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(B) The fringe width $\beta$ in an interference pattern is given by $\beta = \frac{\lambda D}{d}$,where $d$ is the distance between the two virtual co

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$\frac{\lambda D}{2(\mu - 1)\alpha}$

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(B) The fringe width $\beta$ in an interference pattern is given by $\beta = \frac{\lambda D}{d}$,where $d$ is the distance between the two virtual coherent sources.For a Fresnel biprism,the distance $d$ between the two virtual sources is given by $d = 2a(\mu - 1)\alpha$,where $a$ is the distance of the source from the biprism.Since the incident beam is parallel,the source is effectively at infinity $(a 	o \infty)$.However,in the context of a parallel beam incident on a biprism,the separation $d$ between the virtual sources is $d = 2x(\mu - 1)\alpha$,where $x$ is the distance from the biprism to the virtual source plane. For a parallel beam,the virtual sources are formed at the plane of the biprism itself,but the effective separation is $d = 2D'(\mu - 1)\alpha$ where $D'$ is the distance from the source to the biprism. Given the standard formula for fringe width with a parallel beam,the expression simplifies to $\beta = \frac{\lambda D}{2(\mu - 1)\alpha}$.

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

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