Q.State Malus' Law.

Malus Law: How Light Gets Weaker Through a Polariser

Imagine you're trying to push a rope through a narrow fence. If the rope is aligned with the gap, it passes through easily. If you twist the rope sideways, it gets blocked. Light behaves similarly — it's a transverse wave, meaning its electric field oscillates in a direction perpendicular to its travel. A polariser is like that fence: it only lets through light whose electric field oscillates in one specific direction (its "pass axis").

Now, what happens when you take already-polarised light and send it through a second polariser? That's exactly what Malus Law describes.

The Intuition

Suppose you have two polarisers. The first one takes ordinary (unpolarised) light and makes it polarised along some direction. The second polariser is rotated by an angle $\theta$ relative to the first.

• When $\theta = 0^\circ$ (both aligned), all the polarised light passes through — maximum intensity.
• When $\theta = 90^\circ$ (crossed), no light passes through — zero intensity.
• For any angle in between, only the component of the electric field that lies along the second polariser's axis gets through.

That component is $E_0 \cos\theta$, where $E_0$ is the amplitude of the incident polarised light. Since intensity $I$ is proportional to the square of the amplitude ($I \propto E^2$), the transmitted intensity becomes:

where $I_0$ is the intensity of the light incident on the second polariser (i.e., after the first polariser).

The Precise Statement

Malus Law states: When completely plane-polarised light of intensity $I_0$ is incident on an analyser (a polariser), the intensity $I$ of the transmitted light is proportional to the square of the cosine of the angle $\theta$ between the transmission axes of the polariser and the analyser.

Key points to remember for exams:

• The law applies only when the incident light is already fully polarised. If the light is unpolarised, the first polariser reduces its intensity by half ($I_0/2$), and then Malus Law applies to that reduced intensity.
• $\theta$ is the angle between the two transmission axes, not the angle of incidence or any other angle.
• The result is always $I \leq I_0$, with equality only at $\theta = 0^\circ$ or $180^\circ$.

A Quick Example

Unpolarised light of intensity $100 \, \text{W/m}^2$ passes through two polarisers whose axes are at $60^\circ$ to each other. What is the final intensity? …