Answer to Question #194227 in Physics for Praise

Question

Unpolarized light of intensity I_o is incident on the following arrangements of polarizing sheets. The first polarizer is oriented vertically; the second is oriented so it makes an angle of 30^o with the vertical; the third and last polarizer makes an angle of -30^o with the vertical. Determine the intensity of the light transmitted through this group of polarizers as a fraction of I_o.

Answer 1

According to the Malus's law, the intensity of the light after it passes the second polarizer is the following:

$$I_1 = I_0\cos^2\theta_1$$

where $\theta_1 = 30\degree$ is the agle between the first and the second polarizer.

Similarly, the intensity of the light after it passes the third polarizer is the following:

$$I_2 = I_1\cos^2\theta_2$$

where $\theta_2 = 60\degree$ is the agle between the second and the third polarizer.

Finally, obtain:

$$I_2 = I_0\cos^2\theta_1\cos^2\theta_2\\ \dfrac{I_2}{I_0} = \cos^2\theta_1\cos^2\theta_2\\ \dfrac{I_2}{I_0} = \cos^230\degree\cdot \cos^260\degree =\left(\dfrac{\sqrt{3}}{2}\right)^2\cdot\left(\dfrac{1}{2} \right)^2 = 0.1875$$

Answer. The intensity of the light transmitted through this group of polarizers as a fraction of I_o is 0.1875.