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In the Young's double-slit experiment, t...

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

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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