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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

`(1)/(sqrt2)`

B

`(sqrt3)/(2)`

C

`(1)/(2)`

D

`(3)/(4)`

Text Solution

Verified by Experts

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