A ray of light of intensity $I$ is incident on a parallel glass-slab at the point $A$ and it undergoes partial reflection and refraction as shown in the figure. At each reflection, $25\%$of the incident energy is reflected and the rest is transmitted. If the rays $AB$ and $A\prime B\prime$undergo interference, then the ratio $\frac{I_{\max }}{I_{\min }}$ is

• A

  $4\mathrm{ }:1$
• B

  $8\mathrm{ }:1$
• C

  $7\mathrm{ }:1$
• Correct answer
  D

  $49\mathrm{ }:1$

Solution

From figure $I_1=\frac{I}{4}$ and $I_2=\frac{9I}{64}\Rightarrow \frac{I_2}{I_1}=\frac{9}{16}$




By using $\frac{I_{max}}{I_{min}}={\left(\frac{\sqrt{\frac{I_2}{I_1}}+1}{\sqrt{\frac{I_2}{I_1}}-1}\right)}^2={\left(\frac{\sqrt{\frac{9}{16}}+1}{\sqrt{\frac{9}{16}}-1}\right)}^2=\frac{49}{1}$

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