Re: [Phys-L] highway mirage

[Craig Wiegert <wiegert@physast.uga.edu>]

I realize this isn't really appropriate for an introductory physics
class, but it's not too onerous to integrate Snell's Law directly to
determine the path of the ray.

Assume n depends on z, and we'd like to solve for z(x), the trajectory
of the ray through the refractive medium.  Then Snell's law says that

  n_i sin(theta_i) = n(z) sin(theta(z))

and in particular, we can set n_i sin(theta_i) = n_min, the smallest
index of refraction that the ray would "see" as it turns around /
experiences TIR.  This value n_min is a "constant of the motion".

In terms of the path z(x),

  sin(theta(z)) = 1/sqrt(1 + (dz/dx)^2) ,

which, after a bit of algebra with Snell's Law, leads to

  dx/dz = 1/sqrt((n(z)/nmin)^2 - 1) .

In other words, Snell's Law is equivalent to a 1st order DE for (the
inverse function of) the path.

For example, a linear index, say n(z) = n_min + A z, gives rise to a
ray path that's a hyperbolic cosine:

  z(x) = (cosh(k x) - 1)/k
  where k = A/n_min.

If instead the index profile is something like

  n(z) = n_min sqrt(1 + (a z)^2)

then the ray's path is a decreasing exponential, exp(-a x), and the ray
won't experience TIR.

(This should all match John D's "epsilonic" analysis.)

To me, this purely ray optics approach bothered me conceptually... until
I remembered that Snell's Law itself is derived by considering the
behavior of (usually planar) wavefronts.  Is there any way to derive
Snell's Law *without* using wave properties?

  - Craig


On Fri, 11 Apr 2014 13:39:28 -0400
Carl Mungan <mungan@usna.edu> wrote:

> Excellent. I'm satisfied with the argument presented below.
>
> I agree it's not the most sensible way to think 
> about the mirage, and certainly not suitable for 
> a typical classroom. But I'm nevertheless greatly 
> pleased that it *can* be explained using ray 
> optics alone.
>
> I may be an oddity (although I doubt it), but I 
> think that even if a physics principle (for 
> example, ray optics here) is not considered 
> *sensible* in some situation, there is a certain 
> pedagogical virtue to using it anyways, just to 
> see if it can be done and just to satisfy one's 
> confidence in basic physics.
>
> Another example would be the work-kinetic-energy 
> theorem. No, I don't want to start an argument 
> over it, but some people say those principles 
> only apply to point particles and aren't willing 
> to consider applying them to composite systems. I 
> politely disagree. There are often other ways to 
> analyze composite systems. Those other ways may 
> be more "sensible." That doesn't stop me from 
> choosing to use the theorem anyways. One has to 
> be careful. But that's different than saying it 
> can't be done or that it's wrong.
>
> So, much thanks to all who contributed to this 
> discussion. I found it helpful. -Carl
>
> > At each layer boundary, we invoke Snell's law
> >
> >           sin É__i       cos Éø_i
> >          ---------  =  ---------             [1]
> >           sin É__t       cos Éø_i
> >
> >                           1
> >                     =  -------               [2]
> >                         1 + Ɉ
> >
> > Where we assume the complimentary angles (Éø) are small:
> >
> >       Éø_i := 90° - É__i                       [3]
> >            = small
> >
> >       Éø_t := 90° - É__t                       [4]
> >            = small
> >
> > Expanding to lowest order:
> >
> >          1 - Éø^2_i / 2
> >         ---------------  =  1 - Ɉ            [5]
> >          1 - Éø^2_t / 2
> >
> >           Éø^2_(i+1) - Éø^2_i  =  -2Ɉ          [6]
> >
> > Where Éø_i is the angle in "this" layer and we write Éø_(i+1)
> > as a synonym for Éø_t to denote the angle of the ray in the
> > "next" layer.
> >
> > We can use equation [6] as a recursion relation.  Given a
> > value for the angle in the "final" layer we can work backwards
> > to find the angle in all previous layers.  There are four
> > cases to consider, three of which are trivial:
> >   a) If the Éø^2 value in the final layer is zero, the ray is
> >    horizontal, parallel to the layer boundary, and is stuck.
> >   b) If the Éø^2 value in the final layer is 2Ɉ, the angle will
> >    go to zero when the ray tries to leave this layer, and
> >    the ray will thereupon get stuck.
> >   c) For any Éø^2 value greater than 2Ɉ, this wasn't really
> >    the final layer.  The ray will be refracted into the next
> >    layer, and we get to reconsider all these cases, using a
> >    smaller value of Éø^2, smaller by an amount 2Ɉ.
> >   d) For all Éø^2 values in the interior of the interval (0, 2Ɉ),
> >    the ray will be totally internally reflected.
> >
> > Proof by induction on Éø^2.
> >
> > So we are back to something I said earlier:  Except on a set
> > of measure zero, the ray will undergo total internal reflection.
> > And since the layer boundaries are an imaginary construction
> > anyway, I am free to move the boundary half a layer in one
> > direction or the other, so as to make the pathological "stuck"
> > case go away entirely.