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(D) The angular width of the central maximum in a single slit diffraction is given by $	heta = \frac{2\lambda}{d}$,where $d$ is the slit width.Given

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$25$

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(D) The angular width of the central maximum in a single slit diffraction is given by $	heta = \frac{2\lambda}{d}$,where $d$ is the slit width.Given $	heta = 60^o = \frac{\pi}{3}$ radians,but typically for small angles $\sin 	heta \approx 	heta$. However,using the standard formula $\frac{2\lambda}{d} = 2 \sin^{-1}(\frac{\lambda}{d})$. For $60^o$ angular width,$\frac{\lambda}{d} = \sin(30^o) = 0.5$.Thus,$\lambda = 0.5 	imes d = 0.5 	imes 1 \mu m = 0.5 \mu m$.For Young's double slit experiment,the fringe width is $\beta = \frac{\lambda D}{d'}$,where $d'$ is the slit separation.Given $\beta = 1 \ cm = 10^{-2} \ m$,$D = 50 \ cm = 0.5 \ m$,and $\lambda = 0.5 	imes 10^{-6} \ m$.$10^{-2} = \frac{0.5 	imes 10^{-6} 	imes 0.5}{d'}$.$d' = \frac{0.25 	imes 10^{-6}}{10^{-2}} = 0.25 	imes 10^{-4} \ m = 25 	imes 10^{-6} \ m = 25 \mu m$.

Similar Questions

In a Young's double-slit experiment,the slits are separated by $0.12 \, mm$ and the screen is at a distance of $1 \, m$. Find the distance of the $3^{rd}$ dark fringe from the center of the screen in $cm$. Given $\lambda = 6000 \, \mathring{A}$.

The correct curve between fringe width $\beta$ and the distance between the slits $(d)$ is:

In Young's double slit experiment,the central bright fringe can be identified

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