In Young's double slit experiment intensity at a point is `((1)/(4))` of the maximum intensity. Angular position of this point is

Options

A. `sin^(-1) lambda//d`
B. `sin^(-1) lambda//2d`
C. `sin^(-1) lambda//3d`
D. `sin^(-1) lambda//4d`

Answer

C

Explanation

If `delta` is phase difference between waves at this point, then resultant intensity can be written as follows:
`I = 4I_0 cos^2 (delta)/2`
Maximum intensity is `4 I_0` and hence at this point intensity is `1//4` or we can say `I_0`
`implies I_0 = 4I_0 cos^2 (delta)/2`
`implies cos (delta)/2 = 1/2`
`implies delta = 2pi//3`.
Let there be a point at a distance y from the centre, then phase difference at this point can be written as follows:
`delta = (2pi)/(lambda) xx (yd)/D`
`implies (2pi)/3 = (2pi)/lambda xx (yd)/(D)`
`implies y/D = lambda/(3d) " "....(1)`
Let angular separation of this point is `theta`, then
`sin theta = tan theta = y/D`
Using equation (1) we can write the following:
`sin theta = lambda/(3d)`
`implies theta = sin^(-1) (lambda/(3d))`