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Circularly polarized light through a linear polarizer

  1. May 5, 2006 #1
    Anyone know why circularly polarized light, when observed after passing through a linear polarizer, would exhibit minima in intensity? It would seem to me that, since the linear polarizer only allows a particular plane of vibration through, and the circularly polarized light rotates about with a given angular frequency, that integrated over a period you'll just see a constant amplitude independant of the polarizer's angle. So, what gives?
     
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  3. May 5, 2006 #2

    George Jones

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    What is the relationship between circular polariztion and linear polarization?

    Regards,
    George
     
  4. May 5, 2006 #3
    Linearly polarized light has two electrical field components that are pi*n rad out of phase (where n is an integer), while circularly polarized light has two components (pi/2) + n*pi rad out of phase. However, don't most linearly polarizers work by actively cancelling the component of the electrical field perpendicular to its transmission axis, so phase difference wouldn't be the thing driving a linear polarizer? Wouldn't the light only exhibit extrema if the components of the electrical field were not equal, and in fact you were getting an elliptically polarized output from the circular polarizer?

    The way I was thinking about it is, if you have two components, Ex and Ey, and you pass them through a circular polarizer which then adds a phase shift to the Ey component, and then passed the resultant through a linear polarizer, then the field passing through would be equal to the dot product of the circularly polarized electrical field vector and the unit vector of the transmission axis of the linear polarizer. Averaged over a period this yields an intensity of

    <I> = (1/4)*(E_0x^2 + E_0y^2 + (E_0x^2 - E_0y^2)*cos(2*theta))

    Where theta is the angle between the x axis of the incoming light (propagating in the +z direction) and the transmission axis of the linear polarizer, and E0x and E0y are the amplitudes of the x and y components of the electical field. Hence, no minima would be observed if E0x=E0y. Is this analysis incorrect?
     
    Last edited: May 5, 2006
  5. May 5, 2006 #4

    George Jones

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    I'm a little confused by this. It seems that for linearly polarized light, you're considering both electric and magnetic fields, while for circularly polarized light, you're considering only electric fields.

    I think its OK to consider just the electric fields.

    Regards,
    George
     
  6. May 5, 2006 #5
    No, I'm only considering electrical field in both instances. The way my text goes about explaining polarization is by representing the electrical field in terms of two components, an x component and a y component, propogating in the +z direction, so that

    E = E_0x*cos(kz - wt) (i_hat) + E_0y*cos(kz - wt +theta)(j_hat)

    where k has the usual meaning, w is the angular frequency, E_0x and E_0y are the respective amplitudes of the electical field components, and theta is the relative phase between the two electical field components. So, linearly polarized light is just light for which theta = pi*n radians, and circularly polarized light has theta = (pi/2) + n*pi radians.
     
    Last edited: May 5, 2006
  7. May 5, 2006 #6

    George Jones

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    Is your text Hecht and Zajac? If so, and if the newest edition is similar to my old edition, the paragraph previous to the one that conatins your expression explains that the expression results from the superposition of 2 linearly polarized light waves, i.e., both

    [tex]E_{0x} cos \left (kz - wt \right) \hat{i}[/tex]

    and

    [tex]E_{0y}cos \left( kz - wt + \epsilon \right) \hat{j}[/tex]

    are the electric fields of linearly polarized (along different axes) light waves.

    Depending on the values of [itex]E_{0x}[/itex], [itex]E_{0y}[/itex], and [itex]\epsilon[/itex], the superposition can result in: linear polarization; elliptical polarization; circular polarization.

    For example, [itex]E_{0x} = E_{0y} = E_{0}[/itex] and [itex]\epsilon = - \pi/2[/itex] give circular polarization. This means that circularly polarized light results from the sum 2 linearly polarized waves, and this gives

    [tex]
    \begin{equation*}
    \begin{split}
    E &= E_{0} \left[ cos \left(kz - wt \right) \hat{i} + cos \left( kz - wt - \pi/2 \right) \hat{j} \right]\\
    &= E_{0} \left[cos \left(kz - wt \right) \hat{i} + sin(kz - wt)\hat{j} \right]
    \end{split}
    \end{equation*}
    [/tex]


    As you say, after going through a polarizer with transmission axis given by the unit vector [itex]\hat{n}[/itex],

    [tex]
    \begin{equation*}
    \begin{split}
    E' &= E \cdot \hat{n} \hat{n}\\
    &= E_{0} \left[ cos \left(kz - wt \right) cos \theta + sin \left(kz - wt \right) sin \theta \right] \hat{n}\\
    &= E_{0} cos \left(kz - wt - \theta \right) \hat{n}
    \end{split}
    \end{equation*}
    [/tex]

    Again, as you say this gives an average intensity independent of theta.

    Interesting.

    Before doing the calculation, I didn't know the answer.

    Regards,
    George
     
    Last edited: May 5, 2006
  8. May 5, 2006 #7
    Interesting enough, for sure, but unfortunately completely contradictory to the data I took in lab. :yuck: And yes, btw, my text is Hecht's Optics, 2nd ed.
     
    Last edited: May 5, 2006
  9. May 5, 2006 #8

    nrqed

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    I was going to ask you how you knew that there would be a minimum. It seemd clear to me that viewing circularly polarized light through a linear polarizer could not yield a change of intensity as the polarizer is rotated since there is no preferred axis for circularly polarized light!!!

    So if your affirmation was based on data collected from an experiment, the obvious question is: how do you know for sure it was perfectly circularly polarized light?? If it was elliptically polarized light, then for sure there would be a variation of intensity as one rotates the polarizer!
     
    Last edited: May 5, 2006
  10. May 5, 2006 #9

    George Jones

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    Of course! I shouldn't have needed a calculation.

    Regards,
    George
     
  11. May 5, 2006 #10
    I knew there was a minimum because I'd observed it. I would have expected there to be some difference between the components' amplitudes simply because in the real world, materials are non ideal and I would have expected an anistropic crystal to also be anistropic in regards to attenuation of the electric field with distance.

    As far as circularly vs. elliptically polarized, that was the conclusion I came to as well. The confusing part is that the relative angle at which the minimum occurs differs as the circular polarizer itself is rotated and moreover, the angle decreases with increasing angle of the circular polarizer, which from the equation I developed was not anticipated.
     
    Last edited: May 5, 2006
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