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About polarization and polarizer

  1. Jun 12, 2015 #1

    KFC

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    Hi all,
    To me, polarization is the most confusing concepts in optics to me. From the text, it is said that polarization is how the light oriented. A linear polarization means the light is oriented on one direction. But I have few doubts I learn from the book and online materials

    1) In most book and some youtube material, it is said that a linear polarizer is used so an unpolarized light only have one orientation of the light can pass through. My understanding is an unpolarized light has components of all direction. Even some components is not along the transmission direction of the polarizer, but those components could have some projection along the transmission direction. So basically, all components oriented at any direction could pass the liinear polarizer but some direction will all go through some will just a bit, depends on the projection. Is that correct?

    2) In wiki (https://en.wikipedia.org/wiki/Linear_polarization), it explains the linear polarization as linear-polarized along a plane consisting of two orthogonal, in-phase components. My first question is why there will be two components of the electric field? I know electric field component is a vector so it could decomposed into any two components, is this the reasoning why there are two components? If so, can I say linear polarized light is the electic field oriented a specific direction which doesn't not change in time so the decomposed two components are always in-phase?

    3) I read a video in youtube about circular polarization. The guy explain that a circular polarization means two components of the electric fields are not moving at the same speed, one is lag by 1/4 wavelength than the other, so the tip of the synthesized electric field vector is along a helix. If we view from the direction of propagation, we see a circle. Am I understand this correctly?

    4) I remember I read it somewhere about circular polarized light reflected by a mirror, will the direction of circular polarization opposite if we use a mirror to reflect the light in circular polarization? For example, if we have a clockwise circular polarized light inject into a mirror making 30 degree to the light, will the reflected light becomes counterclockwise circular polarized? why?

    5) the last question is about how to check the circular polarization in experiment. I read this online, someone use a quarter-wave plate and a beam splitter cube. Passing the circular polarized light to the quarter-wave plate and left the outgoing light through the beam splitter, the component perpendicular to the direction of propagation should be zero if the incoming light is circular polarized, why is that?

    Sorry for the long question. I am trying to collect as much information as I can to understand the light polarization, I just want to confirm with someone here that my understanding is correct or not.
     
  2. jcsd
  3. Jun 12, 2015 #2

    blue_leaf77

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    1) Yes that's correct.
    2) That's why it's called linear polarization. In general, 180 degree phase difference also amounts to linear polarization.
    3) Probably you have misunderstood what you have heard from that source. As long as there is no birefringence in the medium through which the light propagates, both the x and y components travel with the same speed. One component only lag from the other. If you imagine you can see the tip of the vector, yes it should look like a circle.
    4) If you are being consistent in defining what you call CW or CCW (i.e. if you see the incoming wave by facing it head on then you should also see the reflected one by facing it head on), both the incoming and and reflected waves should have identical polarization state. Note that reflection in general might change the state of polarization as Fresnel formulae for reflectance tell us that the reflection coefficients for the TE and TM components generally differ.
    5) Would be better if you provide the link.
     
    Last edited: Jun 12, 2015
  4. Jun 12, 2015 #3

    A.T.

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    I guess this one?



    Differential speed is only used to create the lag. After passing the quarter-wave plate the components have the same speed again, but retain the λ/4 phase shift.
     
    Last edited: Jun 12, 2015
  5. Jun 12, 2015 #4

    A.T.

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    The clockness is defined wrt to the propagation direction, which swaps during reflection.

    Passing circularly polarized light through a quarter-wave plate will make it linearly polarized, which can be identified with a linear polarizer at different angles.
     
  6. Jun 12, 2015 #5

    blue_leaf77

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    Upon checking in my textbook, that seems to be true and the reason is that because the reflection coefficients for the TE and TM have opposite sign. This actually have been puzzling me for a while because for normal incidence how could we tell which component should be TE or TM?
     
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