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Optics and polarization

  1. Jan 12, 2012 #1
    1. The problem statement, all variables and given/known data
    Hi

    This isn't homework, but something I've wondered about. I post it here, because it might be helpful to others. Say I am looking at the attached optical system:

    I have polarized light in the plane. Then it hits a lambda/4-plate, where the angle between the fast axis and the polarization of the light is denoted Ω. The light then gets reflected on a mirror, and moves through the quarter wave-plate again.

    If Ω=0 degrees, light will just go back the same way it came with the same polarization.
    If Ω=45 degree, light becomes circularly polarized after the wave-plate, changes rotation direction after the reflection and becomes linearly polarized again. In total a phase change of 360 degrees (180 from the reflection and 180 from passing through the waveplate twice). So this is the same as the case with Ω=0? This doesn't make sense to me.

    Can anyone point out where my reasoning is wrong?

    Best,
    Niles.
     

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  3. Jan 12, 2012 #2

    Redbelly98

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    Edit: I have decided this thread is more appropriate for Classical Physics, and moved it there.

    The 180 degree phase change due to two passes through the λ/4 plate refers to the difference in phase between two spatial components of the E-field, the components parallel to and perpendicular to the waveplate's optic axis. Note that the effect is to rotate the (linear) polarization by 90 degrees.

    The 180 degree phase change at the mirror actually flips the E-field by 180 degrees, so that the reflected beam cancels the incident beam right at the mirror surface. There is no change in polarization from this. [EDIT: another way to think of it is that the E-field is retarded (or advanced, doesn't matter in this case) by 180 degrees in time when it reflects from the mirror. Again, no change in polarization. The 180 phase change refers to something entirely different than it does for the waveplate. END OF EDIT]

    Hope that helps.
     
    Last edited: Jan 12, 2012
  4. Jan 13, 2012 #3


    Hi

    Thanks for that. The reflection from the mirror causes an L-state to become an R-state. How can this happen if the individual components are not changed by a phase of 180 degrees wrt. eachother?

    Best,
    Niles.
     
  5. Jan 13, 2012 #4

    Claude Bile

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    My understanding is that "handedness" depends on the relative direction one observes the propagating wave. That is, a wave observed from in front might be right-handed, but observed from behind would be left-handed. Perhaps frustratingly there is no set convention, so when one defines a "handedness", one must also define the direction the wave is being observed.

    Anyway, in light of this, it is no surprise that a reversal in direction should change the handedness of a wave.

    Claude.
     
  6. Jan 13, 2012 #5
    Thanks for that, so the total phase change between the two components is just 180 degrees then. So I guess the arrangement can be used to decide how much of the incoming light is coupled out?
     
  7. Jan 14, 2012 #6

    Redbelly98

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    I don't understand what your are asking -- so far we are (or I was, at least) assuming 100% reflection for the mirror, and 100% transmission by the waveplate. What am I missing? Is there a linear polarizer somewhere?
     
  8. Jan 14, 2012 #7
    Sorry, I meant that if we put a beam splitter cube to the left of the quarter wave-plate we could determine how much light is coupled out.
     
  9. Jan 14, 2012 #8

    Redbelly98

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    Ah, okay.

    Since rotating the waveplate would change how much the final polarization has been rotated, that would change how much of the beam's power is transmitted (or reflected) by a polarizing beam splitter cube. And the polarization is rotated twice as much as the angle between the waveplate's optic axis and the incident polarization.

    (Hope that answers the question.)
     
  10. Jan 14, 2012 #9
    Thanks, that is very kind of you.

    Best wishes.
     
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