Glare off a lake or a road is partly polarized, which is why polarizing sunglasses cut it. At one special angle of incidence, the reflected light is completely polarized. That angle is named for David Brewster, and it follows from Snell's law and one fact about how charges radiate.
Key idea
Light reflected at Brewster's angle $\theta_B$, with $$\tan\theta_B = \frac{n_2}{n_1},$$ is completely polarized parallel to the surface (perpendicular to the plane of incidence). At that angle the reflected and refracted rays are at $90^\circ$ to each other. Like every angle in optics, $\theta_B$ is measured from the normal.
Why it works
The reflected light is made by the electrons of medium 2, which the refracted light shakes back and forth across the direction of the refracted ray. An oscillating charge radiates in every direction except along its own line of motion. Light polarized in the plane of incidence shakes the electrons in that plane, perpendicular to the refracted ray. If the reflected ray happens to point along that line of shaking, that polarization cannot be reflected at all. This happens when the reflected and refracted rays are perpendicular: $\theta_B + \theta_2 = 90^\circ$. Snell's law then gives $$n_1\sin\theta_B = n_2\sin(90^\circ - \theta_B) = n_2\cos\theta_B,$$ so $\tan\theta_B = n_2/n_1$. Only the other polarization, parallel to the surface, is left in the reflection.
Worked examples
Worked example: Glass in air
Find Brewster's angle for light in air reflecting from glass with $n = 1.50$, and the angle of the refracted ray.
- $\tan\theta_B = \frac{1.50}{1.00}$, so $\theta_B = \tan^{-1}1.50 = 56.3^\circ$.
- The refracted ray is perpendicular to the reflected one: $\theta_2 = 90^\circ - 56.3^\circ = 33.7^\circ$.
- Check with Snell's law: $\sin 56.3^\circ/1.50 = 0.555 = \sin 33.7^\circ$.
Answer: $\theta_B = 56.3^\circ$, $\theta_2 = 33.7^\circ$
Worked example: Glare off a lake
How high above the horizon is the Sun when its reflection from a calm lake ($n = 1.33$) is completely polarized?
- $\theta_B = \tan^{-1}(1.33) = 53.1^\circ$ from the normal, which is vertical.
- The Sun's height above the horizon is measured from the surface: $90^\circ - 53.1^\circ$.
Answer: $36.9^\circ$ above the horizon
Worked example: Inside the glass
Light inside a glass block ($n = 1.52$) reaches the glass–water boundary ($n = 1.33$). At what angle is the reflection completely polarized?
- Now $n_1 = 1.52$ (where the light starts) and $n_2 = 1.33$.
- $\tan\theta_B = \frac{1.33}{1.52} = 0.875$, so $\theta_B = 41.2^\circ$.
Answer: $41.2^\circ$ from the normal
Common mistake
Measuring Brewster's angle from the surface instead of from the normal. Every angle in Snell's law and in $\tan\theta_B = n_2/n_1$ is measured from the normal. A glancing angle of $36.9^\circ$ above a lake is $53.1^\circ$ from the normal. Check: for glass in air, $\theta_B$ must be more than $45^\circ$, because $n_2/n_1 > 1$.
Common mistake
Swapping $n_1$ and $n_2$. $n_1$ is where the light comes from. From inside glass to air, $\tan\theta_B = 1/1.50$ and $\theta_B = 33.7^\circ$.
The reflected glare is polarized horizontally (parallel to the water or road), so sunglasses with vertical transmission axes block most of it.