A lamp under water lights up only a round patch of the surface. The edge of that patch is set by the critical angle of Total Internal Reflection; inside it, Snell's law bends each ray as it leaves. Each problem here is one short scene and a chain of questions. Answer them in order: each later question uses what you found before, so a slip early on carries through. Keep full precision between steps and round only at the end.
Key idea
Angles are measured from the normal (here the vertical). $$n_1\sin\theta_1 = n_2\sin\theta_2, \qquad \sin\theta_c = \frac{1}{n}, \qquad r = D\tan\theta_c.$$ With flat layers in between, $n\sin\theta$ is the same in every layer.
Worked example: The bright circle
A lamp lies at the bottom of a pool $2.0\ \text{m}$ deep ($n = 1.33$). Find (1) the critical angle and (2) the radius of the bright circle.
- (1) $\sin\theta_c = 1/1.33$, so $\theta_c = 48.8^\circ$.
- (2) $r = 2.0\tan 48.8^\circ = 2.28\ \text{m}$.
Answer: $48.8^\circ$, $2.28\ \text{m}$
Common mistake
Measuring angles from the surface. Snell's law uses angles from the normal. Check: a ray going straight up (0° from the normal) passes through without bending.