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Deriving formula for final intensity of light through three polarizers.

  1. Apr 3, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that the exit intensity as a function of Io (intensity out of light source) and θ12 (angle of second polarizer compared to the first polarizer) is

    I = (Io/8) sin(2(θ12)))^2

    2. Relevant equations

    Malus' Law
    I = Io (cosθ)^2

    3. The attempt at a solution

    I1 = Io/2
    I2 = I1 (cos(θ12))^2
    = (Io/2) (cos(θ12))^2
    I = I2 (cos(θ23))^2
    = (Io/2)(cos(θ12))^2 (cos(θ23))^2

    I understand that the difference between θ1 and θ13 is 90°, but don't know how to apply this to derive the equation listed above.
     
  2. jcsd
  3. Apr 3, 2012 #2
    I just realized that in the final equation, θ23 = (90 - θ12), so it could be rewritten as I = (Io/2)(cos(θ12))^2 (cos(90-θ12))^2

    Still unsure about the trig involved though.

    Thanks
     
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