Question

Light of original intensity${\mathit{I}}_{\mathbf{0}}$passes through two ideal polarizing filters having their polarizing axes oriented as shown in Fig. E33.28. You want to adjust the angle $\mathit{\varphi }$so that the intensity at point P is equal to ${\mathit{I}}_{\mathbf{0}}\mathbf{/}\mathbf{10}$. (a) If the original light is unpolarized, what should$\mathit{\varphi }$be? (b) If the original light is linearly polarized in the same direction as the polarizing axis of the first polarizer the light reaches, what should$\mathit{\varphi }$be?

Short answer

1. If the original light is unpolarized, therefore$\varphi =63.43°$
2. If the original light is linearly polarized in the same direction as the polarizing axis, therefore$\varphi =71.565°$

Step-by-step solution

Malus’ law

According to Malus' law, the intensity of plane-polarized light that travels through an analyzer varies as the square of the cosine of the angle between the plane of the polarizer and the analyzer’s transmission axes.

No matter how the Polarizing access is oriented, when unpolarized light is incident on a perfect polarizer, the intensity of the transmitted light is exactly half that of the incident and price light. Because the incident light is a random mixture of all states of polarisation, the E field of the incident wave is divided into two components, one parallel to the polarising access and one perpendicular to it. Because the incident light is a random mixture of all states of polarisation, these two components are on average equal to the idol's full size.

If the original light is unpolarized

(a) The polarizer allows just half of the intensity into the analyzer.

If the original light is linearly polarized in the same direction

(b) For linear polarized light