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

Richard Clark wrote:

Richard's following example is similar to a Z0-matched lossless
transmission line example.

1/4WL
1w XMTR---50 ohm ---+---100 ohm---+---200 ohm---+-200 ohm load
feedline feedline feedline
Pfor=3D1w Pfor=3D1.125w Pfor=3D1w
Pref=3D0w Pref=3D0.125w Pref=3D0w

Anyone capable of solving transmission line problems can verify the
above values which are the same as the values in Richard's example
although he doesn't realize it yet.

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

Richard, you failed to admit your errors on the previous example
and failed to appologize to me for all your insults. Now you
present yet another example before the first one was resolved.
And this example contains many of the errors that you made in
the earlier one. It's probably time to shut off your output
and engage in a little input before you embarrass yourself
any farther.

1w | 1/4WL |
laser-----air-----|---thin-film---|---Germanium---...
1st medium | 2nd medium | 3rd medium
n =3D 1.0 n =3D 2 n =3D 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:
(r=B2 + (t=B2 =B7 n2=B2 =B7 cos(theta-t) / n1=B2 =B7 cos(theta-i)=
)) =3D 1

It stands to reason that this can be quickly reduced without need to
use transcendentals for an angle of incidence of 0=B0 (which results in
a refractive angle of 0=B0). All that needs to be known are the
coefficients which for that same angle simplify to
r =3D 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 immateria=
l)

Uhhhhh Richard, the amplitude reflection coefficient, r, is:

r =3D (nt-ni)/(nt+ni) =3D (2-1)/(2+1) =3D 0.3333 =3D (4-2)/(4+2)

You are never going to get it right as long as you cannot even
calculate the reflection coefficient.

r=B2 =3D 0.445

Wrong again! r^2 =3D Reflectance, R =3D (0.3333)^2 =3D 0.1111
or 11.11%

t=B2 =3D 0.445

t^2 doesn't matter. The Transmittance, T =3D (1-R) =3D 0.8889
or 88.89%

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

That's correct but doesn't agree with your r^2 value. Let's say
11.11% or 111.1mW for four digit accuracy.

and
that part of the energy transmitted amounts to 89%
or otherwise expressed as:
890mW

Let's say 88.89% or 888.9mW for four digit accuracy.

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

You are ignoring the wave cancellation energy and getting the
wrong answers. You have proved my point better than I ever could
have.

The first internal reflection is 888.9mW*0.1111 =3D 98.76mW. When
that first internal reflection encounters Medium 1, 87.78mW will
be transmitted through to Medium 1. It is 180 degrees out of phase
with the first internal reflection so the two waves will start to
cancel. Thus 2*87.78 =3D 175.56mW will join the forward wave in
Medium 2. Even after this first re-reflection event, the forward
power in Medium 2 will be 1064.5mW on its way to 1125mW during
steady-state. It is readily apparent that you simply don't know
how to perform this analysis either during the transient state
or steady-state.

The steady-state values are easily known and yours are wrong. Have
you never solved a problem like this before?

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

You calculations are once again wrong. When you add up all the infinite

number of reflections and re-reflections, you will find that the sum of

all the internal transmissions into Medium 1 indeed does equal 111.1mw
thus totally canceling all of the reflections. Assuming a lossless
system, there will be one watt delivered into medium 3. The incident
power necessary for that event is 1w/0.8889 =3D 1.125w. All reflections
from the thin-film are canceled. Here are all the steady-state powers
calculated for you:

1w | 1/4WL |
laser-----air-----|---thin-film---|---Germanium---...
1st medium | 2nd medium | 3rd medium
n =3D 1.0 n =3D 2.0 n =3D 4.0
Pfor=3D1w Pfor=3D1.125w Pfor=3D1w
Pref=3D0w Pref=3D0.125w Pref=3D0w

Note that T*Pref2 =3D 0.8889*0.125 =3D 111.1mW, exactly the amount of
energy required to cancel the initial external reflection of 111.1mW.

