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A ray of light of intensity I is incident on a parallel glass slab at point $A$ 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
From figure $\mathrm{I}_{1}=\frac{\mathrm{I}}{4}$ and $\mathrm{I}_{2}=\frac{9 \mathrm{I}}{64}$
$\begin{aligned} \text { By using } \frac{\mathrm{I}_{\max }}{\mathrm{I}_{\min }} &=\left(\frac{\sqrt{\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}}+1}{\sqrt{\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}}-1}\right)^{2} \\ &=\left(\frac{\sqrt{\frac{9}{16}}+1}{\sqrt{\frac{9}{16}}-1}\right)^{2}=\frac{49}{1} \end{aligned}$
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