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The width of one of the two slits in a Young's double slit experiment is 4 times that of the other slit. The ratio of the maximum of the minimum intensity in the interference pattern is:
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The correct answer is:
$9: 1$
Since, Intensity $\propto$ width of slit $(\omega)$
$\begin{aligned}
& \text { so, } I_1=I, I_2=4 \mathrm{I} \\
& I_{\min }=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2=I \\
& I_{\max }=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2=9 \mathrm{I} \\
& \frac{I_{\max }}{I_{\min }}=\frac{9 \mathrm{I}}{\mathrm{I}}=\frac{9}{1}
\end{aligned}$
$\begin{aligned}
& \text { so, } I_1=I, I_2=4 \mathrm{I} \\
& I_{\min }=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2=I \\
& I_{\max }=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2=9 \mathrm{I} \\
& \frac{I_{\max }}{I_{\min }}=\frac{9 \mathrm{I}}{\mathrm{I}}=\frac{9}{1}
\end{aligned}$
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