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(A) The fringe width $\beta$ is given by the formula $\beta = \frac{\lambda D}{d}$,where $D$ is the distance between the slits and the screen,$d$ is t

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$6000$

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(A) The fringe width $\beta$ is given by the formula $\beta = \frac{\lambda D}{d}$,where $D$ is the distance between the slits and the screen,$d$ is the separation between the slits,and $\lambda$ is the wavelength of light.Given the change in screen distance $\Delta D = 5 	imes 10^{-2} \ m$ and the change in fringe width $\Delta \beta = 3 	imes 10^{-5} \ m$.The change in fringe width is related to the change in distance by $\Delta \beta = \frac{\lambda \Delta D}{d}$.Rearranging for $\lambda$,we get $\lambda = \frac{\Delta \beta \cdot d}{\Delta D}$.Substituting the given values: $\lambda = \frac{(3 	imes 10^{-5} \ m) 	imes (10^{-3} \ m)}{5 	imes 10^{-2} \ m}$.$\lambda = \frac{3 	imes 10^{-8}}{5 	imes 10^{-2}} = 0.6 	imes 10^{-6} \ m$.Converting to $\mathring{A}$: $\lambda = 0.6 	imes 10^{-6} 	imes 10^{10} \ \mathring{A} = 6000 \ \mathring{A}$.

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