Richard Clark wrote:
Let's see, ten billion angstroms equal 1 meter. If we do a simple
conversion we find that your laser light operates at a wavelength of
0.3 millimeters (thicker than a hair).

Yep, I misread the units from a chart in Reference Data for Radio
Engineers. I accidentally picked a point in the low infrared region.
Macular Degeneration is playing havoc with my eyesight. But, FYI,
infrared lasers are readily available.

And EM is still EM no matter what the frequency. If it will make you
happy, let's discuss the familiar helium-neon laser whose wavelength
is 632.8 nm. Incidentally, you said anstroms are not a measure of
frequency. FYI, neither is wavelength, your preferred units. :-)

It really doesn't matter what EM frequency we discuss. We can cause
reflections (glare) at any frequency. But the experiment I previously
discussed used a coherent laser light source normal to the plane of
the thin-film. The two problems solved in the following diagrams are
virtually identical (where 'n' is the index of refraction).

100w laser-----air------|----1/4WL thin-film----|--flat black
n=1.0 n=1.225 n=1.5

100w XMTR--50 ohm coax--|--1/4WL 61.2 ohm coax--|--75 ohm load

The physical reflection coefficient magnitude is the same in both
cases, ~0.1. If one understands how the thin-film eliminates
reflections (glare) then one understands how the match point eliminates
reflections an RF transmission line. How the thin-film works is explained at:

http://www.mellesgriot.com/products/optics/oc_2_1.htm

Incidentally, J. C. Slater, author of _Microwave_Transmission_,
understood all this stuff way back in 1942. Here's a couple of
quotes: "Thus a light wave incident on a discontinuity between
two media, as a surface of separation of air and glass, is reflected.
If the surfaces are separated by a quarter wave film of material
whose index of refraction is the geometric mean of the indexes of
the two media, however, the reflection can be eliminated;"

And for RF transmission lines: "We can get a better understanding
of this device by considering reflections at discontinuities. This
method is useful in considering any impedance-matching device, and
we shall think of it first in connection with the ordinary quarter
wave transformer. The object of this transformer is to eliminate
the reflection that would be present if the impedance Z1 were
connected directly to Zt. The method of eliminating reflections is
based on the INTERFERENCE between waves. Two waves half a wavelength
apart are in opposite phases, and the sum of them, if their amplitudes
are numerically equal, is zero. The fundamental principle behind the
elimination of reflections is then to have each reflected wave CANCELED
by another wave of equal amplitude and opposite phase. In order that
this second wave may have traveled half a wavelength farther than the
first, it is OBVIOUS that it must have gone a quarter wavelength farther
up the line, and correspondingly a quarter of a wavelength back, before
it meets the original reflected wave. In other words, two discontinuities
in characteristic impedance, of such magnitude as to give equal amplitudes
of reflected waves and spaced a quarter of a wavelength apart, will give
no NET reflection and hence will not introduce reflections into the line."

Mr. Slater goes on to provide an example like mine above. So my question
is: If all this stuff was known and published as far back as 1942, why
do you, Richard, reject it in 2004, 62 years later?
--
73, Cecil, W5DXP



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