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Intensity of polarization

  1. May 1, 2006 #1
    I have a quick question about polarization. When I put a polarizer up an unpolarized source, say the polarizer blocks all the polarization in th y direction and so all the photons in the x direction comes out of the polarizer. Is the intesity halved?

    It seems to me that the intensity should be halved, however, as described in http://hyperphysics.phy-astr.gsu.edu/HBASE/phyopt/polcross.html#c2

    why isnt the polarization of the of the three polarizers 1/6 the regular intensity instead of 1/4 as it describes in the example?
  2. jcsd
  3. May 2, 2006 #2
    Follow the “Law of Malus” at the same web site.
    After the light is polarized by filter 1 and then filter 2
    Malus says 1/2 get though – and will be aligned with #2
    Filter 3 is 45o off from #2 so again
    Malus says 1/2 get though – now aligned with #3

    1/2 times 1/2 gives the 1/4 they were talking about
    BUT remember they are comparing to the light coming though filter 1
    NOT the original light.
    Compared to the original assumed un-polarized source it would be 1/8.
    Because, Yes the first filter removed 1/2 the light.

    Another simple one:
    Polarized glasses set their polarization H or V to block predominate glare caused by reflected off flat surfaces like water.
    So what is the convention for defining polarization between Horizontal and Vertical?
    Are Sunglasses polarized V to allow V light though, thus blocking H glare?
    OR is glare V thus Sunglasses are polarized to H so that V is blocked?

    What is the polarized direction aligned with in the EM wave, the E (Electric) or the M (Magnetic)?

    A matter of convention - just what is the convention?
    Maybe someone even knows how the convention was established.
    Example: Ben Franklin or those of his time a generally credited with establishing the charge polarity convention that resulted in the electron being defined as “-" not “+”.
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