How do charges in a wall cancel out incoming electromagnetic waves?

[k9b4]

I think that a wall does not actually stop light, I think the electromagnetic waves exist on the other side of the wall, but the charges in the wall provide their own electromagnetic field, such that charges on the other side of the wall feel both forces at the same time and 'cancel out'.

Is this correct?

 

[elegysix]

I would think of photons as bullets.
If you have a strong enough bullet (enough energy), it will go through the wall. otherwise, not so much.

 

[Nugatory]

I think that a wall does not actually stop light, I think the electromagnetic waves exist on the other side of the wall, but the charges in the wall provide their own electromagnetic field, such that charges on the other side of the wall feel both forces at the same time and 'cancel out'.

Is this correct?

Not quite, but you have the right idea. The waves do not exist on the other side of the wall because the intensity of the electrical and magnetic fields there is zero - there is no wave, just as there are no water waves on the surface of a still pond. However, the intensity is zero for pretty much the reason that you give: the charges in the wall move around in response to the incoming wave in such a way that their electric fields always exactly cancel that of the incoming wave.

 

[Nugatory]

I would think of photons as bullets.
If you have a strong enough bullet (enough energy), it will go through the wall. otherwise, not so much.

Photons do not behave even slightly like bullets, and thinking of light as a stream of photons is almost guaranteed to lead to confusion and misunderstanding. For example: A wall of glass will not stop a beam of visible light (that's why we use glass in windows, right?) but it will stop a beam of ultraviolet light - yet the photons of ultraviolet light have more energy than the photons of visible light.

(So that's what a photon is not. More likely, you want to know what a photon is... Try searching the quantum mechanics forum here with keywords like "photon particle", "photon bullet", "photon grain" and you'll find some posts about that).

 

[k9b4]

Not quite, but you have the right idea. The waves do not exist on the other side of the wall because the intensity of the electrical and magnetic fields there is zero - there is no wave, just as there are no water waves on the surface of a still pond. However, the intensity is zero for pretty much the reason that you give: the charges in the wall move around in response to the incoming wave in such a way that their electric fields always exactly cancel that of the incoming wave.

Cool, thanks for explaining.