A parallel beam of light of intensity I is incident on a glass...

Let the intensity of the incident light be $I$.

Intensity of light reflected from the upper surface is:
$I_1 = 25\% \times I = \dfrac{I}{4}$

Intensity of light refracted into the glass plate is:
$I_t = I - \dfrac{I}{4} = \dfrac{3I}{4}$

At the lower surface, $50\%$ of this light is reflected. Thus, the intensity of light reflected from the lower surface is:
$I_2 = 50\% \times \dfrac{3I}{4} = \dfrac{3I}{8}$

This light emerges from the upper surface to interfere with the first reflected ray. Neglecting any further reflection at the upper boundary, the intensities of the two interfering rays are $I_1 = \dfrac{I}{4} and$I_2 = \dfrac{3I}{8}$.<br><br>The ratio of maximum to minimum intensity in the interference region is:<br>$\dfrac{I_\max}{I_\min} = \left( \dfrac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} \right)^2$<br><br>Substituting the values of$I_1$and$I_2$:<br>$\dfrac{I_\max}{I_\min} = \left( \dfrac{\sqrt{\dfrac{I}{4}} + \sqrt{\dfrac{3I}{8}}}{\sqrt{\dfrac{I}{4}} - \sqrt{\dfrac{3I}{8}}} \right)^2 = $\left( \dfrac{\dfrac{1}{2} + \sqrt{\dfrac{3}{8}}}{\dfrac{1}{2} - \sqrt{\dfrac{3}{8}}} \right)^2$

Answer: \left(\dfrac{\frac{1}{2} + \sqrt{\frac{3}{8}}}{\frac{1}{2} - \sqrt{\frac{3}{8}}}\right)^2