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Circular Polization type in an EM wave.

  1. Dec 18, 2011 #1
    My bad for the spelling fail in the title.

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

    I am utterly confused with the mathematics in a section of my notes, I swear that it's wrong - and it related to a piece of homework which I cannot understand.

    The electric field vector in a z-directed monochromatic EM wave is given by;

    http://img213.imageshack.us/img213/4768/58814275.jpg [Broken]

    It says that if [itex]\alpha[/itex] = [itex]\pi[/itex]/2, [itex]\theta[/itex] = 0 and [itex]E_{0,x}[/itex] = [itex]E_{0,y}[/itex]

    Then

    [itex]E_{x}[/itex] = [itex]E_{R}[/itex]cos(kz-wt),

    [itex]E_{y}[/itex] = [itex]E_{R}[/itex]sin(kz-wt)

    and [itex]E_{R}[/itex] = [itex]\frac{1}{\sqrt{2}}E_{0,x}[/itex]

    3. The attempt at a solution


    I do not see how the component in y direction is right. Taking the parameters [itex]\alpha[/itex] = [itex]\pi[/itex]/2, [itex]\theta[/itex] = 0, [itex]E_{0,x}[/itex] = [itex]E_{0,y}[/itex]

    we get

    [itex]E_{0} = \frac{1}{\sqrt{2}}[E_{0,x} \hat{x} + E_{0,x} e^{i(\pi/2)} \hat{y}][/itex]

    [itex]e^{i(\pi/2)} = i[/itex]

    [itex]E_{0} = \frac{1}{\sqrt{2}}[E_{0,x} \hat{x} + iE_{0,x} \hat{y}][/itex]

    Subbing this in the equation for overall E.

    [itex] E = Re(\frac{1}{\sqrt{2}}[E_{0,x} \hat{x} + iE_{0,x} \hat{y}]e^{i(kz-wt)}) [/itex]

    [itex]E_{R}[/itex] = [itex]\frac{1}{\sqrt{2}}E_{0,x}[/itex]

    [itex] E = Re(E_{R}[\hat{x} + i\hat{y}]e^{i(kz-wt)}) [/itex]

    [itex] E = Re(E_{R}[\hat{x} + i\hat{y}][cos(kz-wt)+isin(kz-wt)]) [/itex]

    [itex] E = E_{R}[\hat{x} cos(kz-wt) - \hat{y}sin(kz-wt)] [/itex]


    So then surely for [itex]\alpha[/itex] = [itex]\pi[/itex]/2

    We have

    [itex]E_{x}[/itex] = [itex]E_{R}cos(kz-wt)[/itex]

    [itex]E_{y}[/itex] = -[itex]E_{R}sin(kz-wt)[/itex]

    So the [itex]E_{y}[/itex] direction is minus instead?

    Similarly for [itex]\alpha[/itex] = [itex]-\pi[/itex]/2 We'd have;

    [itex]E_{x}[/itex] = [itex]E_{R}cos(kz-wt)[/itex]

    [itex]E_{y}[/itex] = [itex]E_{R}sin(kz-wt)[/itex]

    Is this correct?

    [itex]\alpha[/itex] = [itex]\pi[/itex]/2 should correspond to left circular polarization, and [itex]\alpha[/itex] = [itex]-\pi[/itex]/2 to right.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
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