The complete text referenced here can be accessed at:

http://eznec.com/misc/food_for_thought/

The first one is "Forward and reverse power" and is much
easier to read if it is copied-and-pasted into Word.

The first comment is just a nit: "(the current isn't
transformed by the half wavelength line either.)"

A 1/2 wavelength of transmission line reverses the
phase of the current, certainly a 180 degree
"transformation".

The main point centers around this false statement:
"While the nature of the voltage and current waves when
encountering an impedance discontinuity is well understood,
we're lacking a model of what happens to this "reverse power"
we've calculated."

Absolutely false! We are not lacking a model. We have time-
tested models that people with closed minds simply refuse to
consider. One such model is the s-parameter analysis presented
in HP's App Note 95-1, available on the web. Quoting: "Another
advantage of s-parameters springs from the simple relationship
between the variables a1, a2, b1, and b2, and various *POWER
WAVES*. ... s-parameters are simply related to power gain and
mismatch loss, quantities which are often of more interest
than the corresponding voltage functions." (emphasis mine)

The model is there. Concrete brains refuse to take a look
and instead call it "gobbledegook" (sic).

There's another model that agrees 100% with an s-parameter
analysis and it is from the field of optics, covered in
detail in _Optics_, by Hecht. Non-glare glass works the same
way as a 1/4WL matching section in a transmission line.
Again the model is there, just ignored by gurus on this
newsgroup which leads to demonstrably wrong conclusions
on their parts.

What happens to the energy in EM light waves
has been known and understood for many decades and cannot
be understood without an understanding of the laws of physics
governing interference between EM waves. Likewise, what happens
to the energy in EM RF waves cannot be understood without an
understanding of those same laws of physics.

There are many posters to r.r.a.a who have no clue about the
laws of physics governing interference between EM waves. That
ignorance is the entire problem with this discussion. Solution:
Alleviate the ignorance.

Quoting again from Roy's web page:
'ANY MODEL PRESENTED TO ACCOUNT FOR WHAT HAPPENS TO "FORWARD" AND
"REVERSE" POWER AT TRANSMISSION LINE ENDS HAD BETTER GIVE RESULTS THAT
AGREE WITH THE ABOVE TABLE.'

And, of course, the two above power/energy models agree exactly with
Roy's table but that doesn't seem to matter one iota. They are still
"gobbledegook" (still sic).
--
73, Cecil, http://www.qsl.net/w5dxp

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On Thu, 30 Jun 2005 11:05:57 -0500, Cecil Moore
wrote:
Non-glare glass works the same
way as a 1/4WL matching section in a transmission line.
One trivial example that you thoroughly bumbled through every mistake
possible, arriving at no answer before abandoning.

Richard Clark wrote:

Cecil Moore wrote:

Non-glare glass works the same
way as a 1/4WL matching section in a transmission line.

One trivial example that you thoroughly bumbled through every mistake
possible, arriving at no answer before abandoning.

It was an example for Jim Kelley to respond to.
He declined to discuss it, nobody else (including you)
responded with any technical content, so the thread
was abandoned. If you have anything technical to
contribute, feel free to fire it up again.

That happens a lot on this newsgroup. When someone
realizes that he is about to be proven wrong in
public, he simply goes away - human nature.
--
73, Cecil, http://www.qsl.net/w5dxp

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Cecil Moore wrote:
Richard Clark wrote:

Cecil Moore wrote:

Non-glare glass works the same
way as a 1/4WL matching section in a transmission line.


One trivial example that you thoroughly bumbled through every mistake
possible, arriving at no answer before abandoning.


It was an example for Jim Kelley to respond to.
He declined to discuss it, nobody else (including you)
responded with any technical content, so the thread
was abandoned. If you have anything technical to
contribute, feel free to fire it up again.

That happens a lot on this newsgroup. When someone
realizes that he is about to be proven wrong in
public, he simply goes away - human nature.

Cecil, you never actually prove anyone wrong,
you just get excited and irrational whenever
anyone disagrees with you. You're confusing the
fear of being proven wrong with just giving
up in disgust at your silly antics.
73,
Tom Donaly, KA6RUH

Tom Donaly wrote:
You're confusing the
fear of being proven wrong with just giving
up in disgust at your silly antics.

More of the same personal stuff. How about some
technical content for a change?
--
73, Cecil, http://www.qsl.net/w5dxp

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On Thu, 30 Jun 2005 12:14:36 -0500, Cecil Moore
wrote:
More of the same personal stuff. How about some
technical content for a change?

http://www.qsl.net/w5dxp/weblaser.GIF
(which I am sure you will soon hustle off in embarrassment)

For an angle of incidence of 30°

How much power is reflected from Surface A (first incidence)?

How much power is reflected from Surface B (first incidence)?

How much power is transmitted through all interfaces?

