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. :-)
This is one of the places where wave particle dualism gets some people in trouble. Many are taught that it means that light can be a wave and a particle, and that phrasing can lead to some confusion. I find it more intuitive to just rip the bandaid off quickly and say light is neither a wave nor a particle. It is something which, in some situations, can be well modeled as a wave, and in some situations can be well modeled as a particle, but it is its own thing (which can be well modeled in all known cases using a more complicated concept, a "wavefunction").
You can think of photons getting randomly reflected or transmitted on the boundary, but the truth is that the billiard-ball photon model really isn't very effective at describing what happens at this boundary. This is one of the regions where wave mechanics models the effects very well, while particle models don't do so well. If you use wave mechanics, the idea of a wave getting partially reflected and partially transmitted isn't difficult to believe at all. In fact, it's pretty easy to prove.
Thinking in wave terms at these boundaries also gives correct answers in peculiar situations where the particle model simply falls on its face. Consider the interesting case of an "evanescent wave."
In this setup, the laser and prism are set up at the correct angles to cause "total internal reflection." This means that, by the simple models, 100% of the light should bounce off the side of the prism and into the detector. Indeed, if the prism is in the open air, we do see 100% reflection (well, within the error bars of absorption). However, bring an object close to the prisim (but not touching) and things change. You end up seeing effects from the object, even though 100% of the light was supposed to be reflected!
If you think of light like photons, this is hard to explain. If you look at it as a wave governed by Maxwell's equations, you see that you would violate the law of conservation of energy if there was a "pure" reflection. Instead it creates a reflection and an "evanescent wave" which is outside the prism, and its strength falls off exponentially, which is really hard to explain with particles!
Of course, these too are all simplifications. The real answer to your question is that the wavefunction of the light interacts with the electromagnetic fields of the atoms in the prisim, and the result of that interaction leads to reflection, refraction, diffusion, absorption, and eveansecent waves. However, naturally those equations are a bit harder to understand, so we use the older, simpler models from before quantum mechanics. We just have to be sure to use the one which is most applicable in any given situation, because none of them are quite right.
Best Answer
Look here:
http://en.wikipedia.org/wiki/Fresnel_equations
This is for transparent media, the formulas for metals are somewhat more complicated.