The distance between two coherent sources is | vedclass

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(B) Given:Distance between sources,$d = 0.1 \, mm = 10^{-4} \, m$.Distance of screen from sources,$D = 1.2 \, m$.Fringe-width,$\beta = 6.0 \, mm = 6 \

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$5000$

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(B) Given:Distance between sources,$d = 0.1 \, mm = 10^{-4} \, m$.Distance of screen from sources,$D = 1.2 \, m$.Fringe-width,$\beta = 6.0 \, mm = 6 	imes 10^{-3} \, m$.The formula for fringe-width in Young's double-slit experiment is given by:$\beta = \frac{D \lambda}{d}$Rearranging the formula to solve for wavelength $\lambda$:$\lambda = \frac{\beta \cdot d}{D}$Substituting the given values:$\lambda = \frac{(6 	imes 10^{-3} \, m) 	imes (10^{-4} \, m)}{1.2 \, m}$$\lambda = \frac{6 	imes 10^{-7}}{1.2} \, m$$\lambda = 5 	imes 10^{-7} \, m$Converting meters to $\mathring{A}$s $(\mathring{A})$,where $1 \, \mathring{A} = 10^{-10} \, m$:$\lambda = 5 	imes 10^{-7} 	imes 10^{10} \, \mathring{A} = 5000 \, \mathring{A}$.Thus,the correct option is $B$.

The distance between two coherent sources is $0.1 \, mm$. The fringe-width on a screen $1.2 \, m$ away from the source is $6.0 \, mm$. The wavelength of light used is $\dots \, \mathring{A}$.

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