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

At path difference `lambda`, we get maximum intensity.
` :. I_(max) =I`
` I_R = I_(max) cos^2(phi/2)`
`:. I/4 = I cos^2 (phi/2)`
or ` cos (phi/2)= +-(1/2) `
` :. phi/2 = 60^@ or 120^@`
`:. phi= 120^@ or 240^@ or (2pi)/3 and (4pi)/3`
From the relative, `Deltax = (lambda/(2pi)).phi`
We see that,
`Deltax = lambda/3 and (2lambda)/3`