Some questions about polarization and intensity

[a b]

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.

 

[BruceW]

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

 

[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.

 

[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.

 

[sardar]

I would like to address your questions about polarization and intensity. First, it is important to note that the intensity of an electromagnetic wave is a measure of its power per unit area, and is related to the amplitude of the electric field. So, for a linearly polarized wave, the intensity is given by I=(1/2)(E_0^2)/c, where E_0 is the electric field amplitude and c is the speed of light. This is derived from the Poynting vector, which represents the average power per unit area carried by the wave.

Now, for a circularly polarized wave, the intensity is also related to the electric field amplitude, but the Poynting vector is not constant. This is because the electric field is rotating in a circular motion, and the direction of the Poynting vector changes along with it. Therefore, the intensity for a circularly polarized wave is given by I=(E_0^2)/c. This may seem counterintuitive, but it is a result of the rotating electric field.

Moving on to your questions about polarizers, when a circularly polarized wave is passed through a linear polarizer, the resulting wave will have a lower intensity. This is because the polarizer only allows waves with a specific polarization direction to pass through, and a circularly polarized wave has components in both linear polarizations. So, the intensity will decrease due to the reduction of the electric field amplitude.

Similarly, when a linearly polarized wave is passed through a circular polarizer, the resulting wave will have the same intensity, but with a rotated polarization direction. This is because the circular polarizer only allows waves with a specific circular polarization direction to pass through, and a linearly polarized wave can be thought of as a combination of two circularly polarized waves with opposite directions. So, the resulting wave will have the same intensity, but with a different polarization direction.

I hope this answers your questions and clarifies any confusion about polarization and intensity. Your English is perfectly fine, and I am happy to help with any further questions you may have. Keep exploring and learning about the fascinating world of electromagnetic waves!