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

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(A) Given: Slit separation $d = 2 \, mm = 2 	imes 10^{-3} \, m$,Screen distance $D = 2 \, m$,$\lambda_1 = 12000 	imes 10^{-10} \, m$,$\lambda_2 = 10

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$6$

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(A) Given: Slit separation $d = 2 \, mm = 2 	imes 10^{-3} \, m$,Screen distance $D = 2 \, m$,$\lambda_1 = 12000 	imes 10^{-10} \, m$,$\lambda_2 = 10000 	imes 10^{-10} \, m$.For bright fringes to coincide,the position $x$ must satisfy: $x = \frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d}$.This implies $n_1 \lambda_1 = n_2 \lambda_2$,or $\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1}$.Substituting the values: $\frac{n_1}{n_2} = \frac{10000}{12000} = \frac{5}{6}$.For the minimum distance,we take the smallest integers $n_1 = 5$ and $n_2 = 6$.Now,calculate $x$: $x = \frac{n_1 \lambda_1 D}{d} = \frac{5 	imes 12000 	imes 10^{-10} 	imes 2}{2 	imes 10^{-3}}$.$x = 5 	imes 12000 	imes 10^{-7} \, m = 60000 	imes 10^{-7} \, m = 6 	imes 10^{-3} \, m = 6 \, mm$.

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