# What happens to the wavenumbers of a EM plane wave when it crosses an interface between two dielectric media?

Consider an interface between two dielectric media placed in the plane $z=0$. The coordinate system $\tilde{y}-\tilde{z}$ has the $\tilde{z}-$axis perpendicular to this plane, and the coordinate system $y-z$ has the $z-$axis parallel to the propagation direction of a plane wave hitting the interface. The wavenumbers of this wave in both system are connected through

$$\left[\begin{array}{cc} k_{\tilde{y}}\\k_{\tilde{z}} \end{array}\right] = \left[\begin{array}{cc} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{array} \right] \left[\begin{array}{cc} k_{y}\\k_{z} \end{array} \right].$$

I have that

$$k_{\tilde{z}} = \sqrt{k^{^2}-k^{^2}_{\tilde{y}}}.$$

Since $k_{y} = 0$, because in this coordinate system the $z-$axis is parallel to the propagation direction, this implies that $k_{\tilde{z}} = k\cos\theta$, assuming that $k_{z}=k$.

Now, if the wave interacts with an interface we have that

$$k_{\tilde{z}} \rightarrow q_{\tilde{z}} = \sqrt{n^{^2}k^{^{2}}-k^{^{2}}_{\tilde{y}}},$$ where $n$ is the relative refraction index between both media and the $\tilde{z}$ component is the one affected because it is the one along which there is the discontinuity (the interface). Now, following the same procedure as before, and using the Snell's law,

$$\sin\theta=n\sin\psi,$$

where $\psi$ is the refraction angle, we have that $q_{\tilde{z}} = n\,k\,\cos\psi$. We can verify that all this reasoning is correct by checking the Fresnel's coefficients. Let me check just the reflection coefficient. In terms of the wavenumbers it is

$$r = \frac{\displaystyle k_{\tilde{z}}-q_{\tilde{z}}}{\displaystyle k_{\tilde{z}}+q_{\tilde{z}}},$$ which, with the calculation made, returns the correct coefficient as a function of the angles relevant to the problem:

$$r = \frac{\displaystyle \cos\theta - n\cos\psi}{\displaystyle \cos\theta + n\cos\theta}.$$

My question is why

$$q_{\tilde{z}} = \sqrt{n^{^2}k^{^{2}}-k^{^{2}}_{\tilde{y}}}\,\,\,?$$ Where does this come from? A plane wave has the form $\exp[i\,\mathbf{k}\cdot\mathbf{r}]$ without mention to the refractive index. In fact, when we separate the variables to solve the electromagnetic wave equation the spatial equation,

$$\left(\mathbf{\nabla}^{^{2}}+k^{^2}\right)\mathbf{A}(\mathbf{r}) = 0,$$

does not involve the refractive index at all. So why does the wavenumber changes like this?

Thank you very much.

I have to say, this is probably one of the most complicated approaches to this that I've seen. No wonder you're confused!

The refractive index is an interesting thing. It is a macroscopic quantity that can be defined in a few ways:

• the number that will give the appropriate refraction angle according to Snell's law
• the factor by which the wavelength is shorter inside the material compared to vacuum
• the factor by which the speed of light is slower inside the material compared to vacuum
• the square root of the relative dielectric permittivity times the relative magnetic permeability

There are probably more. These definitions are all consistent in the sense that any of them lead to the others. If you look at the second definition, you'll see that $\lambda=\lambda_0/n$. Since $k=2\pi/\lambda$, it is easy to see that inside a material, $k= n k_0$.

The fact that $k= n k_0$ is usually the starting point to the phase-matching-based derivations of Snell's law like the one you presented. It is the physical principle underlying all of the math you presented, promoting it from meaningless coordinate transformations to physics.

• So the $k$ in the wave equation already assumes implicitly that you are in a medium with refractive index $n$. Thank you very much. So simple, so classy. – Gabu Oct 12 '17 at 3:40
• Indeed. But if you are looking at the wave equation, check out the derivation from Maxwell's equations, which gives you the speed of light in terms of permittivity $\epsilon$ and permeability $\mu$. In a material, since these are different from their vacuum quantities, it means that the speed of light is different, by exactly the factor of $n$. This fact ultimately leads to $k=n k_0$. – Gilbert Oct 12 '17 at 3:57