Question

Unpolarized light travels through 2 linear polarizers. What (minimum) angle should the second polarizer be relative to the first polarizer (the difference between q1 and q2) so that the final intensity of light is 1/3 the original value?

Answer 1

To solve this exercise it is necessary to apply the considerations made in the Malus Law.

This law indicates that the intensity of a linearly polarized ray of light, which passes through a perfect analyzer and vertical optical axis is equivalent to:

$I=I_0 Cos^2\theta_i$

Where,

$I_0$ Light intensity before passing through the polarizer

I = Resulting intensity

$\theta_i$ = Angle between the analyzer axis and the polarization axis

In the case where the light is unpolarized the intensity is considered as the intensity of the linearly polarized light transmitted by the first polarizer is

$I_0 = \frac{I_0}{2}$

Given tat the intensity of light after coming out of the second polarizer is

$I_1 = \frac{I_0}{3}$

Applying Malus's law then we have to,

$I_1=I_0 Cos^2\theta_i$

$\frac{I_0}{3} = \frac{I_0}{2} cos^2 \theta_i$

$\theta_i =cos^{-1} \sqrt{\frac{2}{3}}$

$\theta_i = 35.26\°$