Physics · Polarization

Brewster angle

At Brewster incidence, p reflection vanishes; remaining reflection of unpolarized input is s-polarized.

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Formula

tan θ_{B} = \frac{n _{2}}{n _{1}}

n₁,n₂ dimensionless; θB in degrees or radians from the normal.

Variables & units

Variables & units
Symbol	Meaning	Unit
$θB$	Incidence from normal giving zero p reflection	rad or °
$n1$	Incident medium phase index	dimensionless
$n2$	Second medium phase index at the same frequency	dimensionless

LaTeX

\tan\theta_B=\frac{n_2}{n_1}

QUANTA STUDYquanta-study.de

Interface with normal, incident ray at 56.66°, s-reflected ray and transmitted ray at 33.34°. The right angle is between reflection and transmission, not incidence and reflection. — For the lossless model n₁ = 1.000 → n₂ = 1.520, θB ≈ 56.66°. Reflected and transmitted rays are perpendicular. Rp = 0 applies to p polarization; unpolarized input leaves s reflection.

Brewster angle calculator

Find the Brewster angle from two refractive indices, or a refractive index from the measured angle, for a smooth boundary without absorption.

Calculation model

Find

Refractive index n₁ (incoming side)

Refractive index of the medium the light comes from, e.g. 1.00 for air.

Refractive index n₂ (second medium)

Refractive index of the medium the light hits, e.g. 1.50 for glass or 1.33 for water.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for θB, substitute, evaluate

1. Rearrange for θB
$\theta_B=\arctan\frac{n_2}{n_1}$
2. Substitute values (in SI units)
$\arctan\frac{1.5}{1}$
3. Result
θB = 56.31 °

Result: Brewster angle θB (from the normal)

θB = 56.31 °

With n₁ = 1 and n₂ = 1.5, θB = 56.31 °.

Quick tips

At the Brewster angle the reflected and refracted rays are perpendicular: θB + θ₂ = 90°. From air onto glass that is about 56.3° and 33.7°.
The direction matters: tan θB = n₂ ÷ n₁, with n₁ for the medium the light comes from. From glass into air θB ≈ 33.7°, not 56.3°.

Assumptions and limits

One smooth stationary planar interface between transparent isotropic lossless media with equal magnetic permeability and unequal positive real indices at the same frequency. Angle from the normal. The zero concerns p reflection, not all polarizations. No metal, absorption, coating or multilayer calculation.

Derivation & background

The incidence plane defines p and s. At an ordinary ideal dielectric boundary, Brewster incidence makes p Fresnel amplitude and power reflectance zero. Reflected and refracted rays are perpendicular, not generally incident and reflected rays. For unpolarized input, only partly reflected s power remains. Pure p input crosses the single lossless interface completely; absorption, coatings and additional boundaries require separate accounting.

Exam blueprint

Validity range

One smooth stationary interface of ordinary homogeneous isotropic transparent nonmagnetic media with positive real indices. Angles from the normal. Matched indices do not have a unique zero. Absorbing, magnetic, anisotropic or coated media require extended Fresnel or multilayer models.

Derivation steps

Set p Fresnel reflection amplitude to zero and combine with Snell’s law.

1Define incidence plane from ray and normal: p field lies in it, s field is perpendicular to it.
2For transparent nonmagnetic isotropic media, $r_p=\frac{n_2\cos\theta_1-n_1\cos\theta_2}{n_2\cos\theta_1+n_1\cos\theta_2}$ .
3For $r_p=0$ require $n_2\cos\theta_B=n_1\cos\theta_2$ .
4Also use $n_1\sin\theta_B=n_2\sin\theta_2$ . For unequal indices, both conditions give $\theta_B+\theta_2=90^\circ$ .
5With $\sin\theta_2=\cos\theta_B$ , obtain $\tan\theta_B=\frac{n_2}{n_1}$ and $\theta_B=\arctan\!\left(\frac{n_2}{n_1}\right)$ .
6Thus p reflection vanishes, not s reflection. Reflected and transmitted rays are perpendicular; pure p input creates no reflected ray.

