Returning again with a new example, described in text as:

1w | 1/4WL | |/
laser-----air-----|---Si Crystal---|---Al contact--- |/
1st medium | 2nd medium | 3rd medium |/
n = 1.0 n = 5.8275 n 6

whe
the laser is in the UV/deep Blue (355nm to 360nm)
the second medium is a single crystal of Silicon
the third medium is a reflective aluminum contact.

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 = 0.7071 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 = 0.7071 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 = 0.5000
t² = 0.5000
I presume that the remainder of the math can be agreed to exhibit:
that part of the energy reflected amounts to 50%
and that part of the energy transmitted amounts to 50%.

In this example (a farrago, to say the least), it is overwhelmingly
obvious that when the incident energy strikes the first interface that
half of it is reflected and half of it transmitted (a common
specification for this specie). Even more obvious is that the sum
total of all re-reflections between the interfaces (and it has been
absurdly guaranteed that no energy will ever transit the aluminum
mirror) will combine in the proper phase relationship to "totally"
cancel the first reflection.

Now, what to make of this black mirror? The same conservation of
energy has been maintained as has been exhibited in other threads of
this ilk. That conservation expressed above as both a commentary and
a formula is satisfied and "total" cancellation has been presented.

To wit, and in former expressions:

Observing the conservation of energy at the first interface:
X = 0.5000X + 0.5000X
Observing the conservation of energy at the second interface:
0.5000X = 0.5000X + 0.0000X

0.5000X == 0.5000X

Of course, to put the lie to this example, no mirror is completely
reflective. Some in these "debates" wanted to inject loss, or the
lack of perfection. This should satisfy those who need this smudgy
focus.

To brush aside those anticipated rejections requires only expressing
what fraction the "best" mirror could provide, and simple trim the
wavelength of the source to match the numbers. Let me offer:

We have a mirror that is only 98% reflective (it would exhibit a
refractive index more than 1000 in this example). By slipping the
wavelength 2nm shorter (actually there are other wavelengths that
support these indices, but this is as illustrative as any); then we
can allow for 49.5% reflection, a 50.5% transmission, a subsequent
summation of all reflections with the 98% mirror to once again
exhibit:
Observing the conservation of energy at the first interface:
X = 0.495X + 0.505X
Observing the conservation of energy at the second interface:
0.505X = 0.495X + 0.010X

0.495X == 0.495X

Choose another mirror efficiency and one need only perform like
wavelength adjustments to obtain the necessary energy.

As I've offered in other posts, the "total" cancellation is entirely
driven by very simple optics which resolve to obvious ratios in the
expression of conservation of energy.

And, of course, "total" is not. ;-)

Even with this pencil-whipping out 4 places, the reflection products
that remain (in the 5th place) in this example are vastly brighter
than the sun (for those who see this wavelength). Also note that the
complete exercise reveals through the conservation of energy that
there is a loss (due to the incomplete reflection of the in-efficient
mirror) and that all energies add to the energy applied to the system.

73's
Richard Clark, KB7QHC

Richard Clark wrote:
Returning again with a new example, described in text as:

1w | 1/4WL | |/
laser-----air-----|---Si Crystal---|---Al contact--- |/
1st medium | 2nd medium | 3rd medium |/
n = 1.0 n = 5.8275 n 6

As I've offered in other posts, the "total" cancellation is entirely
driven by very simple optics which resolve to obvious ratios in the
expression of conservation of energy.

Uhhhhhhh Richard, that is not a matched example so, of course,
there is no total cancellation of reflections. Why do you
think a mis-matched system would ever have total cancellation?

What can we do to make it a matched example?
Make n3 = 5.8275^2 = 33.93. I have no idea what medium 3 is
made out of but it also has to reflect 1/2 the energy incident
upon its surface. Now the reflectance (power reflection coefficient)
equals 0.5 and is the same at discontinuity A as it is at
discontinuity B. The 3rd medium will be considered to be infinite
for purposes of simplicity.

1w 1st medium | 1/4WL |
laser-----air-----|---2nd medium---|---3rd medium---...
n = 1.0 | n = 5.8275 | n=33.93
A B

Now it is matched and reflections are totally canceled. The
forward power in medium 1 is 1W. The reflected power in medium 1
is zero. The forward power in medium 2 is 2W. The reflected
power in medium 2 is 1W. The forward power in medium 3 is
1W. The reflected power in medium 3 is zero.

The above example is akin to the following transmission line
example with identical forward and reflected powers.

1/4WL
1W XMTR---50 ohm line---+---291.4 ohm line---+---1698 ohm load
A B
Pfor1=1W-- Pfor2=2W-- PL=1W
--Pref1=0W --Pref2=1W

This is a Z0-matched system with no reflections on the 50 ohm
line. The fact that half the power is reflected at each discontinuity
makes it easy to calculate and recognize.

************************************************** *********************
In a lossless matched system where half the power is reflected at each
discontinuity, the forward power in medium 2 will be exactly double
the forward power in medium 1.
************************************************** *********************

Please read that over until you understand it and then try to figure
out why. Hint: it is not a violation of the conservation of energy
principle. Soon the hole you have dug for yourself will be so deep,
you won't be able to reach your keyboard. :-)
--
73, Cecil http://www.qsl.net/w5dxp


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