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/6`
B. `beta/3`
C. `beta/4`
D. `beta/2`

Answer

C

Explanation

`2I_(0)=4I_(0)cos^(2)((Deltaphi)/2)" here, "Deltaphi=pi/2`
But, `Deltaphi=(2pi)/lamdaDeltax" so, "Deltax=lamda/4`
`(dy)/D=lamda/4" "...(i)`
`(lamdaD)/d=beta" "...(ii)`
Multiplying equation (i) and (ii) we get,
`y=beta/4`