Understanding the Compton Effect: How Light Reflects on a Mirror Explained

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Light reflects off a mirror due to its interaction with the surface, where electromagnetic waves are shorted out, causing the light to reflect rather than penetrate. The classical explanation involves the oscillation of the electric field at the light frequency, which results in a perfect reflection when the wave encounters a conductive surface. Quantum Electrodynamics (QED) further describes this process in terms of probabilities, emphasizing that the reflection occurs based on the least time taken by photons. The discussion clarifies that the image seen in the mirror is indeed a reflection of oneself, despite initial confusion about frequency changes and identity. Understanding both classical and modern physics provides a comprehensive view of light reflection.
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Could you explain me how does Light reflect on a mirror?

I assume that Compton effect on electrons is the only way for light to come back to my eyes but in that case, there would be a change of frequency of the photon (and a change of color).

Does it mean that it is not me in the mirror ?

Am I wrong ?

HELP ME, PLEASE ...
 
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While I can't answer exactly how it happens, I can tell you that it IS IN FACT YOU in the mirror. Doesn't everyone else look the same in the mirror to you as they look normally aside from the inversion?
 
Is it really wolverine in the mirror? How does he get back out? Go back to your basic physics book and review waves, and optics.
Bring your image with you too. -Mike
 
I'm agree with you because of the daily experience, but even if optics can explain how ligth reflect on a mirror, I wonder how can modern physics (QED ...) descibe the reflection ...
 
And at the point of reflection, does light have zero velocity?
 
My physics instructor at Yale, Prof. Beringer, once explained to me that a mirror was to light as a radar was to radio waves. (Or an oven cavity is to microwaves, I guess.)

Electromagnetic radiation impinging upon a mirror - a conductor in the visible range; EMR cancels at a conductor - thus maintains a potential opposite to that of the incident ray, which manifests in turn with the emission of the image's reflected photonic signal.

Using mirror symmetries likewise helps solve many problems in electrodynamics.
 
wolverine

If you think of light as a a wave packet of photons, when it stirkes a surface it has four possible outcomes. First, it might travel right through it, like pure glass. Second, it may not travel through it at all but be totally absorbed by it. Thirdly, it may be reflected totally. And lastly, it may be some combination of all of the above. What happens to the light when it strikes the surface depends on the wavelength of the light and the composition of the material it strikes.

For a 100% reflective mirror, the wavelengths cannot penetrate and they reflect. The easiest way to vinsualize this is using the light ray model. But, using QED, as Richard Feynman so aptly describes in his book, QED, we find that the reflection is all about the probability that a photon travels along this path or that depending on the least time it takes.

Hope this is some additional help to you.

Regards
 
It does , thanks.

I just thought that there were a "classical" explanation, like photons hiting electrons and going back to my eyes or photons passing trough matter without doing any interacion.
Assuming that, I was thinking that R was equal to

Ninteractions
_____________

Ntotal

But it seems that we need to use QED to explain that.
 
Originally posted by wolverine
It does , thanks.

I just thought that there were a "classical" explanation,
Sure there's a classical explanation. Light is an electromagnetic wave. The electric field oscillates at the light frequency. But a metal surface "shorts out" the electric field, effectively pinning it to zero at the surface. The wave therefore has no choice but to reflect. It is analogous to waves on a string. Fix one end of the string in place, and the wave you send down the string reflects perfectly. In more detail: the medium outside the mirror permits waves in any direction. But the only wave that when added to the incoming wave always gives zero electric field at the surface of the mirror, is in identical wave coming backward as a reflection.
 
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