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