Physics - Wave Optics Question with Solution | TestHub

Intensity of the incident light is $I_0$.

Intensity of the ray reflected from the upper surface is given by:
$I_1=25\%\text{ of }{I}_0=\frac{1}{4}{I}_0$

Intensity of the emergent ray after reflection from the lower surface is given by:
$I_2=\frac{3}{8}{I}_0$

The maximum intensity in the interference pattern is:
$I_{max}=(\sqrt{I_1}+\sqrt{I_2}{)}^2$

The minimum intensity in the interference pattern is:
$I_{min}=(\sqrt{I_1}-\sqrt{I_2}{)}^2$

The ratio of maximum to minimum intensity is:
$\frac{I_{max}}{I_{min}}={\left(\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}\right)}^2$

Substituting the values of $I_1$ and $I_2$:
$\frac{I_{max}}{I_{min}}={\left(\frac{\sqrt{\frac{1}{4}{I}_0}+\sqrt{\frac{3}{8}{I}_0}}{\sqrt{\frac{1}{4}{I}_0}-\sqrt{\frac{3}{8}{I}_0}}\right)}^2$

$\frac{I_{max}}{I_{min}}={\left[\frac{\frac{1}{2}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right]}^2$

Answer: ${\left[\frac{\frac{1}{2}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right]}^2$