Rearrangements

Brewster angle

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

Measure from the normal.

Task variant

Unpolarized 20 mW light travels from air n₁ = 1.000 to a single ideal lossless nonmagnetic interface with model glass n₂ = 1.520. Find Brewster angle, refraction angle and reflected power. How does Brewster angle change for n₂ = 1.50?

Approach: $\theta_B=\arctan\frac{n_2}{n_1}$ , then $\theta_2=90^\circ-\theta_B$ .
With n₂ = 1.520 the angles are 56.6593° and 33.3407°. p reflection vanishes; s power reflectance is 0.156692.
Unpolarized input shares mean power equally between the components.
Hence $P_r=\frac{20\,\mathrm{mW}}2\cdot0.156692\approx1.56692\,\mathrm{mW}$ .
For n₂ = 1.50, θB ≈ 56.31°.
This is a balance for the single interface.

Common mistakes

Ignoring units or assumptions stated for the formula: n₁,n₂ dimensionless; θB in degrees or radians from the normal.

Convert units before substitution and verify the validity range; round only at the end.

Exam context

Analyze reflection polarization, Brewster windows, index measurements and glare suppression; check axis direction and real residual reflection.

Formula cluster

Polarization

Related relations: malus-gesetz-polarisation, brechungsgesetz-snellius, grenzwinkel-totalreflexion.

I = I₀ cos²θ

Malus’s law

n₁·sin(α₁) = n₂·sin(α₂)

Snell's Law of Refraction

sin θc = n₂/n₁

Critical angle for total internal reflection

Worked example

Air 1.00 toward model glass 1.50 gives θB≈56.31°. For 1.520 the result is 56.6593°, with refracted angle 33.3407°.20 mW unpolarized input produces 1.56692 mW s-reflected power at this single boundary.

Applications

Analyze reflection polarization, Brewster windows, index measurements and glare suppression; check axis direction and real residual reflection.

How do you calculate with Brewster angle for polarization?

Here is how to work through a typical Brewster angle for polarization (tan θB = n₂/n₁) task step by step:

1
Task
Unpolarized 20 mW light travels from air n₁ = 1.000 to a single ideal lossless nonmagnetic interface with model glass n₂ = 1.520. Find Brewster angle, refraction angle and reflected power. How does Brewster angle change for n₂ = 1.50?
Solution path
1. Approach: $\theta_B=\arctan\frac{n_2}{n_1}$ , then $\theta_2=90^\circ-\theta_B$ .
2. With n₂ = 1.520 the angles are 56.6593° and 33.3407°. p reflection vanishes; s power reflectance is 0.156692.
3. Unpolarized input shares mean power equally between the components.
4. Hence $P_r=\frac{20\,\mathrm{mW}}2\cdot0.156692\approx1.56692\,\mathrm{mW}$ .
5. For n₂ = 1.50, θB ≈ 56.31°.
6. This is a balance for the single interface.

Understand and apply the equation

A suitably rotated polarizing filter can reduce glare from a water surface. Two electric-field orientations reflect differently. At Brewster incidence the ideal model has no p-polarized reflection. First identify that orientation, then calculate the special ray geometry and check how much unpolarized light still reflects.

Before calculating

The incident ray and surface normal span the plane of incidence. p means electric field within that plane; s means perpendicular to it. Both field orientations are transverse to their respective propagation direction. These letters do not name incident and transmitted sides. Without identifying polarization, “no reflection” is incomplete.[1]

The simple tangent formula describes one planar interface between lossless isotropic media with equal magnetic permeability and unequal positive indices. Measure incidence from the normal. Coatings or absorption can replace the zero with a different reflection minimum; they are excluded from this basic calculation.[2][3]

p and s refer to the plane of incidence

The incident ray and interface normal define the plane of incidence. The electric field is inside that plane for p polarization and perpendicular to it for s polarization. Both fields are also transverse to their respective propagation directions. Horizontal or vertical alone does not specify these components without the orientation of the incidence plane. Rotating the geometry rotates the reference directions.[2]

