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

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

You are correct that in this system (1/4-wave matching layer), the
total reflectance goes to zero.

However, your understanding of superposition is wrong. You
CANNOT superimpose POWERS, or even talk about the "power" of various
reflections in the same media. You can only add the wave amplitudes
(electric fields). THEN you take the total amplitude and square that
to get the power.


Tor
N4OGW

On 26 Jul 2005 13:08:53 -0500, wrote:

You are correct that in this system (1/4-wave matching layer), the
total reflectance goes to zero.

However, your understanding of superposition is wrong. You
CANNOT superimpose POWERS, or even talk about the "power" of various
reflections in the same media. You can only add the wave amplitudes
(electric fields). THEN you take the total amplitude and square that
to get the power.

Hi Tor,

This is all taken care of in the equation for energy conservation that
I posted. The solution is valid barring a simple error of subtraction
that should read:
110mW - 98mW
or
12mW
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
1200 TIMES MORE POWER THAN THE SUN!

All should note that if you have only 98mW (expressed in the correct
terms of course as the area terms have been abandoned) available to
offset 110mW (again with the same proviso) then it stands that Total
Cancellation is always impossible. As the two energy levels are not
equal, there is no other answer.

This is not to confuse the solution with the practical reduction. On
the other hand, when the remaining reflection products contains so
much more power than the sun's exposure, it is doubly amusing to see
it expressed as "0."

73's
Richard Clark, KB7QHC

Hi Tor,

This is all taken care of in the equation for energy conservation that
I posted. The solution is valid barring a simple error of subtraction
that should read:
110mW - 98mW

Not sure which indices you are now referring to: n_3=4.0 or 4.04? With
4.0, there is zero reflected light. With 4.04 there is a small amount.

Tor
N4OGW

On 26 Jul 2005 14:54:49 -0500, wrote:

Not sure which indices you are now referring to: n_3=4.0 or 4.04? With
4.0, there is zero reflected light. With 4.04 there is a small amount.

Hi Tor,

It doesn't matter one iota. You can attempt to achieve a perfect
1:2:4 correlation, or any similar relationship that satisfies the
square laws so demanded.

The point of the matter is with such design characteristics fulfilled,
then each interface reflects/transmits in the same proportion. With
each reflecting/transmitting in the same proportion, the second
interface, by the actions of the first, must have less incident upon
it. It then follows that the same proportion of less incident is not
enough to "Totally Cancel" the first interface reflection.

Observing the conservation of energy at the first interface:
X = 0.89X + 0.11X
Observing the conservation of energy at the second interface:
0.89X = 0.792X + 0.098X

0.098X 0.11X

What is even more obvious is that following the second interface, you
have lost roughly 20% of that incident upon the first interface.

There is not enough energy at the second interface, reflected back, to
"Totally Cancel" the energy in the reflection of the first interface.
There can be no other outcome.

That difference yields first interface reflection products that have:
1200 TIMES MORE POWER THAN THE SUN!

73's
Richard Clark, KB7QHC

Richard Clark wrote:

On 26 Jul 2005 14:54:49 -0500, wrote:
Not sure which indices you are now referring to: n_3=4.0 or 4.04? With
4.0, there is zero reflected light. With 4.04 there is a small amount.

It doesn't matter one iota.

Of course it matters. The 1,2,4 combination meets the necessary and
sufficient conditions for the elimination of reflections in medium 1
assuming a steady-state lossless system, i.e. perfectly matched.

The point of the matter is with such design characteristics fulfilled,
then each interface reflects/transmits in the same proportion. With
each reflecting/transmitting in the same proportion, the second
interface, by the actions of the first, must have less incident upon
it.

That is simply not true, Richard. You obviously have not read my
Melles-Griot web posting and/or simply don't understand it. The
forward power in medium 2 is greater than the forward power in
either medium 1 or medium 3. You keep making that same mistake
over and over and over. Why don't you simply take time to
understand the truth?

What is even more obvious is that following the second interface, you
have lost roughly 20% of that incident upon the first interface.

