Richard Clark wrote:
Anyway, the optical analogue has proven to be a bust when I
demonstrated that reflection products do persist. This is not the
place to hash over that again. If you wish, you can consult my
original posting and the follow-ons for details.

But in your "proof", you superposed powers which is a no-no.
When 111.1mW interfers with 87.78mW, the result is not
(111.1-87.78). Since the associated E-fields are 180
degrees out of phase, the power equation must take the
interference into account.

Pref1 = 111.1mW + 87.78mW - 2* sqrt(111.1*87.78)

Pref1 = 111.1mW + 87.78mW - 197.5mW = 1.38mW

You subtract 87.78 from 111.1 and get 23.32. That value is
almost 17 times too high. All your math after that is invalid.
All except 1.38mW of reflections are canceled by that first
internal reflection. Your value of 23.32 is simply wrong.

RF engineers usually convert to voltage, perform the superposition,
and then calculate the total power. One doesn't have that luxury
when dealing with light so the power (irradiance) equations must
be used to obtain the correct results.
--
73, Cecil http://www.qsl.net/w5dxp

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