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

Answer

b

Explanation

`I = I_(max) cos ((2phi)/(2))`
`phi = (2 pi)/(lambda) Delta x`
`I = I_(max) cos^(2) ((pi)/(lambda)) Delta x`
`(I_(1))/(I_(2)) = (cos^(2) ((pi)/(lambda)) Delta x_(1))/(cos^(2) ((pi)/(lambda)) Delta x_(2))`
`(I)/(I // 4) = (cos^(2) pi)/(cos^(2) ((pi)/(lambda)) Delta x_(2))`
`implies cos^(2) ((pi)/(lambda)) Delta x_(2) = (2)/(4)`
`implies cos ((pi)/(lambda)) Delta x = +- (1)/(2)`
`(pi)/(lambda) Delta x = (pi)/(3) , (2 pi)/(3)`
`Delta x = (lambda)/(3) , (2 lambda)/(3)`