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(C) The position of the $n^{th}$ bright fringe is given by $y_n = \frac{n \lambda D}{d}$.For the $9^{th}$ bright fringe,$y_9 = \frac{9 \lambda D}{d}$.

Vedclass · Accepted Answer

$5000$

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(C) The position of the $n^{th}$ bright fringe is given by $y_n = \frac{n \lambda D}{d}$.For the $9^{th}$ bright fringe,$y_9 = \frac{9 \lambda D}{d}$.The position of the $m^{th}$ dark fringe is given by $y'_m = \frac{(2m-1) \lambda D}{2d}$.For the $2^{nd}$ dark fringe,$y'_2 = \frac{(2 	imes 2 - 1) \lambda D}{2d} = \frac{3 \lambda D}{2d}$.The distance between them is $y_9 - y'_2 = 7.5 \, mm = 7.5 	imes 10^{-3} \, m$.$\frac{9 \lambda D}{d} - \frac{3 \lambda D}{2d} = 7.5 	imes 10^{-3}$.$\frac{(18-3) \lambda D}{2d} = 7.5 	imes 10^{-3} \implies \frac{15 \lambda D}{2d} = 7.5 	imes 10^{-3}$.Given $d = 0.5 	imes 10^{-3} \, m$ and $D = 1 \, m$.$\lambda = \frac{7.5 	imes 10^{-3} 	imes 2 	imes 0.5 	imes 10^{-3}}{15 	imes 1} = \frac{7.5 	imes 10^{-6}}{15} = 0.5 	imes 10^{-6} \, m = 5000 \, \mathring{A}$.

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