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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

Answer» Correct Answer - 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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