A fish under water sees the out side world in a circular horizon of radius $r$ and in a cone of semi

Question

A fish under water sees the out side world in a circular horizon of radius $r$ and in a cone of semi vertical angle $	heta$. As the fish goes down more and more.\

Professor PC Thomas Classes · Answer

Critical angle for water is $49^{\circ}$. So the fish under water sees the rising sun and setting sun at $49^{\circ}$ with the vertical and $41^{\circ}$ with the horizontal. The fish under water sees the outside world within a cone of semivertical angle $49^{\circ}$. If d is the depth of the fish under water, the radius of the circular horizon is $\mathrm{r}=\frac{\mathrm{d}}{\sqrt{\mathrm{n}^{2}-1}}$ where n is the refractive index of water. As the fish goes down the radius of the circular horizon goes on increasing, but the semivertical angle of the cone in which the fish sees the outside world remains constant. The area of the circular horizon will be $\pi \frac{d^{2}}{n^{2}-1}$\