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Optics - right handed circular polarization

  1. Jul 3, 2014 #1


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    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. jcsd
  3. Jul 3, 2014 #2


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    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:

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

    Would someone please confirm?
    Last edited: Jul 3, 2014
  4. Jul 3, 2014 #3


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    Can anyone confirm it? Is it correct?
  5. Jul 3, 2014 #4


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    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) ?
    Should that be kz instead?
  6. Jul 3, 2014 #5


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