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
Homework Statement
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
Homework Equations
Malus' Law
I = Io (cosθ)^2
The Attempt at a Solution
I1 =...