JEE MAIN - Physics (2024 - 4th April Evening Shift - No. 13)
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:
$$1: 1$$
$$4: 1$$
$$16: 1$$
$$9: 1$$
Explanation
$$\begin{aligned}
& I_{\max }=\left(\sqrt{4 I_0}+\sqrt{I_0}\right)^2=\left(3 \sqrt{I_0}\right)^2=9 I_0 \\
& I_{\min }=\left(\sqrt{4 I_0}-\sqrt{I_0}\right)^2=I_0 \\
& \frac{I_{\max }}{I_{\min }}=9: 1
\end{aligned}$$
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