In the 2nd medium, P1=3D888.9mW, P2=3DPref2*R =3D 125mW*0.1111 =3D 13.89mW
P1 + P2 + 2*sqrt(P1*P2) =3D 888.9 + 13.89 + 222.2 =3D 1125mW

Hecht's equations are correct after all. What you have proven is that
when
you ignore the interference energy, you will get the wrong answer.
--
73, Cecil http://www.qsl.net/w5dxp

Richard Clark wrote:
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

Richard, if you follow the leads of other gurus on this newsgroup,
you will 'ploink' me, never admit a mistake, and dupe yourself
into believing that you were right all along. Your "Extreme Failures
of Poor Concepts" could have been avoided if you guys had just read and

understood the following two quotes that I have posted multiple times.
What is it about, "... all 'lost' reflected intensity will appear as
enhanced intensity in the transmitted beam.", that you guys don't
understand?

www.mellesgriot.com/products/optics/oc_2_1.htm

"Clearly, if the wavelength of the incident light and the thickness
of the film are such that a phase difference exists between reflections

of p, then reflected wavefronts interfere destructively, and overall
reflected intensity is a minimum. If the two reflections are of equal
amplitude, then this amplitude (and hence intensity) minimum will be
zero."

"In the absence of absorption or scatter, the principle of conservation

of energy indicates all 'lost' reflected intensity will appear as
enhanced intensity in the transmitted beam. The sum of the reflected
and
transmitted beam intensities is always equal to the incident intensity.

This important fact has been confirmed experimentally."

http://micro.magnet.fsu.edu/primer/j...ons/index.html

"... when two waves of equal amplitude and wavelength that are 180-
degrees out of phase with each other meet, they are not actually
annihilated. All of the photon energy present in these waves must
somehow be recovered or redistributed in a new direction, according to
the law of energy conservation ... Instead, upon meeting, the photons
are redistributed to regions that permit constructive interference, so
the effect should be considered as a redistribution of light waves and
photon energy rather than the spontaneous construction or destruction
of light."
--
73, Cecil http://www.qsl.net/w5dxp

Ah but you must avoid upsetting GOD (oops I mean Richard.). A simple answer
about grounding given to a guy a while back put him into apoplectic fits,
necessitating the doubling of his meds. You see, "ground" ISN'T "ground".
That isn't "clever" enough for Richard. He promptly claimed that rf currents
flowing in a ground system were a "bad thing' and proceeded to "explain" it
all to us. I wish my elmer 30 years ago had been as "clever" as Richard. All
these books on the shelves simply are irrelevant, because nothing can be
simple, you see... It must be complicated, even if it IS wrong........

"W5DXP" wrote in message
oups.com...
Richard Clark wrote:
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

Richard, if you follow the leads of other gurus on this newsgroup,
you will 'ploink' me, never admit a mistake, and dupe yourself
into believing that you were right all along. Your "Extreme Failures
of Poor Concepts" could have been avoided if you guys had just read and

understood the following two quotes that I have posted multiple times.
What is it about, "... all 'lost' reflected intensity will appear as
enhanced intensity in the transmitted beam.", that you guys don't
understand?

www.mellesgriot.com/products/optics/oc_2_1.htm

"Clearly, if the wavelength of the incident light and the thickness
of the film are such that a phase difference exists between reflections

of p, then reflected wavefronts interfere destructively, and overall
reflected intensity is a minimum. If the two reflections are of equal
amplitude, then this amplitude (and hence intensity) minimum will be
zero."

"In the absence of absorption or scatter, the principle of conservation

of energy indicates all 'lost' reflected intensity will appear as
enhanced intensity in the transmitted beam. The sum of the reflected
and
transmitted beam intensities is always equal to the incident intensity.

This important fact has been confirmed experimentally."

http://micro.magnet.fsu.edu/primer/j...ons/index.html

"... when two waves of equal amplitude and wavelength that are 180-
degrees out of phase with each other meet, they are not actually
annihilated. All of the photon energy present in these waves must
somehow be recovered or redistributed in a new direction, according to
the law of energy conservation ... Instead, upon meeting, the photons
are redistributed to regions that permit constructive interference, so
the effect should be considered as a redistribution of light waves and
photon energy rather than the spontaneous construction or destruction
of light."
--
73, Cecil http://www.qsl.net/w5dxp

but of course ground isn't ground. there is no such thing as ground, or a
ground potential, or a perfect ground, or anything like that. and yes,
currents flowing in the ground can be a bad thing, but only if you don't
understand them and design to handle them.

