# Optics - right handed circular polarization

1. Jul 3, 2014

### DataGG

1. The problem statement, all variables and given/known data

Write an expression for a light wave circular polarized to the right, traveling in the positive ZZ direction, such that the electric field points in the negative XX direction at z=0, t=0.

2. Relevant equations

Right handed polarization is the same as clockwise, I think..

$E_{0x} = E_{0y}$ throughout the whole thing.

$$\vec{E}(z,t)=E_{0x}sin(kz - wt)\vec{i} + E_{0y}sin(kz - wt + \frac{\pi}{2})\vec{j}$$

3. The attempt at a solution

Well, I'm having a problem.

$$\vec{E}(z,t)=E_{0x}cos(kz - wt)\vec{i} + E_{0y}sin(kz - wt)\vec{j}$$

At $z=0$, we have

$$\vec{E}(0,t)=E_{0x}cos(wt)\vec{i} - E_{0y}sin(wt)\vec{j}$$

So it indeed moves clockwise (co-sine decreases while the sin increases negatively)

At $z=0$ and $t=0$:

$$\vec{E}(0,0)=E_{0x}\vec{i}$$ Which does not point in the negative XX direction.

If I place a minus in the co-sine term, then it'll look like this:

$$\vec{E}(0,t)= - E_{0x}cos(wt)\vec{i} - E_{0y}sin(wt)\vec{j}$$

and that means that the co-sine will decrease negatively and the sin increase negatively. That would be a left handed circular polarization, because it would be counter-clockwise. I think?

So, how do I go about doing this?

Last edited: Jul 3, 2014
2. Jul 3, 2014

### DataGG

I think I solved my problem. No idea how I didn't see this sooner. Well, I actually did, but I dismissed it for some reason..

I think the following equation satisfies everything:

$$\vec{E}(z,t)=-E_{0}cos(kz - wt)\vec{i} - E_{0y}sin(kz - wt)\vec{j}$$

Last edited: Jul 3, 2014
3. Jul 3, 2014

### DataGG

Can anyone confirm it? Is it correct?

4. Jul 3, 2014

### BiGyElLoWhAt

While optics is a bit out of my realm, it looks as though what you have will definitely rotate clockwise. If you're not sure you can check the direction of the derivative at a point on the unit circle. If you want the direction it's pointing in check the direction of your function at that point.

Just a quick question, if you have E(z,t), why are you setting kx = 0 at E(0,t) ?

5. Jul 3, 2014

### DataGG

Oh yes, that's definitely a typo. Where it reads $kx$ it should be $kz$. Thanks for the heads-up, will edit right away!