Question

A ray of light of intensity $I$ incidents on a parallel glass slab at a point A, as shown in the given figure. It undergoes partial reflection and refraction. At each reflection $25\%$ of incident energy is reflected. The rays AB and A' B' undergo interference. The ratio $\frac{I_{max}}{I_{min}}$ is :

• A. 4 : 1
• B. 8 : 1
• C. 7 :1
• D. 49 : 1

Answer 1

The correct option is D 49 : 1

Intensity of AB is 25 % of $I_1=I\times \frac{25}{100}=\frac{I}{4}$

Intensity of A'B' is 75% of $I_1\times 25$%$\times 75$%

$=I\times \frac{75}{100}\times \frac{25}{100}\times \frac{75}{100}$

$=I\times \frac{3}{4}\times \frac{1}{4}\times \frac{3}{4}$

$=\frac{9I}{64}$

$I_{max}=\frac{I}{4}+\frac{9I}{64}+2\sqrt{\frac{I}{4}\times \frac{9I}{64}}$

$=\frac{I}{4}+\frac{9I}{64}+2\left(\frac{3I}{16}\right)=\frac{I}{64}(16+9+24)=\frac{49I}{64}$

$I_{min}=\frac{I}{4}+\frac{9I}{64}-2\left(\frac{3I}{16}\right)=\frac{I}{64}(16+9-24)=\frac{I}{64}$

So, $\frac{I_{max}}{I_{min}}=\frac{49I/64}{I/64}=\frac{49}{1}$