opticsreflectionvisible-light
Can anyone please explain why light reflects at the boundary between two regions with different impedances? This sounds very simple but I got confused when I tried to think of how light and atoms interact with one another at the boundary.
This question can actually be generalised to the reflection of all types of waves. I have to admit I have no understanding of the microscopic detail of reflection.
First, I just want to remind readers that it is NOT true that "more glancing angle always means more reflection". For p-polarized light, as the angle goes away from the normal, it gets less and less reflective, then at the Brewster angle it's not reflective at all, and then beyond the Brewster angle it becomes more reflective again:
Nevertheless, it's certainly true that as the angle approaches perfectly glancing, the reflection approaches 100%. Even though the question asks for non-mathematical answers, the math is pretty simple and understandable in my opinion...here it is for reference. (I don't have any non-mathematical answer that's better than other peoples'.)
The Maxwell's equations boundary conditions say that certain components of the electric and magnetic fields have to be continuous across the boundary. The situation at almost-glancing angle is that the incoming and reflected light waves almost perfectly cancel each other out (opposite phase, almost-equal magnitude), leaving almost no fields on one side of the boundary; and since there's almost no transmitted light, there's almost no fields on the other side of the boundary too. So everything is continuous, "zero equals zero".
The reason this cannot work at other angles is that two waves cannot destructively interfere unless they point the same direction. (If two waves have equal and opposite electric fields and equal and opposite magnetic fields, then they have to point the same direction, there's a "right-hand rule" about this.) At glancing angle, the incident and reflected waves are pointing almost the same direction, so they can destructively interfere. At other angles, the incident and reflected waves are pointing different directions, so they cannot destructively interfere, so there has to be a transmitted wave to make the boundary conditions work. :-)
Electromagnetic radiation will reflect from any change in refractive index.
For light reflecting off snow you have an air/ice boundary, and the refractive index of air is about 1.0 while ice is about 1.3, so you get reflection. Actually you get multiple reflections which is why you get a diffuse scattering and why ice looks white.
For RF waves reflecting off the ionosphere, the ionosphere refractive index is changed by the free elections in it, so there is a refractive index mismatch between the very low pressure air below the ionosphere and the ionosphere itself and this causes reflection. It's a bit hand-waving to claim it's just that the ionosphere has a different refractive index, but that's a good starting point. If you want to look in more detail see http://ecjones.org/physics.html and many other easily Googlable articles.
So I suppose the mechanisms are basically the same.
Best Answer
The way I would recommend you approach this question is to look at the derivation of Fresnel's Laws of Reflection, as is shown in detail at this link. The key here is that there must be a continuity of the electric field across the boundary between the regions of different refractive index - but since such regions have different amounts of polarization for the same electric field (that is pretty much the definition of relative dielectric constant...) the only way that this can happen is if part of the electric field "stays outside" while another part penetrates into the second medium. Which is what we call "partial reflection".