(beta)/(6)`
C. `(beta)/(2)`
D. `(beta)/(4)`

Answer

D

Explanation

`I_(max)=cos^(2)((phi)/(2))` when intensity is `(I_(max))/(2)`
Phase deff. `phi=(2pi)/(lamda)x=(2pi)/(lamda).(yd)/(D)impliesphi=(2piy)/(beta)`
Now `(I_(max))/(2)=I_(max)cos^(2)((2piy)/(2beta))`
`impliescos^(2)((piy)/(beta))=(1)/(2)implies(piy)/(beta)=(pi)/(4)impliesy=(beta)/(4)`