In Young's double-slit experiment,the separat | vedclass

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(D) The position of the $n^{th}$ bright fringe is given by $y_n = \frac{n \lambda D}{d}$.For the $8^{th}$ bright fringe $(n=8)$: $y_8 = \frac{8 \lambd

Vedclass · Accepted Answer

$1.76 	imes 10^{-3} \, m$

Vedclass · Answer

(D) The position of the $n^{th}$ bright fringe is given by $y_n = \frac{n \lambda D}{d}$.For the $8^{th}$ bright fringe $(n=8)$: $y_8 = \frac{8 \lambda D}{d}$.The position of the $n^{th}$ dark fringe is given by $y'_n = \frac{(2n-1) \lambda D}{2d}$.For the $3^{rd}$ dark fringe $(n=3)$: $y'_3 = \frac{(2(3)-1) \lambda D}{2d} = \frac{5 \lambda D}{2d} = 2.5 \frac{\lambda D}{d}$.The separation between the $8^{th}$ bright fringe and the $3^{rd}$ dark fringe is $\Delta y = y_8 - y'_3 = \frac{8 \lambda D}{d} - 2.5 \frac{\lambda D}{d} = 5.5 \frac{\lambda D}{d}$.Substituting the values: $\lambda = 480 	imes 10^{-9} \, m$,$D = 2 \, m$,$d = 3 	imes 10^{-3} \, m$.$\Delta y = \frac{5.5 	imes 480 	imes 10^{-9} 	imes 2}{3 	imes 10^{-3}} = \frac{5.5 	imes 960 	imes 10^{-9}}{3 	imes 10^{-3}} = 5.5 	imes 320 	imes 10^{-6} = 1760 	imes 10^{-6} \, m = 1.76 	imes 10^{-3} \, m$.

Similar Questions

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