"Zombie Wolf" wrote in message
...
Ah but you must avoid upsetting GOD (oops I mean Richard.). A simple
answer about grounding given to a guy a while back put him into apoplectic
fits, necessitating the doubling of his meds. You see, "ground" ISN'T
"ground". That isn't "clever" enough for Richard. He promptly claimed that
rf currents flowing in a ground system were a "bad thing' and proceeded to
"explain" it all to us. I wish my elmer 30 years ago had been as "clever"
as Richard. All these books on the shelves simply are irrelevant, because
nothing can be simple, you see... It must be complicated, even if it IS
wrong........

"W5DXP" wrote in message
oups.com...
Richard Clark wrote:
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

Richard, if you follow the leads of other gurus on this newsgroup,
you will 'ploink' me, never admit a mistake, and dupe yourself
into believing that you were right all along. Your "Extreme Failures
of Poor Concepts" could have been avoided if you guys had just read and

understood the following two quotes that I have posted multiple times.
What is it about, "... all 'lost' reflected intensity will appear as
enhanced intensity in the transmitted beam.", that you guys don't
understand?

www.mellesgriot.com/products/optics/oc_2_1.htm

"Clearly, if the wavelength of the incident light and the thickness
of the film are such that a phase difference exists between reflections

of p, then reflected wavefronts interfere destructively, and overall
reflected intensity is a minimum. If the two reflections are of equal
amplitude, then this amplitude (and hence intensity) minimum will be
zero."

"In the absence of absorption or scatter, the principle of conservation

of energy indicates all 'lost' reflected intensity will appear as
enhanced intensity in the transmitted beam. The sum of the reflected
and
transmitted beam intensities is always equal to the incident intensity.

This important fact has been confirmed experimentally."

http://micro.magnet.fsu.edu/primer/j...ons/index.html

"... when two waves of equal amplitude and wavelength that are 180-
degrees out of phase with each other meet, they are not actually
annihilated. All of the photon energy present in these waves must
somehow be recovered or redistributed in a new direction, according to
the law of energy conservation ... Instead, upon meeting, the photons
are redistributed to regions that permit constructive interference, so
the effect should be considered as a redistribution of light waves and
photon energy rather than the spontaneous construction or destruction
of light."
--
73, Cecil http://www.qsl.net/w5dxp

Cecil, W5DXP wrote:
"What is it about, "...all "lost" reflected intensity will appear as
enhanced intensity in the transmitted beam." that you guys don`t
understand?"

A quarter-wave intermediate impedance line section can perfectly match
different resistances at its ends.

When a match exists at the end of a line, there is no discontinuity and
the line appears as if infinite. There is no reflection from an
impedance match.

It`s a multiplicity of waves which seem gratuitous.

Best regards, Richard Harrison, KB5WZI

On Mon, 1 Aug 2005 11:13:34 -0500, (Richard
Harrison) wrote:

It`s a multiplicity of waves which seem gratuitous.
Amen brother!

Hi Richard,

This statement you respond to is one of those attempts to draw the
topic away from its focus to prove another point of match which
humorously was going to be proven with the original topic.

Now, if that sounds confusing, it is simply par for the course in how
we got here. And to extend the metaphor for par, we started out
Bowling for Dollars and have now ended up putting on the back nine.

Anyway, the optical analogue has proven to be a bust when I
demonstrated that reflection products do persist. This is not the
place to hash over that again. If you wish, you can consult my
original posting and the follow-ons for details.

73's
Richard Clark, KB7QHC

Richard Clark wrote:
Anyway, the optical analogue has proven to be a bust when I
demonstrated that reflection products do persist. This is not the
place to hash over that again. If you wish, you can consult my
original posting and the follow-ons for details.

But in your "proof", you superposed powers which is a no-no.
When 111.1mW interfers with 87.78mW, the result is not
(111.1-87.78). Since the associated E-fields are 180
degrees out of phase, the power equation must take the
interference into account.

Pref1 = 111.1mW + 87.78mW - 2* sqrt(111.1*87.78)

Pref1 = 111.1mW + 87.78mW - 197.5mW = 1.38mW

You subtract 87.78 from 111.1 and get 23.32. That value is
almost 17 times too high. All your math after that is invalid.
All except 1.38mW of reflections are canceled by that first
internal reflection. Your value of 23.32 is simply wrong.

RF engineers usually convert to voltage, perform the superposition,
and then calculate the total power. One doesn't have that luxury
when dealing with light so the power (irradiance) equations must
be used to obtain the correct results.
--
73, Cecil http://www.qsl.net/w5dxp

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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