Question 302: Waves Optics

Question

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

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 $\frac{3}{8} \mathrm{I}_0$.
$$\begin{aligned}
	herefore\quad& \mathrm{I}_1=\frac{\mathrm{I}_0}{4} \
& \mathrm{I}_2=\frac{3}{8} \mathrm{I}_0
\end{aligned}$$
Now, $\frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{I_1}+\sqrt{I_2}ight)^2}{\left(\sqrt{I_1}-\sqrt{I_2}ight)^2}=\left(\frac{\frac{1}{2}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}ight)^2$

Answer

$\left[\frac{\frac{1}{2}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right]^2$

Answer

$\left[\frac{\frac{1}{4}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right]^2$

Answer

$\frac{5}{8}$

Answer

$\frac{8}{5}$

Waves Optics

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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