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In the Young's double slit experiment us...

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

A

`(2n + 1)(lambda)/(2)`

B

`(2n + 1)(lambda)/(4)`

C

`(2n + 1)(lambda)/(8)`

D

`(2n + 1)(lambda)/(16)`

Text Solution

Verified by Experts

The correct Answer is:
B
For half of maximum intensity
`2I_(0)=I_(0)+I_(0)+2I_(0) costheta`
`theta` (Phase difference) `=(pi)/(2),(3pi)/(2),(5pi)/(2)……….`
Path difference is `(lambda)/(4),(3lambda)/(4),(5lambda)/(4)……….((2n+1)/(4)lambda)`
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