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A ray of light of intensity $\mathrm{I}$ is incident on a parallel glass slab at point $\mathrm{A}$ as shown in diagram. It undergoes partial reflection and refraction. At each reflection, $25 \%$ of incident energy is reflected. The rays $\mathrm{AB}$ and $\mathrm{A}^{\prime} \mathrm{B}^{\prime}$ undergo interference. The ratio of $\mathrm{I}_{\max }$ and $\mathrm{I}_{\min }$ is :
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The correct answer is:
$49: 1$
$49: 1$
From figure $I_1=\frac{I}{4}$ and $I_2=\frac{9 I}{64}$
$\Rightarrow \frac{I_2}{I_1}=\frac{9}{16}$
By using $\frac{I_{\max }}{I_{\min }}=\left(\frac{\sqrt{\frac{I_2}{I_1}}+1}{\sqrt{\frac{I_2}{I_1}}-1}\right)^2$
$=\left(\frac{\sqrt{\frac{9}{16}}+1}{\sqrt{\frac{9}{16}}-1}\right)^2=\frac{49}{1}$
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