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)) " " impliescos((phi)/(2))=(1)/(sqrt(2))`
`implies (phi)/(2)=(pi)/(4) " " impliesphi=(pi)/(2)(2n+1)`
`implies Deltax=(lambda)/(2pi)phi=(lambda)/(2pi)(2n+1)=(lambda)/(4)(2n+1)`