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

In a Young's double slit experiment the intensity at a point where tha path difference is `(lamda)/(6)` (`lamda` being the wavelength of light used) is I. If `I_0` denotes the maximum intensity, `(I)/(I_0)` is equal to

A

`(3)/(4)`

B

`(1)/(sqrt(2))`

C

`(1)/(sqrt(3))`

D

`(1)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
1
Phase difference =`(2pi)/(lambda)xx"path difference"`
`i.e,phi=(2pi)/(lambda)xx(lambda)/(6)=(pi)/(3)`
As,`I=I_(max) cos^(2)((phi)/(3))`
or`(I)/(I_(max))=cos^(2)((phi)/(2))`
`(I)/(I_(0))=cos^(2)((pi)/(6)) =(3)/(4)`
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