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(D) The fringe width $\beta$ in Young's double slit experiment is given by the formula: $\beta = \frac{\lambda D}{d}$.Here,$\lambda = 6000 \ \mathring

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$0.2$

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(D) The fringe width $\beta$ in Young's double slit experiment is given by the formula: $\beta = \frac{\lambda D}{d}$.Here,$\lambda = 6000 \ \mathring{A} = 6000 	imes 10^{-8} \ cm = 6 	imes 10^{-5} \ cm$.The distance of the screen from the slits is $D = 40 \ cm$.The fringe width is $\beta = 0.012 \ cm$.Rearranging the formula to solve for the distance between the slits $d$:$d = \frac{\lambda D}{\beta} = \frac{6 	imes 10^{-5} \ cm 	imes 40 \ cm}{0.012 \ cm}$.$d = \frac{240 	imes 10^{-5}}{0.012} \ cm = \frac{2.4 	imes 10^{-3}}{1.2 	imes 10^{-2}} \ cm = 2 	imes 10^{-1} \ cm = 0.2 \ cm$.Therefore,the distance between the slits is $0.2 \ cm$.

In Young's double slit experiment,when the wavelength used is $6000 \ \mathring{A}$ and the screen is $40 \ cm$ from the slits,the fringes are $0.012 \ cm$ wide. What is the distance between the slits in $cm$?

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