Re: REFRACTION - REFLECTION

[David Bowman <dbowman@tiger.gtc.georgetown.ky.us>]

Regarding what Tom Wayburn wrote:
> ...                                        .  Could someone, now, put
> that in other words?   Perhaps we could pretend that light is waves and
> water is stuff.  Or that light is particles and water is too.   But one
> is going to have to cook  up an analogy that is fascinating, compelling,
> and easily remembered.  Perhaps, on the day following the lecture, the
> class could be asked, "How many related that explanation to someone
> else?"

It's been a while since I last taught E & M but as I recall I think that the
explanation goes as follows:  (BTW, I doubt that this explanation will work
for a high school level class.)

Consider the radiation incident on the interface as a simple monochromatic 
plane wave for convenience.  Let subscripts 1 & 2 label quantities referring
to regions 1 & 2 on each side of the interface.  The frequency must be the
same for the wave on both sides of the interface.  This means that
[lambda]_1*n_1 = [lambda]_2*n_2 (since frequency f = c/([lambda]*n)) where n
is the index of refraction.  The wave vector k on each side is related to the
wavelength by |k| = 2*[pi]/[lambda] where |k| is the wave number which is the
magnitude of the wave vector.  This means that |k_1|/n_1 = |k_2|/n_2.  Now
let there be one prime on the component of a wave vector which is
perpendicular to the interfacial surface, and let there be two primes on the
component of the wave vector which is parallel to the interfacial surface.
This means that:
((k'_1)^2 + (k''_2)^2)/(n_1)^2 = ((k'_2)^2 + (k''_2)^2)/(n_2)^2 .  Because
of the required continuity conditions on the boundary surface we must have
that k''_1 = k''_2 = k''(constant).  This is because the pattern of crests
and troughs on the boundary surface must be the same when viewed from both
sides of the surface.  Substituting k'' for both k_1'' and k_2'' and solving
for k'_2 gives: k'_2 =(n_2/n_1)*sqrt((k'_1)^2 - ((n_1/n_2)^2 - 1)*k''^2) .

Suppose that n_1 > n_2 so that the wave is going from a lower index region
(1) into a higher index region (2).  In this case the k''^2 term in the sqrt
is negative.  If the frequency of the wave is fixed then on each side we
have that |k| is fixed as the angles of propagation are varied.  As the angle
of incidence is made more shallow then k_1' is reduced as k'' is increased in
such a way that the magnitude |k_1| remains constant.  At a sufficiently
shallow angle (more shallow than the critical angle) then the positive
term in the sqrt becomes less positive than the negative term becomes
negative.  This makes the argument of the sqrt negative and thus makes k'_2
purely imaginary.  This means that the wave in region 2 does not
propagate in region 2 perpendicular to the interface.  The imaginary
component of the wave number makes the wave amplitude decay exponentially
with distance from the interface so that the wave is evanescent in region 2
with an e-folding penetration depth given by 1/|i*k'_2|.  Thus the
wave cannot penetrate too deeply into region 2 and is reflected.  In region
1 the boundary conditions always allow for k'_1 to be either positive or
negative, and in general the wave is a superposition of a positive k'_1
incident part and a negative k'_1 reflected part.  When the wave is
evanescent with an imaginary k'_2 then the amplitude of both of these
superposed wave components in region 1 have the same amplitude.  This makes
the perpendicular (to the interface) component of the energy flux vector
vanish on both sides of the interface in this case (required by energy
conservation).  In the evanescent case the wave's behavior in region 2 is
just like the wave function of a quantum particle in the classically
forbidden side of a classical turning point.  There is always some
tunnelling-type penetration, but there is no net flux into or through region
2 unless that region has a finite thickness and a transmitted wave on the
other side of the finite thickness barrier is formed.

Refraction is fundamentally a *wave* behavior and, as such, any analogy using
particle-type models such as lawn mowers, tricycles, etc. have nothing to do
with the effect.  If the messy details of the wave nature of the phenomenon
are to be suppressed because they are too advanced for a lower level class
then an appeal to Fermat's principle can be made.  This is because any
wave propagation phenomena obeys Fermat's principle in the short wavelength
limit and then the analysis can be done entirely in terms of rays.  In this
case refraction is an automatic result of changing the speed of the rays
in going from one fixed place to another fixed place while minimizing the
time of travel for those rays.

David Bowman
dbowman@gtc.georgetown.ky.us