Richard Clark wrote:
We begin with the model found at:
http://www.qsl.net/w5dxp/weblaser.GIF

Which has been variously described in text as:
1w | 1/4WL |
laser-----air-----|---thin-film---|---glass---...
1st medium | 2nd medium | 3rd medium
n=1.0 n=1.2222 n=1.4938
Pfor=1w Pfor=1.0101w Pfor=1w
Pref=0w Pref=0.0101w Pref=0w
which, of course, is in error

Employing what has been described as the cogent math:
Imin = I1 + I2 - 2*SQRT(I1*I2) = 0
or:
Ptot = P1 + P2 + 2*Sqrt(P1*P2)cos(theta)
or:
Ptot = P1 + P2 - 2*SQRT(P1*P2)

However, this math is far in advance of the necessary ground work to
first establish I1|P1 and I2|P2 in the first place. It may be noted
that they have been "assigned" values. Unfortunately the derivation
of those values, lacking reference, are either pure whim and fantasy,
or they were found by a formula, as of yet unreported. This will not
pass.

The only assigned value is the one watt into the glass. All
other values are derived by formula. Those formulas are trivial
but I will report them here and now.

The formula either discreetly hidden or completely unknown is a class
of optics math called Fresnel's Equations. I will likewise limit
their discussion. It requires the computation of the transmission
coefficient, and the reflection coefficient both of which are then
expressed in the energy equation. 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 = 0.0999918999... a value that is the limit of an asymptote;

I rounded that value to 0.1 which is certainly within engineering
accuracy. The n1 value is actually 1.222222222222... but I thought
four decimal places would be enough. :-)

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 = 0.9000081000... 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 = 0.00999838....
t² = 0.81001458....

Since we are dealing with powers, there's no need for a transmission
coefficient. T, the transmittance can be calculated from the power
reflection coefficient, R, the reflectance. The reflectance is:

R = r^2 = (0.1)^2 = 0.01 i.e. 1% of the incident power is reflected.
T = (1-r^2) = 0.99 i.e. 99% of the incident power is transmitted.

I chose the index of refractions to obtain those two values which
are very easy to work with - almost no possibility of mistakes.
(I appologize for under-estimating your ability to make mistakes. :-)

I presume that the remainder of the math can be agreed to exhibit:
that part of the energy reflected amounts to 0.999838%
or otherwise expressed as:
9.99838mW
and
that part of the energy transmitted amounts to 99.000162%
or otherwise expressed as:
990.00162mW

Richard, you should know better than to use the amplitude reflection
and transmission coefficients on powers. You should be using the power
reflection coefficient, R, which is the reflectance and is the square
of the amplitude reflection coefficient. The power transmission
coefficient, T, is the transmittance and is one minus the reflectance.
Because of that math blunder, your results are incorrect. I am deleting
the rest of your posting for that reason. Please retry your calculations
using r = 0.1, R = 0.01, and T = 0.99

The amplitude reflection coefficient, r, applies to both E-fields
and H-fields, so you have to square it to get the power reflection
coefficient. rho, the transmission line reflection coefficient,
applies to voltage, not power. You have to square rho to obtain
the power reflection coefficient which is rho^2.

So rounding the amplitude reflection coefficient, r, to 0.1 and
squaring it to get the power reflection coefficient, R = 0.01,
we can see that 1% of the power is reflected at each boundary
in the example above. What incident forward power does it take
to get 1w into the glass?

1/.99 = 1.010101 ... watts

That's exactly the forward power in the thin-film in my example.
(There's that formula that you were demanding.)

And since the entire system is assumed to be loss-less, one watt
steady-state is required from the laser. So where does the extra
power come from in the forward power in the thin-film? The
extra 0.010101 ... watts in the thin film is the reflected power
that is 100% reflected due to wave cancellation between the
externally reflected wave and the internally reflected wave.

The external reflection is simply 1w*0.01 = 0.01w
where 0.01 is the power reflection coefficient, R.

The internal reflection is 0.010101w. To obtain the
amount of power transmitted through the first boundary,
multiply by 0.99, the power transmission coefficient.
Thus the interferring wave becomes 0.010101*0.99 = 0.01w

The magnitudes of the external reflection and the internal
transmission are the same. The thin-film being 1/4WL causes
those two reflected wave components to be 180 degrees out
of phase. The two reflected wave components are cancelled.
The energy components in those two waves join the forward
wave in the thin-film. Quoting the Melles-Groit web page.

"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."
--
73, Cecil http://www.qsl.net/w5dxp


----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups
---= East/West-Coast Server Farms - Total Privacy via Encryption =---