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

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(C) For the bright fringes to coincide,the position $x$ must be the same for both wavelengths:$x = \frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{

Vedclass · Accepted Answer

$0.156$

Vedclass · Answer

(C) For the bright fringes to coincide,the position $x$ must be the same for both wavelengths:$x = \frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d}$This implies $n_1 \lambda_1 = n_2 \lambda_2$,where $n_1$ and $n_2$ are integers.$\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{5200 \, \mathring{A}}{6500 \, \mathring{A}} = \frac{4}{5}$For the minimum distance,we take the smallest integer values: $n_1 = 4$ and $n_2 = 5$.Now,calculate the distance $x$ using $n_1 = 4$ and $\lambda_1 = 6500 \, \mathring{A}$:$x = \frac{n_1 \lambda_1 D}{d} = \frac{4 	imes 6500 	imes 10^{-10} \, m 	imes 1.2 \, m}{2 	imes 10^{-3} \, m}$$x = \frac{4 	imes 6500 	imes 1.2 	imes 10^{-7}}{2 	imes 10^{-3}} = 2 	imes 6500 	imes 1.2 	imes 10^{-4} \, m$$x = 15600 	imes 10^{-4} \, m = 1.56 	imes 10^{-2} \, m = 0.156 \, cm$.

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