In Young's double slit experiment, the intensity of light at a point on the screen where path difference is `lambda` is I. If intensity at another point is I/4, then possible path differences at this point are

Options

A. `lambda//2, lambda//3`
B. `lambda//3, 2lambda//3`
C. `lambda//3,lambda//4`
D. `2lambda//3,lambda//4`

Answer

B

Explanation

For path difference `=Delta`
phase difference `phi = (Delta)/(lambda) xx 2pi`
`:.` for `Delta lambda = phi = 2pi`
`:. I = 4I_(0)cos^(2).(phi)/(2) = 4I_(0)`
for `I' = I//4`
`:. 4I_(0)cos^(2).(phi)/(2) = (4I_(0))/(4)`
`cos.(phi)/(2) = +-(1)/(2)`
`(phi)/(2) = 60^(@), 120`
`phi = 120^(@), 240^(@) = (2pi)/(3), 4pi//3`
`:. Delta =(phi)/(2pi) xx lambda = (2pi)/(3 xx 2pi) xx lambda = lambda//3`
and `Delta = (4pi)/(3 xx 2pi) xx lambda = 2lambda//3`