Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

If the electromagnetic field of an unpolarized plane wave is written as $$\bar{E}(t,\bar{x})=(\bar{E}_{0x}+\bar{E}_{0y}e^{i\delta(t)})e^{i(\bar{k}\bar{x}-\omega t)}$$ $$\bar{B}(t,\bar{x})=\frac{1}{\omega}\bar{k}\times\bar{E}(t,\bar{x})$$ where $\delta(t)$ is a random phase shift, then the intensity of this light is given by the time-average of the norm of the Poynting vector $$I(\bar{x})=\left<\|\bar{P}(t,\bar{x})\|\right>_{t}$$ $$\begin{split}\bar{P}(t,\bar{x})=&\frac{1}{\mu_{0}}\mathcal{R}e(\bar{E}(t,\bar{x}))\times\mathcal{R}e(\bar{B}(t,\bar{x}))\\ =&\frac{1}{\omega}\bar{E}_{0x}^{2}\cos^{2}(\bar{k}\bar{x}-\omega t)\bar{k}+\frac{1}{\omega}\bar{E}_{0y}^{2}\cos^{2}(\bar{k}\bar{x}-\omega t+\delta(t))\bar{k} \end{split}$$ $$\Leftrightarrow I(\bar{x})=\frac{1}{2}c\epsilon_{0}\bar{E}_{0x}^{2}+c\epsilon_{0}\bar{E}_{0y}^{2}\left<\cos^{2}(\bar{k}\bar{x}-\omega t+\delta(t))\right>_{t}$$ Can we simplify this further? Is the remaining average also $1/2$?

share|improve this question
    
You've got issues with your definition. –  user2963 Jul 6 '12 at 16:08
2  
That light is quite polarized. To get unpolarized light you need to average over an ensemble with random polarization. You've simply described some light with a very pure elliptical polarization –  Colin K Jul 6 '12 at 16:27
    
You are right, I introduced the random phase shift to go from elliptical to unpolarized, but this is indeed not enough. But the question still remains, how would you calculate the intensity of unpolarized light? –  Wox Jul 10 '12 at 10:00

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.