Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I'm a high school teacher trying to teach my students (15year olds) about refraction. I've seen a lot of good analogies to explain why the light changes direction, like the marching band analogy, that the light "choose" the fastest way etcetc, and for most of my students these are satisfying ways to explain the phenomenon. Some students, however, are able to understand a more precise and physically correct answer, but I can't seem to find a good explanation of why the lightwaves actually changes direction.

So what I'm looking for is an actual explanation, without analogies, of how an increase/decrease in the speed of a lightwave cause it to change direction.

Thanks a lot

share|improve this question
2  
If you're starting with the given information that the speed of light decreases/increases between mediums, then I think that several (carefully formulated) analogies are accurate for getting from there to Snell's law. It's much more difficult to understand why the speed changes in the first part. That goes into QFT and virtual particles. –  Alan Rominger Feb 5 at 14:49
1  
@AlanSE You can explain refraction and the change in the effective speed of light in a mathematically precise way with classical E&M. Every upper division and graduate text should have it. The only thing is that they take chapters to lay the foundations and develop the tools. I've seen it condensed down to about three pages once. –  dmckee Feb 5 at 15:00
    
@dmckee Is that sufficiently explained in the question on the topic: physics.stackexchange.com/questions/11820/… I had seen this discussed before, and that wasn't really my take-away, but reading some of those more carefully, I see that might be what you're getting to here. –  Alan Rominger Feb 5 at 18:45
3  
Ah...this very early question has several takes on the classical description. –  dmckee Feb 5 at 19:06
1  
@RahulNarain Dmckee is right when he says my answer is Huygens principle in disguise. But I strongly suggest you should write what you think is a good answer too. This question is not only about physical correctness of description but also style of desciption - dare I say even poetry in description - a style that will resonate with an intended audience. So two answers with the same underlying physical principles but differently written are altogether acceptable IMO. That's why there can be so many successful textbooks on the same subject: different styles work for different people. –  WetSavannaAnimal aka Rod Vance Feb 6 at 5:44

4 Answers 4

You could appeal to boundary conditions of Maxwell's equations, or any wave equation for that matter. This is not as abstract as it sounds. See my rough and hurried drawing below:

Refraction at an Interface

The waves in the first medium travel quickly, so their crests are further apart than in the second medium. The frequencies of all waves being the same, the ratio of the spacings is $n_1/n_2$.

Now you can explain that the electromagnetic field has to be continuous across the interface - it can't suddenly jump from one value to another. Therefore, the variations of both waves must align at the interface. You can then fiddle with the geometry a bit to show that the spacing between the intersections of the left hand waves with the interface is proportional to $n_1\sin\theta_1$. The spacing between the intersections of the right hand waves with the interface is proportional to $n_2\sin\theta_2$ (same proportionality constant, to wit $c/\nu$. If, as we have argued, the variations must match up, $n_1\sin\theta_1 = n_2\sin\theta_2$ and you get then Snell's law. You don't of course need to derive Snells law to show that the directions of the waves have to be different if their variations along the interface have to align.

share|improve this answer
3  
This is excellent, and is not a lie-to-children. More of this in high school, please. –  Trevor Alexander Feb 6 at 1:50
1  
@TrevorAlexander Thanks. Are you a teacher? If so, you may be interested in my set of science activities for children. I have written it with primary age in mind (well, actually, not having a sound education background, I've tested them at my daughter's school and found that primary children get something out of them), but they are applicable to all ages. So far I've got five activities (forget the "foreword", it's a bit of a rant, although there's a lovely quote in there that I found). –  WetSavannaAnimal aka Rod Vance Feb 6 at 2:12
    
Unfortunately I am not, but this is great. I will pass it on. –  Trevor Alexander Feb 6 at 2:48
2  
this is exactly how i first heard about refraction (other than light "bending" through a glass I guess - i.e. high school science, 14 y.o) - just a half page diagram deriving snell's law. I think it is completely appropriate to that level (motivation is a bigger factor than ability, imo), and I'm confident you would find it any old textbook for that level - math and pictures is just not the fashionable way to teach physics any more. –  user3125280 Feb 6 at 7:18
    
@user3125280 Very interesting. Do you recall the book and reference? I came up with this many years ago when reading Born and Wolf and understanding the Fresnel equations: Snell's law is simply saying that the transverse component of the wavevector has to be continuous across the boundary to uphold field continuity. –  WetSavannaAnimal aka Rod Vance Feb 6 at 7:25

WetSavannaAnimal's derivation is perfectly correct. But, at the highschool level, it may be more relevant to appeal towards their intuition with mechanics.

You can derive Snell's law by momentum conservation. Consider a boundary that has a surface normal in the z direction. Then, such a boundary can only change the z-component of the momentum of any incoming object. For example, if you throw a ball at a wall (with surface normal in the z direction), then the wall only changes the z-component of the momentum of the ball. Thus, the x,y components of momentum cannot change.

