In the Young's double slit experiment using a monochromatic light of wavelength `lamda`, the path difference (in terms of an integer n) corresponding to any point having half the peak

Options

A. `(2n + 1) (lambda)/(2)`
B. `(2n + 1) (lambda)/(4)`
C. `(2n + 1) (lambda)/(8)`
D. `(2n + 1) (lambda)/(16)`

Answer

b

Explanation

`(I_(max))/(2) = I_(m) cos^(2) ((phi)/(2))`
`implies cos ((phi)/(2)) = (1)/(sqrt2)`
`implies (phi)/(2) = (pi)/(4)`
`implies phi = (pi)/(2) (2n + 1)`
`implies Delta x =(lambda)/(2 pi) phi = (lambda)/(2 pi) xx (pi)/(2) (2n + 1) = (lambda)/(4) (2n + 1)`