\tan\theta_B=\frac{n_2}{n_1}

For ordinary isotropic transparent nonmagnetic media; θB is the incidence angle from the normal in medium 1.[1]

The formula describes one smooth boundary. At Brewster incidence Rp=0, while s polarization remains partly reflected. Statistically unpolarized input contains equal powers in the two components. Its reflected wave is therefore s-polarized but generally much weaker than the incident wave.[1]

Malus’s law can then describe an ideal linear analyzer in the reflected beam. The analyzer input must be known: unpolarized incidence at Brewster produces an s-polarized reflected fraction, whereas pure p incidence ideally creates no reflected beam.[1]

A Fresnel amplitude zero and perpendicular rays

For real indices and equal magnetic permeability, a common sign convention gives $r_p=\frac{n_2\cos\theta_1-n_1\cos\theta_2}{n_2\cos\theta_1+n_1\cos\theta_2}$ . Reversing the sign convention changes neither its zero nor power reflectance $|r_p|^2$ . Brewster incidence requires its numerator to vanish while Snell’s law also holds.[2]

\begin{aligned}&n_2\cos\theta_B=n_1\cos\theta_2\\&n_1\sin\theta_B=n_2\sin\theta_2\end{aligned}

For unequal indices these conditions give $\theta_2=90^\circ-\theta_B$ and hence $\tan\theta_B=\frac{n_2}{n_1}$ .[2]

\begin{aligned}\frac{n_2}{n_1}&=\frac{\sin\theta_B}{\sin\theta_2}=\frac{\cos\theta_2}{\cos\theta_B}\\\sin\theta_B\cos\theta_B&=\sin\theta_2\cos\theta_2\\\theta_B+\theta_2&=90^\circ\\\tan\theta_B&=\frac{n_2}{n_1}\end{aligned}

Unequal indices exclude the equal-angle branch; between 0° and 90° the complementary solution remains.[2]

The reflected and refracted rays are perpendicular at this incidence. The incident and reflected rays are not generally perpendicular. Reflection angle equals incidence angle, so $\theta_B+\theta_2=90^\circ$ makes reflected and transmitted directions orthogonal. If n₁=n₂, there is no reflection at any angle; the formal 45° value is not a unique zero.[2]

Snell’s law checks the corresponding refraction angle. Brewster’s condition additionally comes from the p-polarized Fresnel zero; the refraction equation alone does not determine reflected power fraction.[2][1]

Align an external Brewster window

Air against a specified glass model

Given: n₁=1.000; n₂=1.520; planar uncoated lossless interface.
Find: Find θB and the matching refracted angle.

Approach: The p-polarized Fresnel zero gives Brewster incidence; reflected and transmitted rays are then orthogonal.
$\begin{aligned}\theta_B&=\arctan\frac{n_2}{n_1}\\\theta_2&=90^\circ-\theta_B\end{aligned}$
Insert the index ratio into inverse tangent.
$\theta_B=\arctan(1.520)=56.6593^\circ$
Find the complementary angle and check Snell’s law.
$\begin{aligned}&\theta_2=33.3407^\circ\\&\sin56.6593^\circ=1.520\sin33.3407^\circ\end{aligned}$

Result: θB≈56.66°, θ₂≈33.34°; reflected and refracted directions are 90° apart.

Check: The refracted angle is smaller because n₂>n₁. Both sine expressions are about 0.8354. Index ratio is dimensionless, not 1.520° or a speed.

View diagram: comparing s and p reflectance

Only the p curve reaches zero

QUANTA STUDYquanta-study.de

Single ideal interface. Curves show dimensionless power fractions, not field amplitudes or complete-window transmission.[2]

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How much unpolarized light is reflected?

The Brewster geometry alone does not determine s power. For the same nonmagnetic lossless media, $r_s=\frac{n_1\cos\theta_1-n_2\cos\theta_2}{n_1\cos\theta_1+n_2\cos\theta_2}$ , and $R_s=|r_s|^2$ . At Brewster incidence this reduces to $R_s=\left(\frac{n_1^2-n_2^2}{n_1^2+n_2^2}\right)^2$ . For unpolarized input, total reflected power fraction is $\frac{R_s+R_p}{2}$ .[2]

One interface with 20 mW incident power

Given: n₁=1.000; n₂=1.520; θ₁=θB; Pin=20.0 mW, unpolarized.
Find: Calculate reflected power and polarization.

Approach: Unpolarized input has equal p and s power here. At Brewster incidence only its s component reflects.
$\begin{aligned}P_{i,s}&=\frac{P_i}{2}\\R_p&=0\\R_s&=\left(\frac{n_1^2-n_2^2}{n_1^2+n_2^2}\right)^2\\P_r&=P_{i,s}R_s\end{aligned}$
Calculate s power reflectance; p reflection is zero.
$\begin{aligned}&R_s=\left(\frac{1-1.520^2}{1+1.520^2}\right)^2=0.156692\\&R_p=0\end{aligned}$
Only half of the incident power is s-polarized.
$P_r=\frac{20.0\,\mathrm{mW}}2R_s=1.56692\,\mathrm{mW}$

Result: About 1.567 mW is reflected, entirely s-polarized; about 18.433 mW crosses this ideal interface.

Check: Power balance gives Pr+Pt=20 mW. For purely p-polarized 20 mW input, Pr=0. Fully polarized reflection therefore does not describe every input. A second window surface and absorption were excluded.

Reverse incidence, total reflection and angle conventions

Reverse propagation exchanges n₁ and n₂. For the model above, reverse Brewster incidence is arctan(1/1.520)=33.3407°, the earlier refracted angle. The glass-to-air critical angle is 41.1395°. Brewster zero and total-reflection threshold are distinct: the first suppresses p reflection, the second prevents propagating transmission for either polarization.[2]

An angle from the normal complements the corresponding surface angle to 90°. A mechanical window tilt relative to a table is therefore not automatically incidence angle; draw the geometry. A p ray through a parallel-faced window can satisfy Brewster conditions at both faces, but lateral displacement, thickness and finite aperture still matter.[3]

Infer index from a reflection minimum

\begin{aligned}&\frac{n_2}{n_1}=\tan\theta_B\\&\delta\!\left(\frac{n_2}{n_1}\right)\approx\sec^2\theta_B\,\delta\theta_B\end{aligned}

Use radians for small angle changes in the differential expression.[2]

Ideal p-polarized input produces a zero minimum. Real filters have finite extinction, surfaces scatter and beams contain a spread of incidence angles. Fit a curve around the minimum rather than declaring the darkest sampled point an exact zero. Residual light can arise from polarization leakage without implying changed material indices.[2]

A reflection measurement belongs to a specified wavelength and temperature. Dispersion means broadband light may lack one perfect Brewster setting for all spectral components. The detector must also capture a consistent reflected beam cross-section; geometric clipping can otherwise mimic an optical minimum.[2]

An index ratio with angular bounds

Given: n₁=1.000; measured Brewster angle 57.0°±0.2°; ideal p-reflection minimum with bounded angle deviation.
Find: Find index and bounds due to the angle deviation.

Approach: evaluate the monotonic tangent over the relevant interval. These are tolerance bounds, not automatically a statistical standard uncertainty.
$\begin{aligned}n_2&=n_1\tan\theta_B\\n_{2,\min}&=n_1\tan56.8^\circ\\n_{2,\max}&=n_1\tan57.2^\circ\end{aligned}$
Substitute nominal angle and both endpoints separately.
$\begin{aligned}n_2&=1.539865\\n_{2,\min}&\approx1.528160\\n_{2,\max}&\approx1.551696\end{aligned}$

Result: The inferred index is about 1.5399; angle bounds give approximately 1.5282 to 1.5517.

