Some questions about polarization and intensity

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The discussion centers on the intensity calculations of linear and circularly polarized electromagnetic waves. It clarifies that the intensity of a linearly polarized wave is given by I=(1/2)(E_0^2)/c, while for a circularly polarized wave, the intensity is I=(E_0^2)/c due to the constant Poynting vector. When a circularly polarized wave passes through a linear polarizer, only the component aligned with the polarizer is transmitted, affecting the intensity. Conversely, a linearly polarized wave passing through a circular polarizer retains its intensity but has its polarization direction rotated. The conversation emphasizes the differences in wave behavior and intensity based on polarization type.
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Hello everybody,
I have some questions:

I'm talking only about traveling 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.
 
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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
 
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:
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.
 

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