1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook