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

In Young's double slit experiment the intensity of light on the screen where the where the path difference is `lambda` is k ( `lambda` being the wavelength of light used). The intensity at a point where the path difference is `(lambda)/(4)` will be

A

k

B

`(k)/(4)`

C

`(k)/(2)`

D

zero

Text Solution

Verified by Experts

We know I = `I_(0) "cos"^(2) (pi delta)/(lambda)`
When path difference is `lambda` then I=k= `I_(0) "cos"^(2) pi =I_(0)`
When path difference is `(lambda)/(4)` then I = `I_(0) "cos" ^(2) (pi)/(4)= I_(0)/(2)=(k)/(2)`
The option C is correct.
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