A light source with a wavelength of 5000 , ma | vedclass

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(C) For a single-slit diffraction pattern,the position of the $n^{th}$ minimum is given by the formula: $x_n = \frac{n \lambda D}{a}$,where $a$ is the

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$0.2$

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(C) For a single-slit diffraction pattern,the position of the $n^{th}$ minimum is given by the formula: $x_n = \frac{n \lambda D}{a}$,where $a$ is the slit width.Given:Wavelength $\lambda = 5000 \, \mathring A = 5000 	imes 10^{-10} \, m = 5 	imes 10^{-7} \, m$.Distance of the first minimum $x_1 = 5 \, mm = 5 	imes 10^{-3} \, m$.Distance between slit and screen $D = 2 \, m$.Order $n = 1$.Substituting these values into the formula:$5 	imes 10^{-3} = \frac{1 	imes (5 	imes 10^{-7}) 	imes 2}{a}$$a = \frac{10 	imes 10^{-7}}{5 	imes 10^{-3}}$$a = 2 	imes 10^{-4} \, m = 0.2 	imes 10^{-3} \, m = 0.2 \, mm$.Therefore,the width of the slit is $0.2 \, mm$.

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