Transmission and Reflection

Consider if we have to dielectric media, $n,n^{\prime }$ with the xy plane being the interface between them. Let a wave propogate in the z-direction. What is the transmission and reflection of the wave? If $E=E\exp (i(kz-\omega t))\hat{x}$ then $B=\frac{nE}{c}\exp (i(kz-\omega t))\hat{y}$. The transmitted wave vector is $k^{\prime }=n^{\prime }\frac{\omega }{c}\hat{z}$ and reflected is $k^{″}=-k\hat{z}=-n\frac{\omega }{c}\hat{z}$. Then, $E^{\prime }=E^{\prime }\exp (i(kz-\omega t))\hat{x}$ and $E^{\prime }=E^{\prime }\exp (i(kz-\omega t))\hat{y}$. For the reflected wave, $E^{″}=E^{″}\exp (i(-kz-\omega t))(-\hat{x})$ and $B^{″}=B^{″}\exp (i(-kz-\omega t))\hat{y}$.

From our boundary conditions, $E_{t1}=E_{t2}$. Since our wave is already only tangential, the electric field is continuous. Similarly, $H_t$ is continuous so $H=B/{\mu }_0$ is continuous. From these we get,

$$\begin{array}{}\text{(1)}& E-E^{″}& =E^{\prime }\text{(2)}& E+E^{″}& =\frac{n^{\prime }}{n}{E}^{\prime }\end{array}$$

So,

$$\begin{array}{}\text{(3)}& E^{\prime }& =\frac{2n}{n^{\prime }+n}E=tE.\end{array}$$

This gives us the transmission coefficient of the interface for a normal wave propogation. $t=\frac{2n}{n^{\prime }+n}$. $r=1-T=\frac{n^{\prime }+n-2n}{n^{\prime }+n}=\frac{n^{\prime }-n}{n^{\prime }+n}$.

When $n^{\prime }$ is a denser medium, we get a phase flip in the reflected wave and a smaller transmitted wave. If it is not as dense, the phase is unchanged in the reflected wave and we get a larger transmitted wave.

If we consider the full expression for intensity, $I=|\overrightarrow{S}|=\frac{1}{2}nc{\text{ve}}_0|\overrightarrow{E}{|}^2$. $S_i=S_r+S_t=n|E{|}^2+n|E^{″}{|}^2+n^{\prime }|E^{\prime }{|}^2$.

$1=\frac{S_r}{S_i}+\frac{S_t}{S_i}=R+T$, the reflectance and transmittance. So, $1=\frac{|E^{″}{|}^2}{|E{|}^2}+\frac{n^{\prime }|E^{\prime }{|}^2}{n|E{|}^2}=|r{|}^2+\frac{n^{\prime }}{n}|t{|}^2$.

Now consider if we have an oblique angle of incident light, ${\theta }_i$ from $-z-axis$. So, $\overrightarrow{E}={\overrightarrow{E}}_0\exp (i(\overrightarrow{k}\cdot \overrightarrow{x}-\omega t))$. Then we have reflected ${\overrightarrow{k}}^{″}$ and ${\overrightarrow{k}}^{\prime }$ be the transmitted. Starting out with the true,

$$\begin{array}{}\text{(4)}& c\overrightarrow{B}=n\hat{k}\times \overrightarrow{E}\end{array}$$

We know the tangential components must be continuous, $E\exp (i(\overrightarrow{k}\cdot \overrightarrow{x}-\omega t))+E^{″}\exp (i({\overrightarrow{k}}^{″}\cdot \overrightarrow{x}-\omega t))=E^{\prime }\exp (i({\overrightarrow{k}}^{\prime }\cdot \overrightarrow{x}-\omega t))$. So, $E\exp (i(\overrightarrow{k}\cdot \overrightarrow{x}))+E^{″}\exp (i({\overrightarrow{k}}^{″}\cdot \overrightarrow{x}))=E^{\prime }\exp (i({\overrightarrow{k}}^{\prime }\cdot \overrightarrow{x}))$

Note, $\overrightarrow{k}=k_x\hat{x}+k_z\hat{z}$. Then, consider the phases (remembering continuity causes z-phases to cancel), so $f(x)=g(x,y)=h(x,y)$, thus it must only depend on $x$. So, at $z=0$, $\hat{k}\cdot \overrightarrow{x}={\hat{k}}^{\prime }\cdot \overrightarrow{x}={\hat{k}}^{″}\cdot \overrightarrow{x}$. Hence, $k_x=k_x^{\prime }x+k_y^{\prime }y=k_x^{″}x+k_y^{″}y$. Then, $k_y^{\prime }=k_y^{″}=0$. Thus, they are confined to the same plane. $k_x=k_x^{″}=k\sin {\theta }_i$. $k_x^{″}=k^{″}\sin {\theta }_r=k\sin {\theta }_r$. Therefore, ${\theta }_i={\theta }_r$. $k_x^{\prime }=k_x\Rightarrow k\sin {\theta }_i=k^{\prime }\sin {\theta }_t$. Thus, $n\sin {\theta }_i=n^{\prime }\sin {\theta }_t$.

Considering the pattern the incident light has on the boundary, we see that the reflected and refracted light must have the same pattern at the boundary, a period of $\frac{\lambda }{\sin {\theta }_i}=\frac{2\pi }{k_x}=\frac{\lambda }{\sin {\theta }_t}=\frac{2\pi }{k_x^{″}}$.

Consequences:

1. $\overrightarrow{k}$ confines ${\overrightarrow{k}}^{\prime }$ and ${\overrightarrow{k}}^{″}$ to the plane of incidence (i.e. xz)
2. ${\theta }_i={\theta }_r$
3. $n\sin {\theta }_i=n^{\prime }\sin {\theta }_t$