Question

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

• A. $\left(\frac{\frac{1}{2}-\sqrt{\frac{3}{8}}}{\frac{1}{2}+\sqrt{\frac{3}{8}}}\right)^{^{^2}}$
• B. $\left(\frac{\frac{1}{2}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right)^{^{^2}}$
• C. $\left(\frac{1-\frac{3}{\sqrt{8}}}{1+\frac{3}{\sqrt{8}}}\right)^{^{^2}}$
• D. $\left(\frac{1+\frac{3}{\sqrt{8}}}{1-\frac{3}{\sqrt{8}}}\right)^{^{^2}}$

Answer 1

Answer: (D)

$I_{1}=\frac{I}{4}$ and $I_{2}=\frac{9\,I}{32}$
$\therefore \frac{I_{1}}{I_{2}}=\frac{8}{9}$ or $\sqrt{\frac{I_{1}}{I_{2}}}=\frac{2\sqrt{2}}{3}$
$\therefore \frac{I_{max}}{I_{min}}=\left(\frac{\sqrt{\frac{I_{1}}{I_{2}}}+1}{\sqrt{\frac{I_{1}}{I_{2}}}-1}\right)^{^{^{^{^2}}}}=\left(\frac{2\sqrt{2}+3}{2\sqrt{2}-3}\right)^{2}=\left(\frac{1+\frac{3}{\sqrt{8}}}{1-\frac{3}{\sqrt{8}}}\right)^{^{^2}}$