Deriving formula for final intensity of light through three polarizers.

[t_physics]

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 = 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.

 

[t_physics]

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