Now, if we consider light, the magnitude of it's momentum is equal to $n k_o$ where $k_o$ is the wave-number of light in a vacuum (i.e. angular frequency divided by the speed of light). Thus, (in 2-d) the y component of momentum is $n k_o sin(\theta)$ on the left-side of the wall and it's $n' k_o sin(\theta')$ on the right-side of the wall where the wall's surface normal is perpendicular to the y-direction and the angle's are defined relative to the surface normal. By momentum conservation, these two quantities must be equal to each other and by dividing $k_o$ from both sides, Snell's law is derived.

share|improve this answer
1  
The result is correct, but the reasoning is satisfactory only for matter particles crossing thin space slab of potential gradient. For light, it is not clear that the momentum of light is proportional to $nk$, and furthermore it is not clear what part of momentum is carried away by the body into which light penetrates. –  Ján Lalinský Feb 5 at 18:19
    
@JánLalinský - the reasoning is applicable for any object having momentum, and any force constrained to ect only in a certain direction. Why a medium applies a force to light is already a different question :) –  nbubis Feb 5 at 18:26
    
@JánLalinský I agree with with you, however the OP asked " why the lightwaves actually changes direction" and in that sense I think my answer makes this clear. As for the details of light having momentum to beginwith, the student would just have to take this as a given unless they are familiar with Maxwell's Equations along with Special Relativity. –  mcFreid Feb 5 at 18:28
    
mcFreid, for light refraction you cannot formulate the law of conservation of momentum in the $xy$ plane in the way you did above - momentum of the medium in the $xy$ plane should be considered. Momentum of light in the $xy$ plane may not be conserved. –  Ján Lalinský Feb 5 at 20:20

This is a very incomplete answer, but it should put you on the right road. I'm going to assume that you buy the usual argument for why a change in the speed of light generates a bend at the interface and concentrate on the speed of light in a medium.

We start with Maxwell's Equations (expressed here in the differential form and in SI units):

$$ \nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0} \tag{Gauss}$$ $$ \nabla \cdot \mathbf{B} = 0 \tag{Gauss-magnetism}$$ $$ \nabla \times \mathbf{E} = \frac{\partial \mathbf{B}}{\partial t} \tag{Faraday}$$ $$ \nabla \times \mathbf{B} = \mu_0 \left( \mathbf{J} + \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} \right) \tag{Ampere} \,.$$

Notice in Gauss's Law and Ampere's Law the presence of the permittivity of free space $\epsilon_0$ and the permeability of free space $\mu_0$. When you manipulate Maxwell's Equations in a charge and current free region to get the wave equation those constants combine to give the speed of the wave as $$ c = \frac{1}{\sqrt{\mu_0\epsilon_0}} \,.$$

Now, if we are considering a material environment we are no longer in a charge free region. The net charge is zero, but on a microscopic scale the protons and electrons are separated from each other, so there are sources and sinks for the electric field; moreover, the electrons are in motion (nuclei too, but we'll ignore that) so there are sources of curl in the magnetic field.

You might imagine that the effects of all of this on a traveling electromagnetic ware are pretty complicated, but the surprise is that in many case1 we can group the effects by changing the constants to $\epsilon = \kappa \epsilon_0$ and $\mu = (\chi_m +1)\mu_0$2 and otherwise pretending that we are still in a charge and current free region. Showing this is a rather longer development than I have room or time for here. See any upper-division or graduate E&M text.

Both of these new constants are larger than the ones they replace3, which means that when we construct the wave equation we have a new velocity $$ v = \frac{1}{\sqrt{\mu \epsilon}} < c \,.$$


1 Sufficiently weak fields and sufficiently smooth materials, but these conditions include essentially every everyday case.

2 Strictly speaking these new values are frequency dependent $\epsilon_0(f) = \kappa(f) \epsilon_0$ ..., but in this simple discussion I'm going to ignore that. However, this frequency dependence is responsible for the frequency dependence of the index of refraction, which leads to observable phenomena.

3 For normal materials. Work on exotic materials where that statement is too broad is a on-going field of research.

share|improve this answer
2  
Typical 15-yr-olds, at least in USAland, haven't got to calculus yet, let alone read "Div, Grad, Curl, and all That." –  Carl Witthoft Feb 5 at 16:19
3  
@CarlWitthoft I know (in fact only the luckiest US students have access to calculus instruction in high school), but their teacher has asked for an exact explanation, and I want him (or her) to see why all searchs seem to turn up is analogies. For some combinations of (problem) X (level of mathematical preparation) there just isn't a satisfactory exact answer. And I always write the differential forms to save on typing, but the Wikipedia shows the correspondence. –  dmckee Feb 5 at 16:27

A teacher I very highly regard explained it at a high school level to me as follows: (I was already taught something about interference of waves)

You have a light-wave incident on the medium. The light-wave is just an oscillating electric field (plus magnetic field, but we know that electric effects are dominant in such situations). The oscillating electric field sets up induced dipoles in the molecules, which oscillate sympathetically with the field. So all these dipoles, that have been excited by the field emit radiation in all directions. But the radiation from these dipoles interferes.

The constructive interference of the radiation absorbed and re-emitted by all these dipoles happens to be along the refracted ray.

The refractive index is an electrical property of the medium -- it is related to the polarizability of the molecules, or the ability to create an induced dipole. So it is natural that this effect does depend on the refractive index.

I'm yet to put this into rigorous formalism, but I think it can be done. I vaguely remember him referencing this book on Lagrangian Optics: http://books.google.com/books/about/Lagrangian_optics.html?id=OPIzlipW9nwC -- although that might have been in a different context.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.