# Optics – right handed circular polarization

**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.

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

**3. The attempt at a solution**

Well, I’m having a problem.

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

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?

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