Now we venture to new materials, and in this case a solar cell,
described in text as:

1w | 1/4WL |
laser-----air-----|---thin-film---|---Germanium---...
1st medium | 2nd medium | 3rd medium
n = 1.0 n = 2 n = 4.04

where the second medium might be Arsenic trisulfide glass or Lanthanum
flint glass.

When you take the intensity times the area for both the reflected and
refracted beams, the total energy flux must equal that in the incident
beam. That equation appears as:
(r² + (t² · n2² · cos(theta-t) / n1² · cos(theta-i))) = 1

It stands to reason that this can be quickly reduced without need to
use transcendentals for an angle of incidence of 0° (which results in
a refractive angle of 0°). All that needs to be known are the
coefficients which for that same angle simplify to
r = 0.667 a value that is the limit of an asymptote;
it is also invested with either a + or - sign depending
upon the polarization (another issue that was discarded
in the original discussion as more unknown than immaterial)
t = 0.667. a value that is the limit of an asymptote;
here, too, there are polarization issues we will discard as
before. All this discarding comes only by virtue of squaring:
r² = 0.445
t² = 0.445
I presume that the remainder of the math can be agreed to exhibit:
that part of the energy reflected amounts to 11%
or otherwise expressed as:
110mW
and
that part of the energy transmitted amounts to 89%
or otherwise expressed as:
890mW

It follows that at the second boundary there is less power available
due to the conservation of power observed at the first boundary. They
exhibit these results, by percentages:
that part of the energy reflected amounts to 11%
or otherwise expressed as:
98mW
and
that part of the energy transmitted amounts to 89%
or otherwise expressed as:
792mW

Needless to say, that same first interface is going to conserve energy
by the total of refraction and reflection being equal to the energy
incident upon it. I will skip that to allow ALL of this second
reflection to "try" to totally cancel the first reflection:
110mW - 92mW
or
18mW

But how does this "perfect" result fare in the real world where this
1W laser has an un-cancelled reflection remainder? How does it fare
in an application to reduce reflections that the Solar Cell is
sensitive to?

A laser of this power will have a beam size on the order of 1mm². I
typically describe the available power from the sun as being 1000W/M².
This laser then would, in terms of W/M², be quite powerful at
1,000,000W/M². This is more than 1000 times more power than the sun's
exposure to the same target.

However, the sun does not radiate one wavelength of energy. So, if we
were to reduce the amount of sunlight confined to that in the Lased BW
at its operating wavelength; then let's consider that the sun's power
is in a BW of 2000nM and we are talking about (and I will be MOST
generous to offer an absurdly wide) Lased BW of 20nM. It follows that
the Lased power is thus 100,000 times brighter than the sun in the
same BW for the same wavelength. But this still neglects that the
sun's power is not evenly distributed throughout this 2000nM BW.

I am not going to pencil whip this further. Let's simply return to
that un-cancelled power and look at it instead:
18mW
in that same 1mm², which if we cast to the same terms of comparison to
sunlight it becomes 18000W/M² which is 18 TIMES THE POWER OF THE SUN's
total BW emission. As you may guess I am going to use the same BW
correction to find that the un-cancelled reflection products have
1800 TIMES MORE POWER THAN THE SUN!

73's
Richard Clark, KB7QHC