Why does light reflect more intensely when it hits a surface at a large angle?

Question

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?

Answer 1

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. :-)

Answer 2

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.

Answer 3

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.

Answer 4

Well we must differentiate between s and p polarization and we must remember that light excites electric dipoles in the medium. Oscillating electric dipoles have an anisotropic radiation pattern with maximum in the direction peripendicular to the dipole moment vector, and with zero radiation along the axis of the dipole moment vector.

For s-polarization it is simple to answer the question: At large incidence angles (coming from air to a medium) the probability of the light to interact with near surface dipoles is larger due to the fact that it travels a longer path in the near surface region, and since near surface dipoles can scatter out of the medium much likelier than dipoles residing farther away (the oscillating electromagnetic field decreases with the 3rd power of the distance in the near field = within a wavelength distance). The reflected beam is then emerging in the direction of constructive interference of the dipole far fields.

For p-polarization everything that has been written for the s-polarization remains valid with the additional influence of the directional anisotropy (https://en.wikipedia.org/wiki/Dipole#Dipole_radiation) of the dipole scattering. This influence suppresses the radiation in the polarization plane and perpendicular (90°) to the propagation direction fully, thus the reflection initially follows the s-polarization reflectivity but soon starts to decrease with increasing incident angle and vanishes at Brewster's angle (where the angle between the beam in medium = transmitted beam and reflected beam would be 90°). Above Brewster's angle this suppression becomes weaker again and the reflectivity grows quickly and reaches the same value as for the s-polarization at 90° incident angle.

Answer 5

Here is your analogy: How about a game of Red Rover, Red Rover, send Waldyrious on over.

Here is your intuitive explanation:

Consider that your particle (A remarkably photon-like Waldyrious particle, if you will) is represented by a vector. Because we are discussing a photon (ahem, a Waldyrious particle), the speed should be constant cough, and so should the mass be constant cough. Thus your vector will represented only by direction.

Also consider that any barrier represented by a plane with a refractive index other than zero, toward which the particle is careening, has a numerical value associated with it (in our analogy, it is a line of physics professors interlocking leather patched elbows together, tweed jackets and all). This refractive index value shall indicate the energy required to penetrate the barrier (are we talking about brand new fresh 26 year old profs, or the old and dusty variety?). For the sake of easy comparison, let the number be a range between 1, representing the exact energy of a photon, to zero.

The closer to perpendicular the vector is, with regards to the plane, the closer the photon's, I mean Waldyrious', energy imparted toward the barrier is to 1. Imagine that the math is trig (it is not, but you asked for an analogy, not a calculus lesson). The farther from perpendicular - or using your terminology, the larger the angle at which the particle strikes the barrier, the less forward momentum is contributed to break through the barrier - and instead it bounces off...

Run straight and true, Waldyrious. Strike as close to perpendicular as you possibly can, and all of your forward momentum will be imparted toward the ever vigilant wall of profs! Deviate from that path, and you may not penetrate that barrier, and be reflected away.

Answer 6

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.

Answer 7

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.