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The maximum intensity in Young's double ...

The maximum intensity in Young's double slit experiment is `I_(0)` . What will be the intensity of light in front of one the slits on a screen where path difference is `(lambda)/(4)` ?

A

`(I_(0))/(2)`

B

`(3)/(4)I_(0)`

C

`I_(0)`

D

`(I_(0))/(4)`

Text Solution

Verified by Experts

The correct Answer is:
A
corresponding phase difference will be
`phi = ((2 pi)/(lambda)) (Delta x) = ((2 pi)/(lambda)) ((lambda)/(4)) = (pi)/(2)`
or `(phi)(2) = (pi)/(2) , :. I = I_(0) cos^(2) ((phi)/(2))`
`= I_(0) cos^(2) ((pi)/(2)) = (I_(0))/(2)`
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