A parallel beam of light of intensity $I_0$ is incident on a glass plate, $25 \%$ of light is reflected by upper surface and $50 \%$ of light is reflected from lower surface. The ratio of maximum to minimum intensity in interference region of reflected rays is

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Aicharya · Accepted Answer

Given that, $25 \%$ of total intensity of incident light is reflected from upper surface. This implies, if intensity of incident light is $\mathrm{I}_0$, the intensity of light reaching the lower surface of plate will be $\frac{3}{4} \mathrm{I}_0$. As $50 \%$ of this intensity is reflected, the final intensity of light emerging from glass plate will be $\frac{3}{8} \mathrm{I}_0$. $\begin{aligned} \therefore \quad I_1 & =\frac{I_0}{4} \\ & I_2=\frac{3}{8} I_0 \end{aligned}$ Now, $\frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{I_1}+\sqrt{I_2}\right)^2}{\left(\sqrt{I_1}-\sqrt{I_2}\right)^2}=\left(\frac{\frac{1}{2}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right)^2$

A parallel beam of light of intensity $I_0$ is incident on a glass plate, $25 \%$ of light is reflected by…

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