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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.

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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.

Next, essentially you are correct that the limiting ray angle exiting the water (Snell window) can be as close to horizontal as you desire. The problem as stated isn't terribly useful unless you also calculate the relative intensity as a function of viewing angle (or, back to your comment about mirror-surface, the relative intensity of the emanating light as compared to the reflected light from the atmosphere).

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  • $\begingroup$ Thank you, I understand now. But let me ask you another question: for an point object to have an perfect image at some other point, all of it's rays must converge (even if in virtual way) to that point. Because there's always some ray that will be entirely reflected, this means that it's impossible for an object underwater to have an perfect image for an outside observer ? $\endgroup$ – user41887 Mar 5 '14 at 14:18
  • $\begingroup$ @OtávioRapôso not exactly: a "perfect image" just means that all rays which make it to the image plane converge to a point. Lost rays (lost energy) don't affect the image other than the overall brightness. $\endgroup$ – Carl Witthoft Mar 5 '14 at 14:52

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