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Polarizing EM radiation, how does it work?

  1. Jul 16, 2013 #1
    for RF, how does this mechanism work?

    http://i.imgur.com/Crx4Amx.png

    how does the conductor do this? and why does the unpolarized RF not seep through the spaces between the conductive bars?
     
  2. jcsd
  3. Jul 16, 2013 #2

    Cthugha

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    Actually the polarizer does not work as shown on the picture. It is the electric field perpendicular to the wire grid which can pass the polarizer. Electromagnetic fields hitting a conductor will cause the electrons to move along the wire. This is more or less similar to light shining on a metal plate. The result is also the same: Significant portions of the wave get reflected and next to nothing is transmitted. As the conductors are wires, electrons can only move significantly in one direction, while they cannot move much in the perpendicular direction. Therefore, the electric field component which does not cause significant electron movement will be transmitted, while the other will not.

    The waves cannot "pass" the empty space between the bars because we are talking about radio waves here. High frequency radio waves have a wavelength of 1m and more - much larger than the empty space between the bars. The design would not work, if the distance was much larger than the radio wavelength. In that case the radio wave could indeed simply pass through. So the bar distance needs to be small compared to the wavelength of interest. Otherwise the polarizer will not work.
     
  4. Jul 16, 2013 #3
    oh darn!.... i haven't thought about this topic for along time so i forgot.. lol

    alrighty fine, the distance between the bars must be greater than the wavelength, but why? i don't see the relationship between the these two scalars; they are measured on different axes, why does the horizontal direction in the +y direction of the wavelength affect the horizontal direction in the +/-x direction of the distance between the bars?
     
  5. Jul 16, 2013 #4

    Cthugha

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    Smaller actually.

    Hmm, I am not sure I get what you mean. The wavelength is a rough measure of how well you can localize a wave in space. Even with focusing you cannot easily confine it into a space of less than about the length of half a wavelength (in any dimension). So if the bar distance is much larger than the wavelength, the wave literally passes through the openings in between. Only if the distance is smaller, the wave "sees" the polarizer as an effective medium.
     
  6. Jul 16, 2013 #5
    well, radiowaves produced by an antenna will be due to electrons moving back and forth in the antenna wire.

    initially, the charges (at rest) on the antenna wire will already have a static field radiating from it. when the charges start moving back and forth, the acceleration of charges produce the wave in the electric field thereby resulting in an electromagnetic wave. so when i say wavelength, i am strictly referring to the dimension oF wavelength, that which exists along the dimension of radiation that is... is there something i am missing that is happening along te perpendicular horizonal dimension/axis? because i've gotten that explanation plenty of times where someone would tell me that the RF doesn't go through because the wavelength of the RF is sufficiently larger than the spatial distances between the conductive bars. but if you have a wave travelling in a direction, then the wavelength is measured NOT in some perpendicular direction, wavelength is measured in the same dimension as the direction of propogation. so Why is the wavelength a factor in this?
     
  7. Jul 16, 2013 #6

    Cthugha

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    This is a question of diffraction. http://en.wikipedia.org/wiki/Diffraction. For real beams, it is impossible to focus them down to a spot, but the best you can get is an airy disk. Its diameter is given by:
    [tex]d=1.22 \lambda N[/tex],

    where N is the focal length of the lens used divided by the diameter of that length. Collimated beams are typically even broader in space. Assuming a wavelength of at least 1m, it is evident, that such waves will be very broad in real space.

    If it helps, consider a point source: it will radiate into all directions. Now imagine two point sources next to each other and emitting in phase. Draw the resulting field from interference. The emission pattern will still be broad, but somewhat smaller. Now take 3, 5 10, sources next to each other. The directionality will increase. You will find that the point where the wavefront stops spreading badly and becomes somewhat similar to a plane wavefront will appear, if you place enough point sources next to each other, so that the total distance covered by these point sources is on the order of the wavelength o the light.
     
  8. Jul 16, 2013 #7
    thank you so much! you have provided the first helpful answer to this question that i can actually make a connection with, with something i know of.

    ugh.. i'm sorry but would you perhaps be willing to illustrate what you meant by this? you lost me starting from directionality continuing on to the end. could you illustrate what you mean or provide an illustration?
     
  9. Jul 17, 2013 #8

    Cthugha

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    Maybe you can play around a bit with this applet.

    http://micro.magnet.fsu.edu/primer/java/diffraction/basicdiffraction/

    It shows single slit diffraction. You can vary the wavelength of your light and the width of the slit. Check what happens, when the aperture is very large and what happens when the size of your aperture is similar to your wavelength. See how your light field spreads out.

    Now the important thing to note is, that your wave field will always spread out like that when it is distributed over a spatial extent smaller than your wavelength, not only when a slit forces it to, so it naturally broadens: The beam will become wider than the wavelength of your wave. You cannot get smaller beams.
     
  10. Jul 26, 2013 #9
    i am still lost by how polarization via this method works. i think my problem is that i don't understand light. i thought light was a transverse wave of the electric field. So for radio waves, the antenna has oscillating electrons and this produces the oscillating electric field accompanied by the B-field. isn't this nature true for all EM radiation? (not talking about how it's produced i'm referring to the nature of what 'EM radiation', i'm not looking for quantitative specifics; i'm looking for a qualitative description of the nature of light, which i basically just provided.. i'm just asking whether or not this is correct).
     
  11. Jul 31, 2013 #10

    Greetings,
    The wave nature of light can be described as a sinusoidally varying electrical field which gives rise to a correspondingly varying magnetic field which again creates the electric field (they create and replenish each other). So, yeah you are correct as far as classical wave mechanics is concerned.
    Now on the topic of polarization: as the beam is incident of the polarizer, which contains long chain polymers, the electric field vectors parallel to these chain will push the electrons along these chain and thus those waves associated with these vectors will get absorbed. Hence light is polarized.
    As for the distance between two chains or rods I really don't have an idea about the math.
    But in order to be polarized the distance will have to be atleast comparable to wavelength, (according to Feynman the distance should be something around half of the wavelength...)
    Hope it was of any help
    Regards
     
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