Question

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

• A. $\left(\frac{\frac{1}{2} + \sqrt{\frac{3}{8}}}{\frac{1}{2}- \sqrt{\frac{3}{8}}}\right)^{2}$
• B. $\left(\frac{\frac{1}{4} + \sqrt{\frac{3}{8}}}{\frac{1}{2}- \sqrt{\frac{3}{8}}}\right)^{2}$
• C. $\frac{5}{8}$
• D. $\frac{8}{5}$

Answer 1

Answer: (A)

$\left(\frac{\frac{1}{2} + \sqrt{\frac{3}{8}}}{\frac{1}{2}- \sqrt{\frac{3}{8}}}\right)^{2}$

Answer 2

The correct answer is A:$\left(\frac{\frac{1}{2} + \sqrt{\frac{3}{8}}}{\frac{1}{2}- \sqrt{\frac{3}{8}}}\right)^{2}$
According to question, we can derive;
$I_{1}=\frac{I_{0}}{4} \Rightarrow I_{2}=\frac{3}{8} I_{0}$
We know that,
\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}$$(\because I_1=\frac{I_0}{4},I_2=\frac{3}{8}I_0)