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Two wavelengths $\lambda_1$ and $\lambda_2$ are used in Young's double slit experiment. $\lambda_1=450 \mathrm{~nm}$ and $\lambda_2=650 \mathrm{~nm}$. The minimum order of fringe produced by $\lambda_2$ which overlaps with the fringe produced by $\lambda_1$ is $n$. The value of $n$ is _____.
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The correct answer is:
9
$\begin{aligned} & \mathrm{n}_2 \lambda_2=\mathrm{n}_1 \lambda_1 \\ & \frac{\mathrm{n}_2}{\mathrm{n}_1}=\frac{\lambda_1}{\lambda_2}=\frac{450}{650}=\frac{9}{13} \\ & \mathrm{n}_2=9\end{aligned}$
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