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In Young's double slit interference experiment, the slit widths are in the ratio $1: 25$. Then the ratio of intensity at the maxima and minima in the interference pattern is
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2268 Upvotes
Verified Answer
The correct answer is:
$9: 4$
$9: 4$
We know that,
$$
\frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{\frac{\omega_1}{\omega_2}+1}\right)^2}{\left(\sqrt{\frac{\omega_1}{\omega_2}-1}\right)^2}
$$
$I_{\max }$ and $I_{\min }$ are maximum and minimum intensity
$\omega_1$ and $\omega_2$ are widths of two slits
$$
\therefore \frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{\frac{1}{25}}+1\right)^2}{\left(\sqrt{\frac{1}{25}}-1\right)^2}\left(\frac{\omega_1}{\omega_2}=\frac{1}{25} \text { given }\right)
$$
On solving we get,
$$
\frac{I_{\max }}{I_{\min }}=\frac{\frac{36}{25}}{\frac{16}{25}}=\frac{9}{4}=9: 4
$$
$$
\frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{\frac{\omega_1}{\omega_2}+1}\right)^2}{\left(\sqrt{\frac{\omega_1}{\omega_2}-1}\right)^2}
$$
$I_{\max }$ and $I_{\min }$ are maximum and minimum intensity
$\omega_1$ and $\omega_2$ are widths of two slits
$$
\therefore \frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{\frac{1}{25}}+1\right)^2}{\left(\sqrt{\frac{1}{25}}-1\right)^2}\left(\frac{\omega_1}{\omega_2}=\frac{1}{25} \text { given }\right)
$$
On solving we get,
$$
\frac{I_{\max }}{I_{\min }}=\frac{\frac{36}{25}}{\frac{16}{25}}=\frac{9}{4}=9: 4
$$
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