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(B) The fringe width $\beta$ in a biprism experiment is given by the formula $\beta = \frac{D\lambda}{d}$,where $D$ is the distance from the sources t

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$0.1$

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(B) The fringe width $\beta$ in a biprism experiment is given by the formula $\beta = \frac{D\lambda}{d}$,where $D$ is the distance from the sources to the screen,$\lambda$ is the wavelength of light,and $d$ is the separation between the coherent sources.Given:$\lambda = 5000 \ \mathring{A} = 5000 	imes 10^{-10} \,m = 5 	imes 10^{-7} \,m$$\beta = 5 \,mm = 5 	imes 10^{-3} \,m$$D = 1.0 \,m$Rearranging the formula for $d$:$d = \frac{D\lambda}{\beta}$Substituting the values:$d = \frac{1.0 	imes 5 	imes 10^{-7}}{5 	imes 10^{-3}} = 10^{-4} \,m$Converting to millimeters:$d = 10^{-4} 	imes 10^3 \,mm = 0.1 \,mm$.Therefore,the correct option is $B$.

In a biprism experiment,by using light of wavelength $5000 \ \mathring{A}$,$5 \,mm$ wide fringes are obtained on a screen $1.0 \,m$ away from the coherent sources. The separation between the two coherent sources is.....$mm$.

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