The electric field of an electromagnetic wave wiggles across the direction of travel. The direction it wiggles in is the wave's polarization. Light from the Sun or a bulb is unpolarized: a jumble of all directions. Only transverse waves have such a direction, which is the same sideways wiggle you saw on strings in Reflection at Boundaries. A polarizing filter passes only the part of the field along its axis.
Key idea
Unpolarized light through a polarizer: $I = \frac{1}{2}I_0$. Polarized light through a polarizer at angle $\theta$ to its polarization (Malus's law): $$I = I_0\cos^{2}\theta.$$ Only transverse waves can be polarized.
Why it works
A polarizer passes the component of $\vec{E}$ along its axis, $E\cos\theta$. Intensity goes as the field squared, so the intensity drops by $\cos^{2}\theta$. For unpolarized light, the angle takes every value; the average of $\cos^{2}\theta$ over all angles is $\frac{1}{2}$. Sound in air is a longitudinal wave, wiggling along its direction of travel, so there is no sideways direction for a filter to pick: it cannot be polarized.
Common mistake
Thinking no light can get through two crossed polarizers even with a third one between them. With only the crossed pair, nothing passes. Add a filter at $45^\circ$ in the middle and light gets through, because each filter turns the polarization to its own axis. Check by stepping through one filter at a time, using the angle between neighbors.
Worked example: Two polarizers
Unpolarized light of intensity $800\ \text{W/m}^2$ passes through two polarizers whose axes are $30^\circ$ apart. Find the final intensity.
- First polarizer: $I_1 = \frac{1}{2}(800) = 400\ \text{W/m}^2$, now polarized along its axis.
- Second: $$I_2 = 400\cos^{2}30^\circ = 400(0.75) = 300\ \text{W/m}^2.$$
Answer: $300\ \text{W/m}^2$
Worked example: The third polarizer
Unpolarized light of $800\ \text{W/m}^2$ meets polarizers at $0^\circ$, $45^\circ$ and $90^\circ$, in that order. Find the intensity after each.
- After the first: $400\ \text{W/m}^2$.
- The second is $45^\circ$ from the first: $400\cos^{2}45^\circ = 400(0.5) = 200\ \text{W/m}^2$.
- The third is $45^\circ$ from the second: $200(0.5) = 100\ \text{W/m}^2$.
- Without the middle one, the last is $90^\circ$ from the first, and $\cos^{2}90^\circ = 0$.
Answer: $400$, $200$, then $100\ \text{W/m}^2$
Light reflecting off water or a road is partly polarized horizontally. Polarized sunglasses have vertical axes, so they block much of that glare.