New answers tagged

3

As mentioned in comments, a photon is "delocalised", so it feels the whole system. You may imagine a photon as a long-long wave (to have a defined frequency) and as such it interacts with the whole material. More strictly, one can say that the source of photon is the whole set of charges, so the photon is a collective mode of excitation of a given system. ...


4

Let me explain in purely classical terms (not the description of reality, but easy to imagine). You realized that when a ball bounces off the wall, at a certain point, it has no momentum. However, it must still have all the energy of the movement (neglecting losses to environment) - where did that energy go? The ball is formed of many discrete atoms, bound ...


9

Light does not slow down during a reflection. Light is a signal disturbance in electric and magnetic fields. These disturbances propagate through space at a fixed speed $c$ in vacuum. The situation is completely analogous, in a mathematical sense, to a wave pulse that is sent along a string. When the pulse encounters a boundary, it flips direction, and may ...


6

In this case, it would be useful to not consider light in its particle form as photons, but instead to consider it as a wave - see this wikipedia page. Then, the wave is simply reflected from the surface, without us having to consider the kinetics of any particle. The wave, in a vacuum, would continue to propagate at the speed of light, regardless of the ...


1

You'd probably have to know the composition of your "standard glass" to get an idea of the spectrum. Even then it'd be difficult to find any literature specifically related to your glass. If, for example, your sample is a beer bottle or something like that then the manufacturer would surely have that data but might not make it available. It is totally ...


0

To understand this fully, you really need to be thinking of rays as normals to wavefronts. In this representation, a ray's tail gets assigned by a phase factor to represent the total phase of the plane wave. This phase factor advances by $2\,\pi\,\times$ an amount calculated by multiplying the distance advanced along the wavefront, in wavelengths, by the ...


2

I think you are perhaps confusing reflection at a corner with a "corner reflector" (https://en.wikipedia.org/wiki/Corner_reflector). The latter is a type of mirror which reflects light back to where it came from, as shown in your diagrams. The reflection comes from the 3 perpendicular mirrors, rather than at the join between them. There is no "specular" ...


0

From refractiveindex.info you can get n and k for many different materials, as a function of wavelength, and the site even includes a calculator at the bottom with which you can calculate the reflectivity R. All of the provided data are backed up with citations.


2

Your assertion If something was not reflecting photons in any direction, you could then see through it. is not right; you haven't thought of the possibility where the light is absorbed and not re-emitted. An object that absorbs all incident light is not seethrough, but black. It would block one's view of objects behind it, and therefore be very ...


0

Light allows us to see an object in several ways. You've named some but not others. If the photons are absorbed, then we can't see the object. You got that right. But suppose they aren't absorbed, then what? Suppose the photons travel straight through the object unaffected. Then we wouldn't see it either. For example, a clean sheet of glass and ordinary ...



Top 50 recent answers are included