Snell's Law of Refraction I was told that "Snell's law of refraction implies that a light ray in an isotropic medium travels from point a to point b in stationary time." Why is this true?
Thanks.
 A: Let's answer this in reverse.  We'll find a path of stationary time and show it satisfies Snell's law.  
The classic analogy and derivation for this problem are in the Feynman Lectures.  I'll basically be reproducing that argument.  Feynman imagines the material with a lower index of refraction (faster speed) as a beach, and high index of refraction as the ocean.  Point A is on the beach and point B is in the ocean.  You're trying to go from one to the other in the fastest possible time.  The fastest possible time is also a stationary time, simply because minimums are stationary points.
There are many paths you can take from point A to point B.  Here are three examples:

We want to find the place where nearby paths take the same amount of time.  If we have two nearby paths, one spends more time on the beach and the other spends more time in the ocean.  Let's draw two nearby paths, then draw circles with their centers at A and B.  The circles indicate points that are the same distance away, so the little extra bits beyond the circles indicate the extra distance traveled on the beach and in the ocean.

The purple segments are the extra distances traveled on the beach and in the ocean.  We don't want them to be the same distance.  Instead, we want them to take the same amount of time to travel.  That way the extra time spent on the beach cancels the extra time spent in the ocean.
Notice that we can solve this by zooming in closely on the region with the purple lines.  When we zoom in very close and make the purple lines right next to each other, the green circles can be approximated by their tangent lines.  The purple line on top is then perpendicular to the top green circle's tangent and the purple line on bottom is perpendicular to the bottom circle's tangent.  (This follows because radii of circles are perpendicular to tangents.)  So we have two small right triangles, one on the beach and one in the water.
These two triangles have the same hypotenuse, that little horizontal segment between them.  Call it $h$.  
Next we draw a line perpendicular to the beach/ocean interface.  The angle that the path on land make with this line is $\theta_1$ and the angle that the path in the water makes is $\theta_2$.  (It doesn't matter which of the two paths you use to measure this angle; in the limit where the two paths are very close, the angles are the same.)
Then, by trigonometry, we see that the extra distance traveled on the beach is $h\sin\theta_1$ and the extra distance traveled in the water is $h\sin\theta_2$.  The time is takes to travel a distance is proportional to the distance multiplied by the index of refraction, so setting these times equal to each other and canceling out the common $h$, we get
$$n_1\sin\theta_1 = n_2 \sin\theta_2$$
