Problem — Crossed polarisers with one in between

Problem

Three polarisers in a row. The first and third are at $90^\circ$ to each other: without the second, no light passes. Insert a second polariser at $45^\circ$: the light passes again. Explain using Malus’s law.

Solution

Without the central polariser. $P_1$ and $P_3$ are crossed at $90^\circ$. Malus’s law gives

$I=I_1\cos^2 90^\circ=0$

so total darkness.

With $P_2$ at $45^\circ$. Malus’s law is applied to each successive pair. After $P_1$ the light is vertically polarised with intensity $I_1$; after $P_2$:

$I_2=I_1\cos^2 45^\circ=\frac{I_1}{2}$

and the polarisation is now rotated to $45^\circ$. After $P_3$ (which is at $45^\circ$ from $P_2$):

$I_3=I_2\cos^2 45^\circ=\frac{I_1}{4}$

$\ev{I_3=\frac{I_1}{4}}$

The intermediate polaroid “re-projects” the polarisation at $45^\circ$, giving $P_3$ a non-zero component: adding a filter can let more light through, a counterintuitive result.

Collegamenti

Argomenti: Onde elettromagnetiche Concetti: Polarizzazione e legge di Malus · Intensità di un’onda