:-)

On Thu, 30 Jun 2005 11:43:24 -0500, Cecil Moore
wrote:
That happens a lot on this newsgroup. When someone
realizes that he is about to be proven wrong in
public, he simply goes away - human nature.
No doubt this is a set up for abandoning my other question
in this thread
Message-ID:

Richard Clark wrote:

On Thu, 30 Jun 2005 12:14:36 -0500, Cecil Moore
wrote:

More of the same personal stuff. How about some
technical content for a change?


http://www.qsl.net/w5dxp/weblaser.GIF
(which I am sure you will soon hustle off in embarrassment)

For an angle of incidence of 30°

The angle of incidence is always 90 degrees. It is drawn that
way because it cannot be drawn in one dimension. This is typical
of physics textbook drawings. An angle of incidence of 30 degrees
is irrelevant to this particular example.

How much power is reflected from Surface A (first incidence)?

The reflectance is 0.01, so 1%. First incidence 0.01 watts

How much power is reflected from Surface B (first incidence)?

The reflectance is 0.01, so 1%. First incidence is 0.0099 watts
Steady-state after re-reflection is 0.01010101 watts

How much power is transmitted through all interfaces?

One watt net to the "load".
The steady-state forward power in the thin film is 1.010101
watts of which 0.010101 is a steady-state internal reflection
from surface B.
--
73, Cecil, http://www.qsl.net/w5dxp

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On Thu, 30 Jun 2005 13:43:13 -0500, Cecil Moore
wrote:
An angle of incidence of 30 degrees
is irrelevant to this particular example.
Changing the question to suit the answer, Hmm? As if it mattered!

How much power is reflected from Surface A (first incidence)?
The reflectance is 0.01, so 1%. First incidence 0.01 watts
depending upon the polarization r¬ = 1.583% and r|| = 0.5485%
Splitting the difference (1.066%)
0.0107W
or
0.9893W @ 24° available at the next interface
How much power is reflected from Surface B (first incidence)?
The reflectance is 0.01, so 1%. First incidence is 0.0099 watts
depending upon the polarization r¬ = 1.3381% and r|| = 0.7099%
Splitting the difference (1.024%)
0.0101W
or
0.9792W @ 19° available at the next interface
How much power is transmitted through all interfaces?
One watt net to the "load".
Already provided as 0.9792W

**********************************

Since you couldn't answer the original question, let's explore how
accurate your answer to your own question was:

How much power is reflected from Surface A (first incidence)?
The reflectance is 0.01, so 1%. First incidence 0.01 watts
depending upon the polarization r¬ = 0.9999% and r|| = 0.9999%
Splitting the difference (0.9999%)
0.0100W
or
0.9900W @ 0° available at the next interface
How much power is reflected from Surface B (first incidence)?
The reflectance is 0.01, so 1%. First incidence is 0.0099 watts
depending upon the polarization r¬ = 0.9998% and r|| = 0.9998%
Splitting the difference (0.9998%)
0.0099W
or
0.9801W @ 0° available at the next interface
How much power is transmitted through all interfaces?
One watt net to the "load".
Already provided as 0.9801W

But, hey, what's 2% error in a conservation of energy equation? You
can prove anything (especially your absolute proofs) if you simply
discard precision. Pons and Fleishman proved cold fusion by throwing
away fewer digits than you.

Well, for 1W of light and presuming cancellation (you cannot achieve
full cancellation); this leaves 100µW of light reflected from a
non-reflecting layer - which is quite bright.

So, energy is conserved, and there is no such thing as complete
cancellation.

By the way, the math is available from:
Hecht, Eugene, Optics, 2nd Ed, Addison Wesley, 1987
or perhaps you should invest in:
Hecht, Eugene, Optics, Schaum's Outline Series, McGraw-Hill ,1975

Richard Clark wrote:
But, hey, what's 2% error in a conservation of energy equation?

The error is yours. During steady-state, the forward power
in the thin film is 1.0101 watts. 1% of that is 0.0101 watts.
1.0101 - 0.0101 = one watt delivered to the "load". 100%
accuracy is guaranteed because it is a mental conceptual
exercise. To summarize:

Forward power in air is one watt.
Reflected power in air is zero watts.

Forward power in the thin-film is 1.010101010101010101 watts.
Reflected power in the thin-film is 0.10101010101010101 watts.

Power delivered through Surface B is exactly one watt, exactly
the output of the laser. Exactly the difference between the
steady-state forward and reflected power in the thin-film.

Well, for 1W of light and presuming cancellation (you cannot achieve
full cancellation)

That's the great thing about a mental conceptual example. Full
cancellation is guaranteed.

So, energy is conserved, and there is no such thing as complete
cancellation.

True for real world stuff. False for mental conceptual examples
like the one being discussed.
--
73, Cecil, http://www.qsl.net/w5dxp


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