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In the classic metaphor, a light beam bends for the same reason that a wagon getting one wheel stuck in the sand does...the wheels travel at uneven speeds, and the wheel on the smoother surface travels faster.

But the key to the wagon scenario is the axle - if the two wheels were not bound, the faster wheel would sail on, heedless of the other wheel's difficulty. So, if the metaphor is useful at all, there is an axle force binding the photos in a beam of light, which causes it to turn when it hits a different medium non-perpendicularly.

Or is this just a bad metaphor for the rubes?

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3 Answers 3

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First important point is that light always travels in a straight lines in an environment of constant index of refraction:

alt text

What creates the dispersion is the fact that the index of refraction depends on the frequency of the light $n = n(f)$. This also means that phase velocity (defined as $v = {c \over n}$) depends on the frequency. You don't see this in the image but you can imagine that in usual materials (e.g. glass) a wavefront will propagate faster along red tracks than along the violet tracks (equivalently, red tracks are less bent than the violet tracks, as can be seen in the picture). So this will make the wavefront bulging up in the red part of the spectrum (the precise shape depending on the precise material and angle of incidence).

So it's probably just the wavefront the metaphor is talking about. But you can see that wavefront will spread out as it propagates in the material (precisely because there is no "axle" between individual photons). In my opinion it's a pretty bad metaphor except for giving a correct illustration of the relation "denser environment" $\leftrightarrow$ "slower speed".

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I realize that this question has been dead for a while, but I think I'm not understanding the concept of "wavefront", and I think that's the missing concept that unlocks this analogy. –  Chris B. Behrens Sep 13 '11 at 14:44
    
@Chris: See my comment on the other answer. You know how waves interfere, right? Drop two pebbles in a pond, blah blah. Now if you drop a whole row of pebbles close together, or a long straight board, in the pond, you get a line-wave because all other directions cancel out. A wave at any point in time is just like a row of pebbles generating what the wave will be an instant later. So a wave-front is necessarily always moving perpendicular to its front. –  Mike Dunlavey Sep 18 '11 at 17:50
    
"all other directions cancel out"...that was the magic piece of the puzzle that I was missing. Thanks. –  Chris B. Behrens Sep 19 '11 at 20:20

Its a metaphor. But any physical analogy can only be taken so far.

As an alternative instead of a wagon we can consider a marching band. In this case the band marches in formation until it reaches the edge of a muddy patch. If the approach was head on, the band would keep marching in formation albeit at a slower speed. If the encounter is at an angle, then the marchers on the patch move slower than the ones not on it - leading to a "bending" in the ranks. In this case there is no "axle" connecting the marchers but they "refract" nevertheless.

This page has a beautiful applet illustrating this analogy.

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+1 That page is the best explanation. I would only add two points. 1) that what determines a wave's direction of propogation is its self-interference. Otherwise the marchers might keep going in the same direction, just in diagonal lines. 2) The dual-slit experiment, since the question was about photons, and you have to understand how photons are carried forth on their waves. –  Mike Dunlavey Sep 18 '11 at 17:42
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@Mike - your answer is really what the question was about. Even with the marching band analogy, the formation refracts because the band is trying to stay in formation. What I didn't understand was the mechanism by which the photons stay in formation...self-interference. –  Chris B. Behrens Oct 7 '11 at 16:33
    
@Chris: Thanks. I didn't really understand it myself until I thought about it. –  Mike Dunlavey Oct 7 '11 at 16:45

This is a bad metaphor for rubes, as you say. The axle gives a conserved quantity across the transition for a wagon, the distance between the wheels is constant as the transition is happening. So the bending goes to 90 degrees when the boundary becomes parallel to the direction of cart propagation. This is not what happens with light, light refracts at a finite angle less than 90 degrees for light coming in nearly parallel to the surface, this is the angle past which you have total internal reflection in the medium.

The reason is that the perpendicular horizontal distance along the wavefronts is not the quantity that is conserved when light enters a stationary medium, like it would be if photons had little axles (they don't). The quantity that is conserved across the refraction transition is the frequency of the light, the energy of the photons. The reason is that the material provides a time-independent propagation environment, and in a time-independent background, the energy is conserved. Classically, the modes for a time-independent medium are found by separation of variables with a fixed frequency in time, and this is saying the same thing but without using quantum mechanics to relate conservation of frequency to conservation of energy.

So the analog of the axle length in this case is the time between wave-crests crossing a given point. This means that the wavelength in the medium is reduced by the index of refraction (to keep the frequency of crests crossing a given point constant), so that the outside wavelength is $\lambda$ and the interior wavelength is $\lambda/n$.

If medium surface lies parallel to the x-axis, and the incoming light wave crests make an angle of $\theta$ with respect to the x-axis ($\theta=0$ is crests parallel to the x-axis, so the light is coming head on, and no refraction), then the distance between the points where successive wave crests hit the medium boundary is $\lambda/\sin(\theta)$. In the interior of the medium, the same argument tells you that it's $\lambda/n\sin(\alpha)$ where $\alpha$ is the angle of the crests with respect to the x-axis in the medium, so in order for the crests inside to match the crests outside, you need

$$ \sin(\theta) = n \sin(\lambda)$$

and this is Snell's law for the case where the n outside is 1. You can always consider the wave-speed outside to be 1, so the quantity n is the ratio of the speed of waves inside to outside, and so it isn't really a special case. Also note that the same law holds for sound refraction, or any wave.

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