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#1
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Cecil Moore wrote:
Jim Kelley wrote: The momentum in reflected waves changes direction upon reflection. What part of that do you need to have explained? What causes 100% reflection when the power reflection coefficient (reflectance) is only 0.5? If you would just work the problem the hard way, you would see where you're misconception lies. Any given wave front will never reflect 100% from a surface which is only 50% reflective, no matter how vicious your insults become, how may URLs you cut and paste, or how furiously you wave your hands. But when you work the problem as has been suggested you will see how energy gets from source to load. It does not rely on macroscopic layman's explanations or mathematical shortcuts in order to get there. The only energy "lost" by partial reflection in the process is that which is reflected back toward the source or stored in the system during the transient period. The sum of all the partial reflections equals the energy stored in the system (less the portion of energy admitted to the load or reflected back to the source). After the transient period, no energy is reflected back to the source, and the energy entering the system from the source equals the energy existed the system through the load. I know that you understand the difference between potential and kinetic energy in mechanics. Please try to consider that the concepts are no less valid in electromagnetism. (Note that we even use the word 'potential' to describe voltage.) 73, ac6xg |
#2
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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 |
#3
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![]() 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 |
#4
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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 |
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