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Polarized Light - basic stuff

  1. Sep 1, 2006 #1
    I don't understand this question I guess, because I am not getting the right answer (the internet-site is not accepting it).

    Three polarizing sheets are stacked. The first and third are crossed; the one between has its polarizing direction at 31.0(deg) to the polarizing direction of the other two. What fraction of the intensity of an originally unpolarized beam is transmitted by the stack?

    My Work)
    First the equations that will be used:
    [tex] I = \frac{1}{2}I_0 [/tex]
    [tex] I=I_0 \cos^2 \theta [/tex]

    So unpolarized light comes in through the first filter.
    Lets label the light coming out as [itex] I_1 [/itex]

    [tex] I_1 = \frac{1}{2}I_0 [/tex]

    Next [itex] I_1 [/itex] is polarized and travels through filter two. This filter is at an angle of 31(deg) to filter one. Thus,

    [tex] I_2 = I_1 \cos^2 (31) [/tex]

    Next [itex] I_2 [/itex] is polarized and travels trhough filter three. This filter is at an angle of ???. I think this is where I'm making the mistake. So the question says that filter two has an angle of 31 degrees to filter-one and filter-two. I'll just call this angle [itex] \alpha [/itex] for now, since everything I've tried is not working :(

    [tex] I_3 = I_2 \cos^2 \alpha [/tex]

    Now combining the equations yields:
    [tex] I_3 = \frac{1}{2}I_0 \cos^2(31) \cos^2(\alpha) [/tex]

    Now it wants a fraction so,
    [tex] fra=\frac{I_3}{I_0}=\frac{1}{2} \cos^2(31) \cos^2(\alpha) [/tex]

    Am I doing this correctly? What about that pesky angle?
  2. jcsd
  3. Sep 1, 2006 #2
    When the question says, "The first and third are crossed"

    that means when the polarizing direction of filter-one is orthogonal to filter-two.

    I just thought they couldn't be parallel. :(

    grr... that was annoying.

    so the angle turned out to be [itex] \alpha = 59 [/itex] (deg)
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