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

• Correct answer
  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

  $\frac{8}{5}$

Solution

According to question, we can draw the following diagram
$$ I_1=\frac{I_0}{4} \Rightarrow I_2=\frac{3}{8} I_0 $$ We know that, $$ \frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{I_1}+\sqrt{I_2}\right)^2}{\left(\sqrt{I_1}-\sqrt{I_2}\right)^2}=\left(\frac{\frac{1}{2}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right)^2 $$

Hence, the correct answer is option: 'A'.