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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Cecil Moore wrote:
Forward power in the thin-film is 1.010101010101010101 watts.
Reflected power in the thin-film is 0.10101010101010101 watts.

Darn, missed a zero. Should be 0.010101010101010101 watts.
--
73, Cecil, http://www.qsl.net/w5dxp


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On Thu, 30 Jun 2005 15:52:23 -0500, Cecil Moore
wrote:
Darn, missed a zero. Should be 0.010101010101010101 watts.
How could it possible matter?

On Thu, 30 Jun 2005 15:47:19 -0500, Cecil Moore
wrote:
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.

Hi Tor,

So, do you see by this example what I describe as the devil's in the
details? Absolute proof of complete cancellation using referenced
math, which is then abstracted to another field to prove imponderables
(mental conceptual examples).

And what purpose is served through this prostitution of citations?
That thin films completely cancel reflections? - they do not. That
optical interfaces exhibit complete transmission? - they do not. What
makes the proofs of these unproven phenomenon necessary such as to
dismiss the correct answers? Dare we follow the thread of a suspect
theory to arrive at other, equally flawed steps along the way?

73's
Richard Clark, KB7QHC

Richard Clark wrote:
And what purpose is served through this prostitution of citations?

Well Richard, since you are being so picky, I am
going to have to insist that you prove that you
exist. :-)
--
73, Cecil, http://www.qsl.net/w5dxp


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