Confusion regarding Snell's window and critical angle As far as I know, the critical angle exists only if light passes from a medium with a greater index of refraction to one with a smaller index of recreation. However, when it comes to Snell's window, a critical angle (48.6°) exists, even though light is passing from a medium with an index of refraction of 1 (air) to a medium with an index of refraction of 1.33 (water). How is this possible, or am I completely misinterpreting things?

 A: Maria,
You are correct in your confusion - authors who write about this topic may be confusing two similar things that occur side-by-side.
When you look up from the bottom of a pool, there are a lot of weird things you see.  One of the features is Snell's Window: the entirety of the world above the water seems condensed into a circle, as seen below:

The Snell's Window part of this image doesn't have anything to do with total internal reflection.  It's just what you get when 180 degrees of reality are squeezed into less of an angle.  The diagram that you posted above explains it in more detail.
If you look at the dark area outside the "window", this area is where total internal reflection can take place (provided there's enough light bouncing around inside of the pool in the first place).  In the photograph above, the photographer seems to be submerged in a fairly well shaded pool - hence the area around Snell's window is basically black.  In a more wide open pool you would see colors (but no image) in this region - caused by light going through total internal reflection while bouncing around inside the pool.
A: 
As far as I know, the critical angle exists only if light passes from a medium with a greater index of refraction to one with a smaller index of recreation.

The critical angle you mention here is the angle that causes total internal reflection. Say light travels in a medium $n_1$ and hits the boundary between two mediums at an angle $\theta_1$ to the normal. It then travels in medium $n_2$ at an angle $\theta_2$ to the normal. Then Snell's Law tells us:
$$n_2\sin\theta_1=n_2\sin\theta_2 \implies \sin\theta_2 = \frac{n_1}{n_2}\sin\theta_1$$
Suppose we are freely able to change the angle $\theta_1$. Notice that if $n_1>n_2$ there will be an angle $\theta_\text{1,critical}$ where $\sin\theta_2$ reaches its maximum value of one which is achieved when $\theta_2=90^\circ$, that is the light exits the first medium parallel to the surface. After this critical angle, light can no longer pass through the boundary and will start totally reflecting.
In short, the critical angle here refers to the maximum angle that allows transmission.

However, when it comes to Snell's window, a critical angle (48.6°) exists, even though light is passing from a medium with an index of refraction of 1 (air) to a medium with an index of refraction of 1.33 (water). How is this possible, or am I completely misinterpreting things?

Notice here, the critical angle does not refer to the angle of the incident beam. Instead, it refers to the maximum angle of the transmitted beam (which happens to occur when the incident beam comes parallel to the water surface).
These two situations are actually very well linked together. In your diagram, instead of having the light beams come into the water: try the reverse and have light beams originate from the shark and shoot around it. The ray diagrams will make the exact same shape. This is known as the reversibility of light.
To summarize, the two scenarios are equivalent. You just have to be careful when defining what the critical angle is. 
