Looking for the actual reason of refraction explained precisely without analogies

Question

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

Answer 1

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:

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.

Answer 2

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):

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

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 cases¹ we can group the effects by changing the constants² to $\epsilon = \kappa \epsilon_0$ and $\mu = (\chi_m +1)\mu_0$ and otherwise pretending that we are still in a charge and current free region.³ 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 replace⁴, which means that when we construct the wave equation we have a new velocity $$ v = \frac{1}{\sqrt{\mu \epsilon}} < c \,.$$

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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 Here $\kappa$ is the dielectric constant of the material and $\chi_m$ is it's magnetic susceptibility.

⁴ For normal materials. Work on exotic materials where that statement is too broad is a on-going field of research.

Answer 3

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.

Answer 4

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.