# Refraction problem - limiting radius of observation

I solved this problem but I don't know if the solution is correct. Please, correct me if I'm wrong.

There are two mediums, 1 and 2, with absolute refraction indexes $n_1$ < $n_2$, and an observer in medium 1 with height H (measured from surface to his eyes) . What is the limiting radius R that the observer can look at the surface and see an object located in medium 2? (as a function of $n_1$, $n_2$ and H)

Let $\theta_c$ be the critical angle of refraction. Then light that comes from medium 2 leaves the surface only for incident angles $\theta_2 < \theta_c$. Call $\theta_2 = \theta_c - \epsilon$, with $\epsilon > 0$. The refrated angle will be, by Snell's Law: $$\theta_1 = \sin^{-1}(\frac{n_2}{n_1}\sin(\theta_2))$$ Such that: $$R = \frac{H}{\tan(\frac{\pi}{2} - \theta_1)} = H\tan{\theta_1} = H\tan{(\sin^{-1}(\frac{n_2}{n_1}\sin(\theta_2)))}$$ But in the limit $\epsilon \rightarrow 0$: $$\theta_2 \rightarrow \theta_c = \sin^{-1}(\frac{n_1}{n_2}\sin(\frac{\pi}{2})) \Rightarrow \theta_2 \rightarrow \sin^{-1}(\frac{n_1}{n_2})$$ And therefore: $$R \rightarrow H\tan(\sin^{-1}(1)) = H\tan(\frac{\pi}{2}) \Rightarrow R \rightarrow \infty$$

But this conclusion seems absurd to me since you can only see the mirror of an lake at long distances.

Two comments: First, you see a mirror surface on a lake at shallow angles, true, but only because there is light in the atmosphere ( $n_1$ in your equations). For your problem, assume there is no such light source, so only light emanating from the water ($n_2$) is of concern.