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Linearly polarized light on a quarter wave plate

  1. Nov 8, 2012 #1
    1. The problem statement, all variables and given/known data
    A linearly polarized beam propagates in the z-direction with its E-field
    oscillating in the y-direction. It is incident on a quarter wave plate (QWP) located in the
    x-y plane at the origin.

    a. How should the fast and slow axes of the QWP be oriented if the beam emerging
    from it is to be right hand circularly (RCP) polarized? State the directions.

    b. Assume the light wave incident on the QWP can be written as

    [itex] E(z, t)= E_{0}\widehat{y} e^{i(kz -\omega t)}[/itex]. What is the complex unit vector that describes the
    polarization of the wave leaving the QWP? Write this down explicitly in terms of [itex]\widehat{x}[/itex] and [itex] \widehat{y}[/itex] . Write down an explicit form of the wave.

    c. Now another identical QWP is placed in the beam following the first one, and its
    fast and slow axes are oriented in exactly the same manner as the first one. Write
    down a form for the light wave that emerges from the second QWP.



    2. Relevant equations
    Not sure what should be here. There's always the form of the incident E-field: [itex] E(z, t)= (E_{f}\widehat{f} + E_{s}\widehat{s})e^{i(kz -\omega t)}[/itex] where the f and s axes are the quarter-wave plate's fast and slow axes.


    3. The attempt at a solution
    a. For this one I just sorta reasoned it out geometrically. I said the fast and slow axes had to be at 135 degrees and 225 degrees (second and third quadrant). I'm not sure if this is right though. I think this makes it right handed polarized from the point of view of the receiver.

    b. I'm not entirely sure what to do about the vector describing the polarization. I know it has to be rotating, so my best guess would be: [itex] (\widehat{x} + \widehat{y})e^{-i(\omega t)}[/itex]. I'm not sure what exactly defines the "polarization vector". Is it just the vector that points in the plane of the E-field at any given time?

    I think since the fast and slow axes make a 45 degree angle with polarization of the light (y axis) then the form of the emitted wave should just be: [itex] E(z, t)= \frac{E_{0}}{\sqrt{2}} (\widehat{x} + \widehat{y})e^{i(kz -\omega t)}[/itex]
    Although I could be oversimplifying things.

    c. This one is really confounding me. I think that if the polarizer was of the opposite handidness, then there would be zero emitted light (or I could be wrong on this too). However, since it's an identical polarizer, I'm not really sure what would happen. My instinct is that nothing would happen to the emitted light, but I don't have much to back this up with.

    Thanks for the help.
     
  2. jcsd
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