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A parallel coherent beam of fight falls ...

A parallel coherent beam of fight falls on fresnel biprism of bipism of refractive index `mu` and angle `alpha` .The fringe width on a screen at a distance `D` from biprism will be (wavelength =`lambda`)

A

`(lambda)/(2(mu-1)alpha)`

B

`(lambdaD)/(2(mu-1)alpha)`

C

`(D)/(2(mu-1)alpha)`

D

none

Text Solution

Verified by Experts

The correct Answer is:
A
`beta=(lambdaD)/(d)`
`=(lambda(a+b))/(2a(mu-1)A)`
`=(lambda)/(2(mu-1)A)(1+(b)/(a))[a=oo]`
`=(lambda)/(2(mu-1)A)`
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