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A parallel beam of light of intensity $I_0$ is incident on a coated glass plate. If $25 \%$ of the incident light is reflected from the upper surface and $50 \%$ of light is reflected from the lower surface of the glass plate, the ratio of maximum to minimum intensity in the interference region of the reflected light is
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Verified Answer
The correct answer is:
$\left(\frac{\frac{1}{2}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right)^2$
According to question, we can draw the following diagram
$$
I_1=\frac{I_0}{4} \Rightarrow I_2=\frac{3}{8} I_0
$$
We know that,
$$
\frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{I_1}+\sqrt{I_2}\right)^2}{\left(\sqrt{I_1}-\sqrt{I_2}\right)^2}=\left(\frac{\frac{1}{2}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right)^2
$$
$$
I_1=\frac{I_0}{4} \Rightarrow I_2=\frac{3}{8} I_0
$$
We know that,
$$
\frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{I_1}+\sqrt{I_2}\right)^2}{\left(\sqrt{I_1}-\sqrt{I_2}\right)^2}=\left(\frac{\frac{1}{2}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right)^2
$$
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