Check: The bounds bracket the nominal value and are slightly asymmetric because tangent is nonlinear. Taking inverse tangent reproduces 56.8°, 57.0° and 57.2°. Additional polarization effects have not been included as angle uncertainty.

When the simple tangent formula is insufficient

Metals and strongly absorbing media have complex indices and often only a nonzero p-reflection minimum. Magnetic, anisotropic or coated media require appropriate boundary conditions or multilayer models. Do not apply the tangent of a real index ratio indiscriminately to those systems. Brewster’s relation is not a universal antireflection prescription for every surface.[3]

Reflected and transmitted beams may have different cross-sections and normal energy-flux components. Turning transmission amplitude into power requires index and cosine factors, not just squaring amplitude. The s-reflection calculation above needs no such extra factor because incident and reflected beams occupy the same medium.[2]

Five exercises checking polarization and limits

Calculate · Foundation

n₁=1.00 toward n₂=1.33: calculate θB and θ₂.

Show hint

Use inverse tangent and complementary angle.

Learning goal: Find Brewster and transmitted angles.

Show solution and calculation

Approach: use indices in propagation order.
The ideal interface gives θB≈53.0612° from inverse tangent.
The refracted angle is θ₂≈36.9388°.

Check: the angles add to 90° and $\sin\theta_B=1.33\sin\theta_2$ . p reflection vanishes, not automatically every reflected component.

\begin{aligned}\theta_B&=\arctan\frac{1.33}{1.00}\approx53.0612^\circ\\\theta_2&=90^\circ-\theta_B\approx36.9388^\circ\end{aligned}

Interpret · Application

A p-polarized 8 mW beam hits an ideal lossless air-glass interface exactly at θB. What is reflected?

Show hint

The zero applies to the p component.

Learning goal: Interpret zero reflection for pure p input.

Show solution and calculation

Approach: input contains only p polarization, whose ideal reflection coefficient is zero at Brewster incidence.
No power reflects; 8 mW crosses this single lossless interface.

Check: reflected and transmitted powers sum to 8 mW. No s-reflected beam can appear because the input contains no s component.

P_r=R_pP_i=0

Interpret · Challenge

For n₁=n₂=1.40, tanθB=1 formally gives 45°. Is this the only reflection-free angle?

Show hint

Check the complete Fresnel coefficients.

Learning goal: Recognize the index-matched degenerate case.

Show solution and calculation

Approach: inspect the full Fresnel coefficients rather than just evaluating tangent.
Matching index and permeability give an optically matched interface, so both reflection components vanish at all nongrazing incidence angles.
The formal 45° value is not an isolated Brewster zero.
Claiming that angle alone is special is therefore incorrect.

R_s=R_p=0\quad(n_1=n_2)

Interpret · Challenge

An unpolarized 8 mW beam hits a single ideal nonmagnetic lossless interface from n₁=1 to n₂=√3 at Brewster incidence. Find reflected polarization and power using the article’s s-amplitude coefficient.

Show hint

Unpolarized means equal mean s and p power; only p has zero reflection.

Learning goal: Separate the p-reflection zero from total reflected power.

Show solution and calculation

Approach: θB=60° and θ₂=30°.
For s, $r_s=-\frac12$ , hence $R_s=\frac14$ .
Incident s power is 4 mW, giving 1 mW reflected.
The reflected beam is purely s-polarized.

Check: 7 mW crosses this single lossless interface, not all 8 mW. Squaring removes the amplitude sign for reflectance.

\begin{aligned}r_s&=\frac{1\cos60^\circ-\sqrt3\cos30^\circ}{1\cos60^\circ+\sqrt3\cos30^\circ}=-\frac12\\P_r&=\frac{8\,\mathrm{mW}}2\left|r_s\right|^2=1\,\mathrm{mW}\end{aligned}

[1][2]

Error analysis · Application

An absorbing surface has a complex refractive index with real part 2. A solution computes $\arctan 2\approx63.435^\circ$ and claims exactly zero p-reflection there. Is that zero established? Describe the required calculation instead.

Show hint

A complex index cannot simply be replaced by its real part.

Learning goal: Avoid applying the real Brewster formula to absorbing media.

Show solution and calculation

Approach: the real tangent relation assumes suitable lossless dielectric media.
The real part alone determines neither location nor depth of a p-reflection minimum.
The full complex index, incident medium and appropriate Fresnel conditions are needed.

Check: only a genuinely zero complex reflection coefficient would give Rp=0. A commonly nonzero reflection minimum is not an established Brewster zero.

\begin{aligned}R_p&=|r_p|^2\\R_p=0&\ \Longleftrightarrow\ r_p=0\end{aligned}

[3][2]

What to take away

Brewster incidence makes p reflection vanish, not every reflected component. Unpolarized input leaves an s-reflected fraction. The refracted angle complements Brewster incidence to 90°. Reverse propagation uses the reciprocal index ratio. Residual measured reflection may arise from polarization, angular spread or surfaces rather than a changed index alone.[1][2]

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

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This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

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An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 2, 2026Added relevant connections; preserved existing derivations, calculations, exercises, figures and model limits.
October 1, 2026Explained p/s polarization before calculation; expanded Brewster derivation and power checks; added index measurement and an optional Fresnel plot.
October 1, 2026Added prerequisites and model assumptions; clarified derivation steps and arithmetic checks; included an additional original example, diagram and summary.

Cite this reference page

Quanta. Brewster angle for polarization. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/brewster-winkel

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Frequently asked questions about Brewster angle for polarization

What does Brewster angle mean for polarization?+

Brewster incidence is the angle where p-polarized reflection vanishes at an ideal dielectric interface. The p direction lies in the plane defined by ray and normal; s is perpendicular to that plane. Unpolarized input gives purely s-polarized reflected light, but only part of its original power. Purely p-polarized input produces no reflected ray there. Geometrically, reflected and refracted rays are perpendicular. It would be incorrect to assign that special orthogonality generally to the incident and reflected rays. The simple tangent relation applies to positive real indices of ordinary isotropic nonmagnetic transparent media.

How do I check Brewster angle and reflected power?+

Number indices in propagation order and take inverse tangent of n₂ divided by n₁. Air 1.000 toward model glass 1.520 gives 56.6593° from the normal. Refracted angle is 33.3407°, completing 90° with incidence. Also check Snell’s law: both index-weighted sine expressions must agree. A power check requires Fresnel coefficients; p reflection is zero and s reflectance here is 0.156692. Unpolarized 20 mW input therefore gives 1.56692 mW reflection. This is not a complete two-surface window balance. Identify whether the task asks for angle, field amplitude or power fraction before comparing different kinds of quantities.

Under what assumptions does the simple Brewster formula apply?+

The simple Brewster angle concerns one smooth interface between ordinary isotropic transparent nonmagnetic media with real indices. Coated surfaces need multilayer models, while magnetic or anisotropic media require corresponding boundary conditions. Strongly absorbing media and metals have complex indices and may show a nonzero minimum rather than vanishing reflection. Real filters also do not produce perfect p polarization, so residual light can remain at the measured minimum. A beam with finite angular width contains several incidence directions simultaneously. Small residual reflection therefore does not automatically disprove the tangent relation or demonstrate a different material index.

How do I check units and angular scale?+

The index ratio is dimensionless and inverse tangent returns an angle from the normal. Use the correct degree or radian mode. A larger n₂ gives Brewster incidence above 45°; a smaller n₂ gives it below 45°. Reversing propagation swaps indices: the 1.520 glass-air model gives 33.3407° instead of 56.6593°. That reverse Brewster angle lies below its 41.1395° critical angle. Differential uncertainty for the index ratio uses angle changes in radians. If indices match, the ideal matched boundary reflects nothing at every angle; formal 45° is then not a special isolated Brewster zero.

