Recent content by 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

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 =...