Light of wavelength lambda = 5000 mathring{A | vedclass

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(A) For a single slit diffraction pattern,the position of the $n^{th}$ minima is given by the formula: $x_n = \frac{n \lambda D}{d}$.Here,$n = 1$ for

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$0.1$

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(A) For a single slit diffraction pattern,the position of the $n^{th}$ minima is given by the formula: $x_n = \frac{n \lambda D}{d}$.Here,$n = 1$ for the first minima.Given values are: $\lambda = 5000 \ \mathring{A} = 5000 	imes 10^{-10} \ m = 5 	imes 10^{-7} \ m$,$D = 1 \ m$,and $x_1 = 5 \ mm = 5 	imes 10^{-3} \ m$.Substituting these values into the formula:$5 	imes 10^{-3} = \frac{1 	imes 5 	imes 10^{-7} 	imes 1}{d}$$d = \frac{5 	imes 10^{-7}}{5 	imes 10^{-3}} \ m$$d = 10^{-4} \ m = 0.1 \ mm$.Therefore,the width of the slit is $0.1 \ mm$.

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