1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Optics - right handed circular polarization

  1. Jul 3, 2014 #1

    DataGG

    User Avatar
    Gold Member

    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

    DataGG

    User Avatar
    Gold Member

    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

    DataGG

    User Avatar
    Gold Member

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

    BiGyElLoWhAt

    User Avatar
    Gold Member

    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

    DataGG

    User Avatar
    Gold Member

    Oh yes, that's definitely a typo. Where it reads ##kx## it should be ##kz##. Thanks for the heads-up, will edit right away!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted