Brewster's Angle

Equivalent forms

\theta_B = \arctan\left(\frac{n_2}{n_1}\right)

\theta_B + \theta_r = 90°

A tangent ratio captures the precise moment when reflected and refracted rays become perpendicular — geometry dictating polarization.

Symbol	Name	SI unit	Dimension	Range
$theta_B$	Brewster's angleoutput Angle of incidence at which reflected light is fully polarized	rad	1	0 – 1.5708
$n_1$	Refractive index (medium 1) Refractive index of the incident medium	dimensionless	1	1 – 2.5
$n_2$	Refractive index (medium 2) Refractive index of the second medium	dimensionless	1	1 – 2.5

Unit systems

Symbol	SI	CGS	Imperial
$theta_B$	rad	rad	deg

Where it holds

Valid for dielectric (non-conducting) surfaces. Does not apply to metallic surfaces where free electrons prevent full polarization of reflected light.

Dimensional analysis

\tan (\theta _B) \to [1]

(dimensionless)

n_{2}/n_{1} \to [1]/[1] = [1]

Both sides are dimensionless.

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Discovery

David Brewster · 1815

Brewster discovered empirically that reflected light is fully polarized when the reflected and refracted rays are perpendicular.

Try this

Why do polarized sunglasses cut the glare off a lake but not off a metal car hood?

Find the incidence angle at which reflected light from water (n=1.33) is perfectly polarized. Use theta_B = arctan(n2/n1).

Research status: stable

Real-world applications

Laser cavities: Brewster windows eliminate reflection losses for p-polarized light.
Polarized photography: photographers use polarizing filters to cut glare at near-Brewster angles.
Ellipsometry: measuring Brewster's angle determines the refractive index of thin films.
Fiber optic connectors: angled polishing $at \sim 8^{\circ}$ reduces back-reflection.

Common misconceptions

All reflected light is polarized at Brewster's angle — only the reflected light is fully s-polarized; the transmitted beam is partially polarized.
Brewster's angle works for metals — metals have complex refractive indices; reflected light is never fully polarized.
Polarized sunglasses use Brewster's angle — they use dichroic filters, but they work because glare from horizontal surfaces is s-polarized near Brewster's angle.

Experimental verification

Brewster measured polarization of reflected light from glass surfaces (1815). Modern: verified using polarimeters and laser sources at dielectric interfaces. Brewster windows in gas lasers exploit this — they transmit p-polarized light with zero reflection loss.

Derivation

At Brewster's angle, the reflected and refracted rays are perpendicular:

\theta _B + \theta _r = 90^{\circ}

.

From Snell's law:

n_{1}\cdot \sin (\theta _B) = n_{2}\cdot \sin (\theta _r) = n_{2}\cdot \sin (90^{\circ} - \theta _B) = n_{2}\cdot \cos (\theta _B)

.

Dividing both sides by

\cos (\theta _B)

:

n_{1}\cdot \tan (\theta _B) = n_{2}

, giving

\tan (\theta _B) = n_{2}/n_{1}

.

Physically, the reflected ray cannot contain the p-polarization component because the oscillating dipoles in the surface would have to radiate along their own axis, which is impossible.

Limiting cases

n_1 = n_2

⟶

\theta _B = 45^{\circ}

Same medium:

\arctan (1) = 45^{\circ}

, but no interface exists to reflect.

n_2 \to \infty

⟶

\theta _B \to 90^{\circ}

Extremely dense medium requires near-grazing incidence.

n_2 \to n_1

⟶

\theta _B \to 45^{\circ}

As media become similar, Brewster angle approaches 45 degrees.

What if…

What if you approach from the denser medium?

Brewster's angle from glass to air: $\theta _B = \arctan (1.0/1.52) = 33.3^{\circ}$ . The two Brewster angles (from each side) are complementary: they sum to $90^{\circ}$ .

What if the surface is a metal?

Metals have complex refractive indices. The reflected light is elliptically polarized at all angles — there is no true Brewster angle for conductors.

What if

n_{1} = n_{2}

?

$\tan (\theta _B) = 1$ , $so \theta _B = 45^{\circ}$ . But with no actual interface (same medium), there's no reflection at all.

1

Brewster's angle for water surface

Given ·

n 1:: 1
n 2:: 1.33

Find · theta_B

Steps

Apply $\tan (\theta _B) = n_{2}/n_{1}$
$\tan (\theta _B) = 1.33/1.0 = 1.33$
$\theta _B = \arctan (1.33) = 0.926\,\mathrm{rad}$
Convert: $0.926 \times 180/\pi = 53.06^{\circ}$

Result ·

\theta _B = \arctan (n_{2}/n_{1}) = \arctan (1.33/1.0) = \arctan (1.33) = 53.06^{\circ} = 0.926\,\mathrm{rad}

2

Brewster's angle for glass

Given ·

n 1:: 1
n 2:: 1.52

Find · theta_B

Steps

$\tan (\theta _B) = n_{2}/n_{1} = 1.52/1.0 = 1.52$
$\theta _B = \arctan (1.52) = 0.989\,\mathrm{rad} \approx 56.66^{\circ}$
At this angle, reflected light from glass is 100% s-polarized

Result ·

\theta _B = \arctan (1.52) = 56.66^{\circ} = 0.989\,\mathrm{rad}

Snell's Law→Law of Reflection→