How is Brewster angle combined with other optical relations?+

Use Snell’s law to determine Brewster’s transmitted angle and verify the perpendicular-ray geometry. Fresnel coefficients provide remaining s reflection and actual power accounting. Malus can then describe a linear analyzer in the reflected beam if its input polarization is known. A parallel Brewster window needs both interfaces treated in their respective propagation order. Pure p light can ideally have zero reflection at both, while displacement, finite aperture and absorption remain separate issues. Do not equate Brewster with total reflection: one suppresses a reflection component, while the other prevents propagating transmission above a different threshold.

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Symbol

Meaning

Unit

θB

Incidence from normal giving zero p reflection

rad or °

n1

Incident medium phase index

dimensionless

n2

Second medium phase index at the same frequency

dimensionless

Derivation & background

Exam blueprint

Validity range

Derivation steps

Set p Fresnel reflection amplitude to zero and combine with Snell’s law.

1Define incidence plane from ray and normal: p field lies in it, s field is perpendicular to it.
2For transparent nonmagnetic isotropic media, $r_p=\frac{n_2\cos\theta_1-n_1\cos\theta_2}{n_2\cos\theta_1+n_1\cos\theta_2}$ .
3For $r_p=0$ require $n_2\cos\theta_B=n_1\cos\theta_2$ .
4Also use $n_1\sin\theta_B=n_2\sin\theta_2$ . For unequal indices, both conditions give $\theta_B+\theta_2=90^\circ$ .
5With $\sin\theta_2=\cos\theta_B$ , obtain $\tan\theta_B=\frac{n_2}{n_1}$ and $\theta_B=\arctan\!\left(\frac{n_2}{n_1}\right)$ .
6Thus p reflection vanishes, not s reflection. Reflected and transmitted rays are perpendicular; pure p input creates no reflected ray.

How do you calculate with Brewster angle for polarization?

Here is how to work through a typical Brewster angle for polarization (tan θB = n₂/n₁) task step by step:

Task

Solution path

Approach: $\theta_B=\arctan\frac{n_2}{n_1}$ , then $\theta_2=90^\circ-\theta_B$ .
With n₂ = 1.520 the angles are 56.6593° and 33.3407°. p reflection vanishes; s power reflectance is 0.156692.
Unpolarized input shares mean power equally between the components.
Hence $P_r=\frac{20\,\mathrm{mW}}2\cdot0.156692\approx1.56692\,\mathrm{mW}$ .
For n₂ = 1.50, θB ≈ 56.31°.
This is a balance for the single interface.

Understand and apply the equation

Before calculating

p and s refer to the plane of incidence

\tan\theta_B=\frac{n_2}{n_1}

For ordinary isotropic transparent nonmagnetic media; θB is the incidence angle from the normal in medium 1.[1]

A Fresnel amplitude zero and perpendicular rays

\begin{aligned}&n_2\cos\theta_B=n_1\cos\theta_2\\&n_1\sin\theta_B=n_2\sin\theta_2\end{aligned}

For unequal indices these conditions give $\theta_2=90^\circ-\theta_B$ and hence $\tan\theta_B=\frac{n_2}{n_1}$ .[2]

\begin{aligned}\frac{n_2}{n_1}&=\frac{\sin\theta_B}{\sin\theta_2}=\frac{\cos\theta_2}{\cos\theta_B}\\\sin\theta_B\cos\theta_B&=\sin\theta_2\cos\theta_2\\\theta_B+\theta_2&=90^\circ\\\tan\theta_B&=\frac{n_2}{n_1}\end{aligned}

Unequal indices exclude the equal-angle branch; between 0° and 90° the complementary solution remains.[2]

Align an external Brewster window

Air against a specified glass model

Given: n₁=1.000; n₂=1.520; planar uncoated lossless interface.
Find: Find θB and the matching refracted angle.

