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You can perfectly well have total internal reflection in
a graded-index medium with a smooth index profile; no
steps are required. Optical fibers (e.g. transatlantic
communication fibers) rely on this.
http://en.wikipedia.org/wiki/Graded-index_fiber
That's a helpful example, but I'm still not seeing how it works. Let's make up some specific numbers:
Say the road is at z=0 and the top of the relevant air layer is at z=1 in normalized units. And let's make a linear index profile, with n=1 at the road and n=1.1 at the top. Now let's start with an incident ray just below the top making an angle of 80 degrees relative to the vertical (ie. the normal to the road). Then Snell's law says the ray reaches an angle of 90 degrees when n=1.083, which occurs at a height of z=0.83. That ray travels parallel to the road.
I don't see how it bends upward, within a *strict* ray (of zero cross-sectional size) approximation. I agree that if I treat the wavefront as having a finite extent, then I can consider the top of the wavefront to do something different than the bottom, and perhaps that's the explanation. (I think that's the explanation of graded-index fiber, but certainly correct me if I'm wrong.) I consider that to be a non-ray-optics explanation, however.
So again my question is: Can we explain the highway mirage using geometric ray optics alone? If not, okay, but textbooks certainly give the impression that one can. -Carl