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Re: [Phys-L] highway mirage



I've got some bad news and some good news.

The bad news is that almost everything I wrote about
interface reflection 04/10/2014 12:04 PM is incorrect.

It remains true that whenever there is an interface between
media with different indices, there will be a reflection.
HOWEVER ... if you have N layers each with a small index
change, the amount of reflection gets small. On a
per-layer basis, it gets small /faster/ than linearly
in 1/N, so that even if you add up the contributions
from all the layers, the overall reflection (from this
mechanism) goes away in the large-N limit.

I should have known this; I've even done this
calculation before, in the context of anti-reflection
coatings for lenses.........

You may be wondering, whatever happened to "check the
work". Well, I don't like to make lame excuses, but I
did do some checking before posting. I checked it using
a lowest-order expansion that wasn't sensitive enough to
detect the error. Sorry.

I have no excuse, lame or otherwise, for not seeing the
connection to AR coatings.

=====

The good news is that there's a simpler explanation for
what's going on.

At each layer boundary, we invoke Snell's law

sin θ_i cos α_i
--------- = --------- [1]
sin θ_t cos α_i

1
= ------- [2]
1 + ε

Where we assume the complimentary angles (α) are small:

α_i := 90° - θ_i [3]
= small

α_t := 90° - θ_t [4]
= small

Expanding to lowest order:

1 - α^2_i / 2
--------------- = 1 - ε [5]
1 - α^2_t / 2

α^2_(i+1) - α^2_i = -2ε [6]

Where α_i is the angle in "this" layer and we write α_(i+1)
as a synonym for α_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 α^2 value in the final layer is zero, the ray is
horizontal, parallel to the layer boundary, and is stuck.
b) If the α^2 value in the final layer is 2ε, the angle will
go to zero when the ray tries to leave this layer, and
the ray will thereupon get stuck.
c) For any α^2 value greater than 2ε, 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 α^2, smaller by an amount 2ε.
d) For all α^2 values in the interior of the interval (0, 2ε),
the ray will be totally internally reflected.

Proof by induction on α^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.

Let's be clear: The ray does not bend "smoothly and asymptotically"
toward horizontal. To approach an asymptote, the step sizes would
need to get smaller and smaller as we go from layer to layer,
approaching the final layer. It seems to me this is the opposite
of what happens. The α^2 steps are equally spaced, which means
the α steps are unequally spaced, and get dramatically /larger/
(on a per-layer basis) as we approach the final layer.

There are geometrical reasons why it makes sense for α steps to
get bigger on a per-layer basis. We expect the ray to be turning
at a steady rate, a steady rate of change in α per unit distance
in the X direction ... but as the angles get shallower the ray
spends more time per layer, more X distance per unit Z distance.

So the math agrees with the qualitative physical and geometrical
intuition.

I *still* don't recommend the ray+layers model. The wave
mechanical (physical optics) model is vastly simpler and in
every way better, and lays a good foundation for future work
in optics, quantum mechanics, et cetera. However, if you insist
on using the ray+layers model, I reckon it can be made to work.