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A parallel coherent beam of light falls ...

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

A

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

B

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

C

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

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
A

`beta=((a+b)lamda)/(2a(mu-1)alpha)=((1+(b)/(a))lamda)/(2(mu-1)alpha)`
For parallel beam, `a=infty,` so `beta=(lamda)/(2(mu-1)alpha)`
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