Question

vb9-5) Three polarizers are placed parallel to each other. 
The first and third are intersected, and the second's pass axis is at an angle of 45° to the other two's pass axes. 
What is the fraction of the intensity that is transmitted when natural light enters perpendicular to the first polarizer?

Answer 1

Natural light is unpolarized, and when an unpolarized light is made to pass through a polarizer, it's intensity gets halved.

We will need to use Malus Law which states that

I = I' cos²($\theta$)

where

I' = initial intensity

I = final intensity

$\theta$ = angle between the polarizer axes.

Let I0 = Initial intensity.

POLARIZER-1

After passing through Polarizer 1 the light becomes polarized, and the intensity = I0 / 2

POLARIZER- 2

Angle it makes with polarizer 1 = 45o

So, from Malus law Intensity = (Io /2) cos2 (45) = I0 / 4

POLARIZER- 3

Angle it makes with polarizer 2 = 45o

So, from Malus law Intensity = (Io /4) cos2 (45) = I0 / 8.

Hence, 1/8 th of the initial intensity is able to be finally transmitted. Hence, the fraction of the intensity of the natural light that is transmitted is 1/8 times of the initial intensity.