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(A) Let the $n_1$-th bright fringe of $\lambda_1$ coincide with the $n_2$-th bright fringe of $\lambda_2$.For bright fringes,the position $x$ is given

Vedclass · Accepted Answer

$6$

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(A) Let the $n_1$-th bright fringe of $\lambda_1$ coincide with the $n_2$-th bright fringe of $\lambda_2$.For bright fringes,the position $x$ is given by $x = \frac{n\lambda D}{d}$.Since the positions coincide,we have $\frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d}$,which implies $n_1 \lambda_1 = n_2 \lambda_2$.Rearranging gives $\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{10000}{12000} = \frac{5}{6}$.Thus,the minimum integer values are $n_1 = 5$ and $n_2 = 6$.Now,calculate the distance $x$ using $n_1 = 5$:$x = \frac{n_1 \lambda_1 D}{d} = \frac{5 	imes 12000 	imes 10^{-10} 	imes 2}{2 	imes 10^{-3}}$.$x = \frac{5 	imes 12 	imes 10^{-7} 	imes 2}{2 	imes 10^{-3}} = 60 	imes 10^{-4} = 6 	imes 10^{-3}\, m$.$x = 6\, mm$.

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