A ray of light of intensity ' I ' is incident on a parallel glass slab at a point 'A' as shown in figure. It undergoes partial reflection and refraction. At each reflection $25 \%$ of incident energy is reflected. The rays $A B$ and $A^1 B^1$ undergo interference. The ratio $I_{\text {max }} / I_{\text {min }}$ is

Given an incident ray of intensity $I_0$ reflecting between two surfaces with reflection coefficient $R = \frac{1}{4}$ and transmission coefficient $T = \frac{3}{4}$, the intensities of the two interfering rays are calculated.

The first ray reflects directly from the upper surface with intensity $I_1 = R I_0 = \frac{1}{4} I_0$.

The second ray transmits through the upper surface, reflects from the lower surface, and transmits out again, with intensity $I_2 = T R T I_0 = \frac{3}{4} \cdot \frac{1}{4} \cdot \frac{3}{4} I_0 = \frac{9}{64} I_0$.

For interference between two waves, the intensity ratio is given by $\frac{I_{\text{max}}}{I_{\text{min}}} = \left(\frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}}\right)^2$. Substituting $\sqrt{I_1} = \frac{1}{2}\sqrt{I_0}$ and $\sqrt{I_2} = \frac{3}{8}\sqrt{I_0}$ yields:

$\frac{I_{\text{max}}}{I_{\text{min}}} = \left(\frac{\frac{1}{2} + \frac{3}{8}}{\frac{1}{2} - \frac{3}{8}}\right)^2 = \left(\frac{\frac{7}{8}}{\frac{1}{8}}\right)^2 = 7^2 = 49$

The maximum to minimum intensity ratio is 49:1.