Richard Clark wrote:
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 A n = 2 B n = 4.04

I have added points 'A' and 'B' to Richard's diagram.

Let's make the math simple by having the index of refraction,
n, of the 3rd medium be equal to 4.0. The power reflection
coefficient is 0.1111 and the power transmission coefficient
is 0.8889. All powers will be stated in mW.

Since n2 = sqrt(n1*n4), the system will be reflectionless
in the 1st medium. Proof of that assertion will be left
to the reader but that is a necessary and sufficient
condition for a lossless system.

Here is the transient state of the system an instant
after the first internal reflection from point B arrives
back at point A.

1st medium 2nd medium 3rd medium
A B
1000mW----| |
|----888.9mW----|
111.1mW----| |----790.1mW
|----98.76mW----|
87.79mW----| |
|----10.97mW

Since the 2nd medium is 1/4WL, in the 1st medium, the
111.1mW external reflection will be 180 degrees out
of phase with the 87.79mW internal transmission leaving
very little of the external reflection uncanceled. A
quick calculation indicates 1.37mW left uncanceled.

When the second internal reflection from point B
arrives back at point A, more wave cancellation
will occur. After very few iterations of reflections,
the 111.1mW external reflection will be canceled.
As in the earlier example, the result is flat black
for reflections in the 1st medium.

Richard Clark sez:
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!

1800 TIMES MORE POWER THAN THE SUN from flat black????
Quick, Richard, get a patent on that process.

Someone sent me an email wondering if Richard C. has a death
wish. :-)
--
73, Cecil http://www.qsl.net/w5dxp