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(C) For bright fringes in a Young's double slit experiment, the position is given by $y = \frac{n \lambda D}{d}$.Let the $n^{th}$ bright fringe of $\l

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$4.5$

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(C) For bright fringes in a Young's double slit experiment, the position is given by $y = \frac{n \lambda D}{d}$.Let the $n^{th}$ bright fringe of $\lambda_1 = 900 \,nm$ coincide with the $m^{th}$ bright fringe of $\lambda_2 = 750 \,nm$.Since $\lambda_1 > \lambda_2$, the $n^{th}$ fringe of $\lambda_1$ will coincide with the $(n+1)^{th}$ fringe of $\lambda_2$ at the minimum distance.So, $\frac{n \lambda_1 D}{d} = \frac{(n+1) \lambda_2 D}{d}$.$n 	imes 900 = (n+1) 	imes 750$.$900n = 750n + 750$.$150n = 750 \Rightarrow n = 5$.The distance $y = \frac{n \lambda_1 D}{d} = \frac{5 	imes 900 	imes 10^{-9} 	imes 2}{2 	imes 10^{-3}}$.$y = 5 	imes 900 	imes 10^{-6} \,m = 4500 	imes 10^{-6} \,m = 4.5 	imes 10^{-3} \,m = 4.5 \,mm$.

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