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(B) 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

$600$

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(B) 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 to the screen,and $d$ is the slit separation.When the screen is moved by $\Delta D$,the change in fringe width $\Delta \beta$ is given by $\Delta \beta = \frac{\lambda}{d} \Delta D$.Given: $\Delta D = 5 	imes 10^{-2} \,m$,$\Delta \beta = 3 	imes 10^{-3} \,cm = 3 	imes 10^{-5} \,m$,and $d = 1 \,mm = 10^{-3} \,m$.Rearranging the formula for $\lambda$: $\lambda = \frac{\Delta \beta \cdot d}{\Delta D}$.Substituting the values: $\lambda = \frac{3 	imes 10^{-5} 	imes 10^{-3}}{5 	imes 10^{-2}} = \frac{3 	imes 10^{-8}}{5 	imes 10^{-2}} = 0.6 	imes 10^{-6} \,m$.Converting to nanometers: $\lambda = 600 	imes 10^{-9} \,m = 600 \,nm$.

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