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 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;
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....
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

It follows that at the second boundary there is less power available
due to the conservation of power observed at the first boundary. The
optics are hardly remarkable at this second boundary as they exhibit
nearly the same results, by percentages.

It then further follows that at the second boundary the amount of the
original energy passing through the first medium (air) has been
reduced to allow for a reflection returning to the first interface of
only 9.89841mW. 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:
9.99838mW - 9.89841mW
or
99.968µW

Now, admittedly, this appears all quite trivial and a strain on the
precision to say the least. However the "initial conditions" have
been stated to be an unrealistic example, or a perfect example, or a
thought experiment, or... what have you. As a consequence this minute
excess of energy stands in contrast to the expressed desire to prove
"total" cancellation. I need not remind the reader that this could
only get worse in the real world.

But how does this "perfect" result fare in the real world where this
1W laser has an un-cancelled reflection remainder? How does it fare
in an application to reduce reflections that the eye is sensitive to?
We can presume that to be either 510nM or 555nM, or somewhere in
between.

A laser of this power will have a beam size on the order of 1mm². If
you were to shine it against a white target that was exposed to the
noon day sun, on the equator, at either equinox, we could compare it
to the power of the sun's illumination. In this case, and being quite
generous, I typically describe the available power from the sun as
being 1000W/M². This laser then would, in terms of W/M², be quite
powerful at 1,000,000W/M². This is more than 1000 times more power
than the sun's exposure to the same target.

However, the sun does not radiate one wavelength of energy. So, if we
were to reduce the amount of sunlight confined to that in the Lased BW
at its operating wavelength (noted above); then let's consider that
the sun's power is in a BW of 2000nM and we are talking about (and I
will be MOST generous to offer an absurdly wide) Lased BW of 20nM. It
follows that the Lased power is thus 100,000 times brighter than the
sun in the same BW for the same wavelength. But this still neglects
that the sun's power is not evenly distributed throughout this 2000nM
BW.

I am not going to pencil whip this further. Let's simply return to
that un-cancelled power and look at it instead:
99.968µW
in that same 1mm², which if we cast to the same terms of comparison to
sunlight it becomes 99.968W/M² which is a tenth of the power of the
sun's total BW emission. But we are talking about brightness, and we
do not perceive the total BW of the sun's emission. As you may guess
I am going to use the same BW correction to find that the un-cancelled
reflection products are
TEN TIMES BRIGHTER THAN THE SUN!

Consider that this is being exceedingly generous (to subdue the
brightness of the Laser and its reflections) in contrast to the
actuality of even greater brilliance due to narrower BW and the actual
power distribution of the sun's energy across its spectrum of
emission.

Now, there is every chance I slipped a decimal point here or there in
my rushing through the numbers. I will leave that to others to decide
and exhibit my mistakes. If they are sufficiently in error, we can
revisit the conversions and equivalencies in viewing this un-canceled
reflection product. I bet it will still be quite brilliant and not
the blackness so prophesied.

Vision is so taken for granted, that few who venture into the field of
optics have the discipline to divorce their senses from the numbers at
the right time, or the common sense to re-invest sight into what would
be observed.

73's
Richard Clark, KB7QHC

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


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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

Richard, it would be nice if you retracted your hyperbole since
by now you have realized that everything above is correct, given
the previously stated boundary conditions. You have certainly
successfully demonstrated "The Failure of Poor Concepts in
Discussing Thin Layer Reflections".

I am going to use the same BW correction to find that the un-cancelled
reflection products are TEN TIMES BRIGHTER THAN THE SUN!

There's a magnitude of difference between the *amplitude* reflection
which you mistakenly used and the *power* reflection coefficient,
which you should have used. You will find that the two reflected waves
are exactly the same magnitude (0.01w) and 180 degrees out of phase and
thus cancel to a magnitude of zero in the direction of the source. As I
said before, the reflections from that thin-film are flat black. The
energy components from the wave cancellation join the forward wave
in the thin film.

P1 = 1w*0.99 = 0.99w
P2 = 0.0101w*0.01 = 0.000101w
P1 + P2 + 2*sqrt(P1*P2) = 0.99w + 0.000101w + 0.02w = 1.0101w
--
73, Cecil http://www.qsl.net/w5dxp


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