Refraction

Let’s consider what happens when a wave crosses a boundary between materials. We considered this in 1D, and we will now consider it in 2D or 3D.

Assume we have a boundary between two materials at $x=0$ , and an incident wave with wave vector $\mathbf k$ in the $x,y$ plane coming from the left. Just like in the 1D case, when the wave reaches the boundary we will have a reflected and a transmitted wave.

Let’s consider the boundary conditions. Just as in the 1D case, the waves on either side of the boundary must have the same frequency $\omega$ , otherwise there would be discontinuities at the boundary. If the boundary were a membrane, for instance, a discontinuity might correspond to the membrane breaking. Whatever the medium and wave though, the waves on either side need to be continuous.

Consider the following figure showing the wavefronts of an incident and transmitted waves. Because the wavelengths are different on either side of the boundary, the angle of the incident and transmitted wave must be different so that the boundary is continuous

This gives us the following boundary conditions.

\begin{align*} \psi_\text{incident} &= Ae^{i(\mathbf k \cdot \mathbf r - \omega t)} \\ \psi_\text{reflected} &= ARe^{i(\mathbf k_r \cdot \mathbf r - \omega t)} \\ \psi_\text{transmitted} &= ATe^{i(\mathbf k_t \cdot \mathbf r - \omega t)}. \end{align*}

At $x=0$ , $\psi_L=\psi_R$ , since the boundary doesn’t break, so

\begin{align*} Ae^{i(k_y y - \omega t)} + ARe^{i(k_{ry}y-\omega t)} &= ATe^{i(k_{ty}y-\omega t)}. \end{align*}

The only way this can be true for all $y$ is if $k_y=k_{ry}=k_{ty}$ . We can now relate the angular frequency to the wave vectors. Let’s assume both materials are not dispersive, then the dispersion relation is $\omega = v_p |\mathbf k|$ .

\begin{align*} |\mathbf k_r| &= \frac{\omega}{v_L} = \sqrt{k_{ry}^2 + k_{rx}^2} \\ |\mathbf k| &= \frac{\omega}{v_L} = \sqrt{k_{y}^2 + k_{x}^2} \\ |\mathbf k| &= |\mathbf k_r| \end{align*}

Since $k_y = k_{ry}$ , we know $k_x^2=k_{rx}^2$ . In this case since the boundary is along the $y$ axis, the reflection is flipped in $x$ , so $k_{rx} = -k_x$ .

Looking at the transmitted wave, we know that the $y$ component $k_{ty} = k_y$ is the same as the incident wave. We can use this to relate the incident angle $\theta$ to the transmitted angle $\theta'$ .

\begin{align*} \sin\theta &= \frac{k_y}{|\mathbf k|} = \frac{k_y v_L}{\omega} &\quad \frac{k_y}{\omega} &= \frac{\sin \theta}{v_L} \\ \sin\theta' &= \frac{k_{ty}}{|\mathbf k_t|} = \frac{k_{ty} v_R}{\omega} &\quad \frac{k_{ty}}{\omega} &= \frac{\sin \theta'}{v_R} \end{align*}

Which gives us the following relation

\frac{\sin\theta}{v_L} = \frac{\sin\theta'}{v_R}.

Refraction of light

In the case of light, we define an index of refraction $n := c/v \approx \sqrt{\epsilon / \epsilon_0}$ , (since $\mu \approx \mu_0$ for transparent materials) giving us the relation known as Snell’s law

n_1\sin\theta_1 = n_2\sin\theta_2.

This allows us to relate the transmitted angle to the incident angle. We will now consider how to relate the amplitudes of the incident, reflected, and transmitted waves. The exact behavior at the boundary will depend on the polarization of the incident wave. We can consider the incident light as a superposition of two polarization states; we will consider each separately.

P-polarization

The P-polarized component of the incident wave is at some angle from the boundary, its electric field shown in the figure.

We further break down this electric field into the components perpendicular and parallel to the boundary. First consider the perpendicular component (here the boundary is the plane in/out of the page).

We apply Gauss’s law at the boundary over the volume shown with cross sectional area $A$ and depth $d$ . In the limit as $d \to 0$ the enclosed charge $\rho$ goes to zero,

\oiint \epsilon \mathbf E \cdot \mathrm d \mathbf A = \epsilon_1 A \mathbf E_{P\perp1} - \epsilon_2 A\mathbf E_{P\perp2} = 0.

Which gives us $\epsilon_1 \mathbf E_{P\perp1} = \epsilon_2 \mathbf E_{P\perp2}$ .

Now consider the parallel component.

We apply Faraday’s law along the contour highlighted above. In the limit as $d \to 0$ , the magnetic flux $\Phi_B$ goes to zero,

\oint \mathbf E \cdot \mathrm d\mathbf s = l \mathbf E_{P\parallel1} - l \mathbf E_{P\parallel2} = -\frac{\mathrm d}{\mathrm dt} 0 = 0.

Which gives us $\mathbf E_{P\parallel1} = \mathbf E_{P\parallel2}$ .

We now use these conditions in combination with Snell’s law to relate theincident $E_{P,i}$ , reflected $E_{P,r}$ and transmitted $E_{P,t}$ amplitudes of the (P-polarized) waves.

From the perpendicular condition we find

\begin{align*} \epsilon_2 E_{P,t} \sin \theta' &= \epsilon_1 (E_{P,i} \sin \theta - E_{P,r} \sin \theta) \\ E_{P,i} - E_{P,r} &= \frac{\epsilon_2}{\epsilon_1} \frac{\sin\theta'}{\sin\theta} E_{P,t} = \frac{\epsilon_2 n_1}{\epsilon_1 n_2} E_{P,t} = \beta E_{P,t}, \\ \text{where } \beta &:= \frac{\epsilon_2 n_1}{\epsilon_1 n_2}. \end{align*}

And from the parallel condition,

\begin{align*} E_{P,t} \cos\theta' &= E_{P,i} \cos\theta - E_{P,r} \cos\theta \\ E_{P,i} - E{P,r} &= \frac{\cos\theta'}{\cos\theta} E_{P,t} = \alpha E_{P,t}, \\ \text{where } \alpha &:= \frac{\cos\theta'}{\cos\theta}. \end{align*}

Which finally allows us to relate the amplitudes

\begin{align*} E_{P,r} &= \frac{\alpha-\beta}{\alpha+\beta} E_{P,i} =: R E_{P,i} \\ E_{P,t} &= \frac{2}{\alpha+\beta} E_{P,i} =: T E_{P,i} \end{align*}

in terms of the reflection and transmission coefficients $R,T$ .

Notice a few things about this solution (for p-polarization only):

$\theta$ such that $\alpha = \beta$ is called Brewster’s angle, and is the case where $R=0$ and no light is reflected. This happens when $\theta + \theta' = \pi/2$ .
When $\theta = 0$ (perpendicular to the boundary), $\alpha = 1$ and $\beta = \frac{\epsilon_2 n_1}{\epsilon_1 n_2} = \frac{n_2}{n_1}$ , so ( R = frac{n1-n2}{n1n2}, quad T=frac{2n1}{n1n2}. )

Note the symmetry with the one dimensional case.

When $\theta = 90\degree$ (parallel to the boundary), $\alpha \to \infty$ , $R \to 1$ , $T \to 0$ , so there is no transmission.

S-polarization

The S-polarized component of the incident wave is in the plane of the boundary, its electric field shown pointing in and out of the page in the figure.

Considering the magnetic fields we can do a similar derivation to what we did above.

TODO: actually do it.