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Polarized e/m wave and magnetism

  1. Apr 28, 2005 #1


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    Ok i was wondering something. If you send some light through a polarizer... how does the magnetism get through? I know you never find E without its M... but if you can block light with a polarizer (the light coming at a different angle), why would the magnetic part come out with the light that actually made it through the polarizer (seeing as how hte magnetic part is perpendicular)
  2. jcsd
  3. Apr 28, 2005 #2


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    Why would it be absorbed in the first place...?What is the polarizer made of...?

  4. Apr 28, 2005 #3
    The polariser only blocks the electric field component.
  5. Apr 28, 2005 #4

    Meir Achuz

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    The convention that the polarization is in the directioon of E is just a convention.
    It could just as well have been chosen to be the B direction, but E was chosen.
    The E and B fields in a wave need each other. If one is absorbed the other must also decrease in proportion. Polarizers that work by absorption can absorb either the E or B field. Then the other field also is reduced because of the interconnection.
  6. Apr 28, 2005 #5


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    A polarizer has to block either both the E and B fields, or neither of them.

    A polarizer has electrons that are constrained so that they can oscillate back and forth only along a certain direction. If the incoming E field is aligned along that direction, it causes the electrons to oscillate. As the electrons oscillate, they radiate an electromagnetic wave that has its E field aligned in the same direction, and of course a B field perpendicular to it. The "new" wave is out of phase with the incoming one, so they cancel, both E and B.

    If the incoming wave's E field is perpendicular to the allowed direction of oscillation of the electrons, the electrons don't oscillate. There are no "new" E and B fields to cancel the incoming ones, so the incoming fields go right on through.

    An ordinary kitchen cooking rack made of parallel metal rods makes a good polarizer for microwaves. To let the waves through, you have to orient the rods perpendicular to the E field.
  7. Apr 28, 2005 #6
    Couldn't it be so that: given that, statistically speaking, one half of the photons (regardless of lambda) are spin +one and other half are spin -one and further, that the population penetrating a single polarizer is also 50:50. Now imagine that E is associated with say, +1 spin, and B is associated with -1 spin. When a second polarizer is coupled to the first, the transparency of both E and B photons pass freely through both when the grids are parallel but when the grids approach perpendicular the transparencies of both E and B are extinguished. Its just a thought. Cheers, Jim
  8. Apr 28, 2005 #7

    Hans de Vries

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    This is the best anwser. One could add that the magnetic field has always
    the relation [itex] \vec{B}\ =\ \hat{R}\ \times \vec{E}[/itex], where [itex] \hat{R}[/itex] is the unit vector from the source of the
    radiation. The magnetic field is not an extra degree of freedom.
    (see also Jackson 14.13)

    The magnetic field B is stil very useful as a seperate field because the
    effects of many sources from all directions may be can be added to a single
    effective value B. The E components of multiple sources for instance may
    cancel each other out, while the B components can add up.

    Regards, Hans.
  9. Apr 28, 2005 #8


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    Could you explain how this happens. I'm not clear on why B fields can't cancel.


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