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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 + É
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 - É [5]
1 - Éø^2_t / 2
Éø^2_(i+1) - Éø^2_i = -2É [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 2É, 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 2É, 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 2É.
d) For all Éø^2 values in the interior of the interval (0, 2É),
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.