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(A) The fringe width $\beta$ in a Young's double-slit experiment is given by $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is th

Vedclass · Accepted Answer

$6000$

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(A) The fringe width $\beta$ in a Young's double-slit experiment is given by $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance between the slits and the screen,and $d$ is the separation between the slits.Since $\beta \propto D$,the change in fringe width $\Delta \beta$ due to a change in distance $\Delta D$ is given by $\Delta \beta = \frac{\lambda \Delta D}{d}$.Given: $\Delta \beta = 3 	imes 10^{-5} \ m$,$\Delta D = 5 	imes 10^{-2} \ m$,and $d = 10^{-3} \ m$.Rearranging the formula for wavelength: $\lambda = \frac{\Delta \beta \cdot d}{\Delta D}$.Substituting the values: $\lambda = \frac{(3 	imes 10^{-5} \ m) 	imes (10^{-3} \ m)}{5 	imes 10^{-2} \ m} = \frac{3}{5} 	imes 10^{-6} \ m = 0.6 	imes 10^{-6} \ m = 6 	imes 10^{-7} \ m$.Converting to $\mathring{A}$: $\lambda = 6 	imes 10^{-7} 	imes 10^{10} \ \mathring{A} = 6000 \ \mathring{A}$.

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