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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)
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  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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