# Some questions about polarization and intensity

1. Sep 10, 2011

### a b

Hello everybody,
I have some questions:

I'm talking only about travelling waves, not standing waves.
If a linear polarized electromagnetic wave has electric field amplitude E_0, i know that its intensity is given by

I=(1/2) (E_0^2 ) /c

that is the Poynting vector averaged over a period.

But if I have the same E_0 amplitude in a circularly polarized wave, the Poynting vector remains constant so i'd expect to have an intensity

I= (E_0^2)/c

this difference sounds strange to me: am I right or am I making some mistake?

and also, if I send a circularly polarized wave through a linear polarizer, what will it be the intensity of the wave exiting from the polarizer?
And if I send a linearly polarized wave through a circular polarizer (circular dichroic filter), am I right if I say that the exiting wave will have the same intensity but with a linear polarization rotated in respect to the ingoing wave?

Thank you in advance and sorry if my english is not very good.

2. Sep 10, 2011

### BruceW

Actually, the time-average of the Poynting vector of a linearly polarised plane wave is equal to:
$$\frac{\varepsilon_0 c}{2} E_0^2$$

3. Sep 10, 2011

### a b

to BruceW:
I was certainly wrong , but maybe you too, or maybe you were just using a notation that I don't know:
if you define Poynting vector as
$$\vec S= \frac{1}{\mu } \vec E \times \vec B$$

then you have, for linear polarization,

$$<S>= \frac{1}{2} \sqrt{\frac{\varepsilon }{\mu}} E_0^{2}$$

thank you for your correction.
As you see my questions are still valid, correcting the formulas i previously wrote.

Last edited: Sep 10, 2011
4. Sep 10, 2011

### Ken G

A circularly polarized wave can be thought of as a superposition of two linearly polarized waves that are 90 degrees out of phase with each other (which counts as "oppositely polarized" for linear polarization). So you shouldn't expect the Poynting flux of a single linearly polarized wave to equal that of a superposition of two. Thus you are right, but there is no problem with it. Also, when you pass through a linear polarizer, it just selects the component with that linear polarization.