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

The Correct Option is A

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,
\(\frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)^{2}}{\left(\sqrt{I_{1}}-\sqrt{I_{2}}\right)^{2}}=\left(\frac{\frac{1}{2}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right)^{2}\)\((\because I_1=\frac{I_0}{4},I_2=\frac{3}{8}I_0)\)