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A simulation of reflecting light that can be found here
Light will reflect on shiny metals, but why, and why is the reflection phase shifted 180 degrees compared to the incoming light?
The electrical component of the incoming light will influence the electrons in the metal and move them in the direction towards the “positive end” of the field. The positive nuclei of the metals will basically remain (on average) stationary though. (This due to their larger mass, and that they are locked into the crystal structure of the metal.) This will cause a field opposing the incoming field.
If the frequency of the incoming field is low enough the electrons will have time to move enough to, almost completely, oppose the incoming field. The moving electrons will now produce a wave that is 180 degrees out of phase compared to the incoming wave. This will propagate into the metal as well as outward. The wave going into the metal will interfere destructively with the incoming wave, and no signal will thus propagate through the metal. The metal will be opaque. The reflected wave will cause a standing wave on the outside of the metal. A standing wave does not transport any net energy, and this is due to that the incoming wave transport energy to the right, and, if 100% of the light is reflected, the reflected wave will transport the same energy to the left. The net energy transported is thus 0.
A simulation of the above can be found on this page.Up a level : Waves
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