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

Answer

B

Explanation

The intensity I is given as
`I=I_(0)` `cos^(2)(phi)/(2)` Where `I_(0)` is the peak intensity
Here `I=(I_(0))/(2)`, `:'` `(I_(0))/(2)=I_(0)` `cos^(2)(phi)/(2)` `:'` `phi=(pi)/(2)(2n+1)`
For a phase difference of `2pi` the path difference is `lamda`
`:'` For a phase difference of `(2n+1)` `(pi)/(2) the path difference is `(2n+1)(lamda)/(4)`. option (b) is correct