But you have gained the energy involved in the wave cancellation
in medium 1. The forward power in medium 2 is higher than it is
in medium 1. You don't actually believe the BS by some gurus on
this newsgroup that standing waves don't contain any energy, do you?

There is not enough energy at the second interface, reflected back, to
"Totally Cancel" the energy in the reflection of the first interface.
There can be no other outcome.

Only in your mind, Richard, only in your mind. The outcome in reality
is complete cancellation of the reflections in medium 1 assuming
1,2,4 indices of refraction.

That difference yields first interface reflection products that have:
1200 TIMES MORE POWER THAN THE SUN!

Somebody get the net!
--
73, Cecil http://www.qsl.net/w5dxp

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Richard Clark wrote:

wrote:
However, your understanding of superposition is wrong. You
CANNOT superimpose POWERS, or even talk about the "power" of various
reflections in the same media. You can only add the wave amplitudes
(electric fields). THEN you take the total amplitude and square that
to get the power.

This is all taken care of in the equation for energy conservation that
I posted.

Tor, by misunderstanding my posting, you have just fed the monster. :-)
--
73, Cecil http://www.qsl.net/w5dxp

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Tor, by misunderstanding my posting, you have just fed the monster. :-)

I see :O Well, I can't get everyone to understand basic physics.

Tor
N4OGW

wrote:
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


You are correct that in this system (1/4-wave matching layer), the
total reflectance goes to zero.

However, your understanding of superposition is wrong.

Powers obey the equation P1 + P2 + 2*(P1*P2)cos(theta)
when the fields are superposed. 2*(P1*P2)cos(theta) is
known as the "interference" term. That interference term
automatically adjusts the power equation IN ORDER TO AVOID
SUPERPOSITION OF POWERS. That's what I did.

You CANNOT superimpose POWERS,

I would not even dream of superposing powers. Perhaps you
don't understand the role of the interference term in the
power equation which is to AVOID the superposition of
powers.

P1 + P2 would be superposing powers. I didn't do that.

P1 + P2 + 2*sqrt(P1*P2)cos(theta) is NOT superposing powers.

or even talk about the "power" of various
reflections in the same media. You can only add the wave amplitudes
(electric fields). THEN you take the total amplitude and square that
to get the power.

Eugene Hecht would be surprised to hear you say such since he
talked about the powers (irradiance) of various reflections in
the same media.

I did not superpose powers. I used Hecht's equations for powers
(irradiance). When E-field1 is superposed with E-field2, powers
obey the equation:

Ptotal = P1 + P2 + 2*sqrt(P1*P2)cos(theta)

where (theta) is the angle between the E-fields, P1 is the power
associated with E-field1, and P2 is the power associated with
E-field2. Reference: _Optics_, by Hecht, equation 9.14

Since the above is a steady-state matched situation, the forward
power in second medium involves total constructive interference:

So Ptotal = P1 + P2 + 2*sqrt(P1*P2)

where P1 = Pfor1*T and P2 = Pref2*R
T is the Transmittance and R is the reflectance
These are very similar to the S-parameter power equations.

Reference "total constructive interference" in _Optics_, by Hecht
equation 9.15. The only difference in irradiance and power is
the unit-area which, for our example, is arbitrary.
--
73, Cecil http://www.qsl.net/w5dxp

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A B
1000mW----| |
|----888.9mW----|
111.1mW----| |----790.1mW
|----98.76mW----|
87.79mW----| |
|----10.97mW
^^^^^

I would not even dream of superposing powers. Perhaps you
don't understand the role of the interference term in the

Then don't draw diagrams like above labeled with powers of
multiple reflections bouncing around in the same media

Eugene Hecht would be surprised to hear you say such since he
talked about the powers (irradiance) of various reflections in
the same media.

Only when dealing with the interference of TWO waves.
He works with electric field when discussing anti-reflection
layers. See 9.7.1

I'm not saying the power equation you give is wrong, but it
is correct for just two waves. With more than two waves, it
obviously becomes more complicated


Tor
N4OGW