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Cecil Moore wrote:
Please apply the same irradiance equation that optical physicists were using before you were born. Any physics professor should be able to accomplish that simple task. If optical physicists were in error in using that equation 100 years ago, please explain their error. So far no response to this simple request. The graphic of the non-reflected glass example is at http://www.w5dxp.com/thinfilm.gif We know that the reflections are 100% canceled during steady-state. The problem is: With the given data, calculate the magnitude of the total reflection back toward the source immediately after the first internal reflection arrives back at the thin-film to air surface at time t3 and is superposed with the external reflection. The two superposed waves are 180 degrees apart. The external reflection is 0.01 watts at a reference angle of zero deg. This is the normal reflection from the thin-film surface The first internal reflection is 0.009801 watts at 180 degrees. This is the first reflection from the glass. These two waves superpose at t3. The irradiance equation, using P for power density, is: Ptotal = P1 + P2 + 2*SQRT(P1*P2)cos(A) where (A) is the angle between the two waves and Ptotal is the total power density of the reflections toward the source after the two waves are superposed. Ptotal = 0.01 + 0.009801 + 2*SQRT(0.01*0.009801)(-1) Ptotal = 0.019801 - 0.0198 = 0.000001 watt There is 0.0198 watts of destructive interference which, according to the conservation of energy principle, must result in 0.0198 watts of constructive interference in the opposite direction. The magnitude of the reflection toward the source drops from 0.01 watt to 0.000001 watt at the arrival of the first internal reflection from the glass. That's a five magnitude reduction in the reflections at the time of the arrival of the first internal reflection. The reflections back toward the source are eventually completely eliminated during steady-state. There was no vector math and no need to calculate the magnitudes and phases of the electric and magnetic fields - just a straight forward calculation to get the answer. As long as the source remains in place, the steady- state cancellation of the reflections toward the source is permanent. It is difficult to analyze how that could happen unless the internal reflections interact with the external reflection resulting in wave cancellation. Everyone is invited to prove the above calculation to be incorrect. Hint: if one actually performs a vector analysis, the magnitude of reflection result will be exactly the same as above. The above method does not invalidate or replace a vector analysis. Unlike a vector analysis, it simply shows where the energy goes. Optical physicists knew this a century ago - most RF engineers have yet to learn it. -- 73, Cecil http://www.w5dxp.com |
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