A small lamp hangs 1.8 m below the surface of still water in a pool (n = 1.33), as in the picture…

Step 1. The critical angle. $\sin C = \dfrac{1.00}{1.33} = 0.752$, so $C = 48.8°$.

Step 2. The radius. The ray to the edge goes from the lamp up to the surface at the angle $C$ to the vertical. So $\tan C = \dfrac{r}{h}$, and we want $r$, so $r = h\tan C = 1.8\ \text{m} \times \tan 48.8° = 1.8\ \text{m} \times 1.14 = 2.05\ \text{m}$.

Step 3. The area. $A = \pi r^2 = \pi \times (2.05\ \text{m})^2 = 13.2\ \text{m}^2$.