Jim Kelley wrote:
Any given wave front will never reflect 100%
from a surface which is only 50% reflective,

That's all you have to say, Jim, to defeat your argument.
If you would stop refusing to perform a simple calculation
involving my example at:

http://www.w5dxp.com/thinfilm.GIF

you would understand. When the internal (0.009801w)
wave reflection arrives at t3 and interferes with the
(0.01w) external reflection wave, what is the resulting
reflected power back toward the source. When you calculate
the results and realize that it is not 0.01 - 0.009801 watts,
you will begin to understand the nature of interference.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:

Jim Kelley wrote:

Any given wave front will never reflect 100% from a surface which is
only 50% reflective,


That's all you have to say, Jim, to defeat your argument.

I am quite content to agree to disagree on that point if you wish.
Any exceptions to it that you would try to make could only derive from
fiction.

73, ac6xg

Jim Kelley wrote:

Cecil Moore wrote:

Jim Kelley wrote:
Any given wave front will never reflect 100% from a surface which is
only 50% reflective,

That's all you have to say, Jim, to defeat your argument.

I am quite content to agree to disagree on that point if you wish. Any
exceptions to it that you would try to make could only derive from fiction.

I seriously doubt that you are blind to the contradictions
in your argument and are simply hoping to slip them through
while no one is paying attention. Suffice it to say, it is
impossible for a 50% reflective surface to cause 100% reflections
without help from interference in the form of wave cancellation
due to permanent destructive interference. Anybody who understands
the process of anti-reflective thin-film coatings understands
the process of destructive interference redistributing the energy
in the direction that allows for constructive interference.

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

"Clearly, if the wavelength of the incident light and
the thickness of the film are such that a phase difference
exists between reflections of p, then reflected wavefronts
interfere destructively, and overall reflected intensity is
a minimum. If the two reflections are of equal amplitude,
then this amplitude (and hence intensity) minimum will be
zero." (Referring to 1/4 wavelength thin films.)

"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."

http://micro.magnet.fsu.edu/primer/j...ons/index.html

"... when two waves of equal amplitude and wavelength that are
180-degrees ... out of phase with each other meet, they are not
actually annihilated, ... All of the photon energy present in
these waves must somehow be recovered or redistributed in a new
direction, according to the law of energy conservation ... Instead,
upon meeting, the photons are redistributed to regions that permit
constructive interference, so the effect should be considered as
a redistribution of light waves and photon energy rather than
the spontaneous construction or destruction of light."
--
73, Cecil http://www.w5dxp.com