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I mean, what is happening at a microscopic level to cause this behavior? Here's what I got from Wikipedia:

  1. On Reflection (physics)#Reflection of light it says that "solving Maxwell's equations for a light ray striking a boundary allows the derivation of the Fresnel equations, which can be used to predict how much of the light reflected, and how much is refracted in a given situation."
  2. On Specular reflection#Explanation it says that "for most interfaces between materials, the fraction of the light that is reflected increases with increasing angle of incidence $\theta_i$" (but doesn't explain why)
  3. Finally, on Reflection coefficient#Optics, it says basically nothing, redirecting the reader to the Fresnel equations article.

What I'm trying to find, instead, is a basic level explanation that could provide an intuition on why this happens, rather than analytic formulations or equations to calculate these values. Is there a good analogy that explains this behavior?

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Not only is it reflected more intensely, but sometimes there is no smooth (non-scattered) reflected image at all at a straight angle; then when you're observing at a grazing angle the surface could appear mirror-like. This is noticed when you're making optics (e.g. telescope mirrors) and you're polishing the glass: when you're working with fine abrasives such as 5 micron Al2O3, the mirror surface seems matte at a straight angle, but appears smooth and reflective at a grazing angle. – Florin Andrei Jul 8 '11 at 22:30
up vote 8 down vote accepted

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:image from wikipedia

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 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. :-)

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Thanks, Steve! I'd accept this answer, since it seems to have received the blessing of the community as the most correct explanation of the phenomenon. However, I must admit that I cannot, with my current knowledge, understand everything in it. Then again, perhaps explaining this at a layman level would require a much longer text, to introduce all the relevant concepts. I'll try to become more familiarized with the theory behind this and maybe later if I can understand it better, I'll select it as the best answer. I feel that's would be more honest and aligned with the scientific spirit. – waldyrious Jul 11 '11 at 13:55
Update: revisiting the question without all those thoughts in my head (I was preparing for an exam at the time) allowed me to realize that your explanation was actually quite intuitive and simple. I'll accept it now. Thanks, and sorry for the delay! :) – waldyrious Jan 7 '12 at 15:22

Well, imagine shooting a piece of glass with a gun. If you shoot at a glancing angle, it is more likely to ricochet off the glass without damaging it. This is because the impulse required to reflect the bullet is smaller for shallow grazing angles, as most of the bullets momentum is parallel to the interface. Now obviously the physics of reflecting a photon is very different than reflecting a bullet, but the analogy is that the impulse required to reflect the photon becomes smaller, and it becomes relatively "easier" for the medium to supply that small impulse than to let the photon go through.

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This makes sense and would be my first intuition, but I expected that while the result is similar to macroscopic bouncing, the principles behind it would be potentially very different. I'd love if you could expand this analogy with a bit more detail on what happens microscopically (if that's possible when adopting this metaphor). – waldyrious Jul 8 '11 at 17:48

user1631's analogy holds. You can think of the reflection as the surface generating an electromagnetic wave which cancels the incoming wave on the non-reflecting side and reflects the wave on the reflecting side.

At a high angle of incidence this wave needs to have a high intensity. At a low angle of incidence, the wave needs a lower intensity.

At higher intensities, more of this "reflecting wave" is going to be transmitted through the surface or turned into wasted energy.

In other words, at low angles, you only need to change the direction of the EM vector a little bit. At high angles, it needs to change direction completely.

How does reflection work? might help.

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The linked question was very interesting. Thanks for the pointer! – waldyrious Jul 9 '11 at 0:01

Perhaps it is because in the case of grazing incidence a lot more scatterer are in the path of the photon. At least this is why I think we observe X-ray reflection at all and were able to build Chandra and Rosat.

But I would like to hear a better explanation as well.

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REFLECTION under two different angles

The area of the intersection of the ray with the surface at A is greater than the area of intersection at B.

To a greater area correspond more atoms to reflect the light.

EDIT add

A justification:

The equations related to graphs on Steve answer (the reflected coefficients $R_s and R_p$) are the Fresnel equations when I read there $(n_1/n_2\cdot\sin\theta_i)^2$ I see a proportionality to the area.
The angles are in relation to the normal and sin² traces an area. A sensivity of the equations in relation to this factor is apparent in the graphs ($\theta_i$), because $\sin\theta_i, \sin\theta_i^2$ will 'follow' the shape of ($\theta_i$)
I think that my viewpoint is justified.

EDIT add end

the reflected rays are absent in the pic. Its not relevant.

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That's a very interesting way to look at it, and makes a lot of sense! Is this description physically correct, or just an analogy? – waldyrious Jul 8 '11 at 23:59
One problem with this explanation: Shine a pencil-thin laser beam and a very wide flashlight beam at the same surface at the same angle, with both having the same intensity. Both beams are reflected the same amount, even though the flashlight takes up a much larger area. – Steve B Jul 9 '11 at 4:19
I found now that the user whoplisp gave a similar answer. Also the answer of Steve is more correct. The 'whys' must be found in the continuity of Maxwell equations at the boundary conditions, and treat in separate the parallel and the perpendicular components of E field. The medium is made of scatterers (atoms) that absorb and re-emit photons sequentially in time. The time of the interaction is not null (how much of it?). To be processed at the surface leads to less absorption then at deeper layer (more heat). The number of available scatterers is part of the solution, imo. – Helder Velez Jul 9 '11 at 12:50
@Steve, same intensity = same number of photons, probably not all available area is good for scatter a particular photon. For each one there must exist a cross section (related to the wave length?). If you ask if I'm 100% sure of my answer I've to say no, I'm not. Even so, I do not know how to disregard the number of scatterers. @all There exist yet another parameter that I think that may be relevant: after reflection how vary, with the angle, the volume above the surface that is available to the interference phase?. I'm making an hypothesis without prior elaboration. It may be misleading. – Helder Velez Jul 9 '11 at 13:38
@all I included in the answer a justification using Fresnel coefficients. (ignore the above mention to volume because I can not trace it in any equation). – Helder Velez Jul 9 '11 at 15:19

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