Re: [Phys-L] highway mirage

[Carl Mungan <mungan@usna.edu>]

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 ɐ_i       cos Éø_i
>          ---------  =  ---------             [1]
>           sin ɐ_t       cos Éø_i
>
>                           1
>                     =  -------               [2]
>                         1 + Ɉ
>
> Where we assume the complimentary angles (Éø) are small:
>
>       Éø_i := 90° - ɐ_i                       [3]
>            = small
>
>       Éø_t := 90° - ɐ_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.

--
Carl E Mungan, Assoc Prof of Physics  410-293-6680 (O) -3729 (F)
Naval Academy Stop 9c, 572C Holloway Rd, Annapolis MD 21402-1363
mailto:mungan@usna.edu     http://usna.edu/Users/physics/mungan/