Three polarizing sheets are placed in a stack with the polarizing directions of the first and third perpendicular to each other. what angle should the polarizing direction of the middle sheet make with the polarizing direction of the first sheet to obtain maximum transmitted intensity when unpolarized light is incident on the stack?

c) 45°

Explanation:

Given unpolarized light with an intensity = 1o

When this unpolarized light passes through a polarizer, the intensity of light reduces to half =Io/2 and this light becomes polarized.

Afterwards, the light passes a 2nd polarizer at an angle θ from the first.

From Malus’s law:

I2 = I1 cos2 θ

I2 = cos2 θ

When the light goes through a 3rd polarizer at (90-θ) it becomes;

I3 = I2 * cos2 (90-θ)

I3 =(Io/2)*cos2 θ* cos2(90-θ)

But I2 = (Io/2)*cos2 θ*sin2 (θ) = (IO/8)* sin2 (2θ)

From Sine functions;

Sin2(2θ) = 1

Sin (2θ) = 1

2θ = arcSin 1 = 90 °

Θ = 45°

We can eliminate a) and e) because light doesn’t get through.

the answer is b, have negligible mass