# Intensity of elliptically polarized light

1. Jul 13, 2012

### Wox

The time averaged norm of the Poynting vector of this electromagnetic field (elliptically polarized light):
$$\begin{split} \bar{E}(t,\bar{x})=&(\bar{E}_{0x}+\bar{E}_{0y}e^{i \delta})e^{\bar{k}\cdot\bar{x}-\omega t}\\ \bar{B}(t,\bar{x})=&\frac{1}{\omega}(\bar{k}\times\bar{E}(t,\bar{x})) \end{split}$$
with $\bar{E}\perp\bar{B}\perp\bar{k}$, becomes (as I calculated in SI-units $J/(m^{2}s)$)
$$I(\bar{x})=\left<\left\|\bar{P}(t,\bar{x})\right\|\right>=\frac{c\epsilon_{0}}{2}(\bar{E}_{0x}^{2}+2\bar{E}_{0x}\cdot\bar{E}_{0y}\cos\delta+\bar{E}_{0y}^{2})$$
I have been trying to verify this, but I can't find a source that explicitly discusses this. For a linear polarized beam, $\delta=0$ so that $I(\bar{x})=\frac{c\epsilon_{0}(\bar{E}_{0x}+\bar{E}_{0y})^{2}}{2}$, which is correct. For general elliptical polarization I found this link which basically says that
$$I(\bar{x})=E_{x}E_{x}^{\ast}+E_{y}E_{y}^{\ast}= \bar{E}_{0x}^{2}+\bar{E}_{0y}^{2}$$
which can't be right (as it doesn't work for linear polarized light). Does anyone know of a proper reference for this? Or even better, can someone verify my solution?

2. Jul 13, 2012

### Wox

Ok, this is embarassing. I didn't see that $\bar{E}_{0x}\cdot\bar{E}_{0y}=0$ which fixes the problem.