Approach: The p-polarized Fresnel zero gives Brewster incidence; reflected and transmitted rays are then orthogonal.
$\begin{aligned}\theta_B&=\arctan\frac{n_2}{n_1}\\\theta_2&=90^\circ-\theta_B\end{aligned}$
Insert the index ratio into inverse tangent.
$\theta_B=\arctan(1.520)=56.6593^\circ$
Find the complementary angle and check Snell’s law.
$\begin{aligned}&\theta_2=33.3407^\circ\\&\sin56.6593^\circ=1.520\sin33.3407^\circ\end{aligned}$

Result: θB≈56.66°, θ₂≈33.34°; reflected and refracted directions are 90° apart.

Check: The refracted angle is smaller because n₂>n₁. Both sine expressions are about 0.8354. Index ratio is dimensionless, not 1.520° or a speed.

View diagram: comparing s and p reflectance

Only the p curve reaches zero

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How much unpolarized light is reflected?

One interface with 20 mW incident power

Given: n₁=1.000; n₂=1.520; θ₁=θB; Pin=20.0 mW, unpolarized.
Find: Calculate reflected power and polarization.

Approach: Unpolarized input has equal p and s power here. At Brewster incidence only its s component reflects.
$\begin{aligned}P_{i,s}&=\frac{P_i}{2}\\R_p&=0\\R_s&=\left(\frac{n_1^2-n_2^2}{n_1^2+n_2^2}\right)^2\\P_r&=P_{i,s}R_s\end{aligned}$
Calculate s power reflectance; p reflection is zero.
$\begin{aligned}&R_s=\left(\frac{1-1.520^2}{1+1.520^2}\right)^2=0.156692\\&R_p=0\end{aligned}$
Only half of the incident power is s-polarized.
$P_r=\frac{20.0\,\mathrm{mW}}2R_s=1.56692\,\mathrm{mW}$

Result: About 1.567 mW is reflected, entirely s-polarized; about 18.433 mW crosses this ideal interface.

Reverse incidence, total reflection and angle conventions

Infer index from a reflection minimum

\begin{aligned}&\frac{n_2}{n_1}=\tan\theta_B\\&\delta\!\left(\frac{n_2}{n_1}\right)\approx\sec^2\theta_B\,\delta\theta_B\end{aligned}

Use radians for small angle changes in the differential expression.[2]

An index ratio with angular bounds

Given: n₁=1.000; measured Brewster angle 57.0°±0.2°; ideal p-reflection minimum with bounded angle deviation.
Find: Find index and bounds due to the angle deviation.

Approach: evaluate the monotonic tangent over the relevant interval. These are tolerance bounds, not automatically a statistical standard uncertainty.
$\begin{aligned}n_2&=n_1\tan\theta_B\\n_{2,\min}&=n_1\tan56.8^\circ\\n_{2,\max}&=n_1\tan57.2^\circ\end{aligned}$
Substitute nominal angle and both endpoints separately.
$\begin{aligned}n_2&=1.539865\\n_{2,\min}&\approx1.528160\\n_{2,\max}&\approx1.551696\end{aligned}$

Result: The inferred index is about 1.5399; angle bounds give approximately 1.5282 to 1.5517.

When the simple tangent formula is insufficient

Five exercises checking polarization and limits

Calculate · Foundation

n₁=1.00 toward n₂=1.33: calculate θB and θ₂.

Show hint

Use inverse tangent and complementary angle.

Learning goal: Find Brewster and transmitted angles.

Show solution and calculation

Approach: use indices in propagation order.
The ideal interface gives θB≈53.0612° from inverse tangent.
The refracted angle is θ₂≈36.9388°.

Check: the angles add to 90° and $\sin\theta_B=1.33\sin\theta_2$ . p reflection vanishes, not automatically every reflected component.

\begin{aligned}\theta_B&=\arctan\frac{1.33}{1.00}\approx53.0612^\circ\\\theta_2&=90^\circ-\theta_B\approx36.9388^\circ\end{aligned}

Interpret · Application

A p-polarized 8 mW beam hits an ideal lossless air-glass interface exactly at θB. What is reflected?

Show hint

The zero applies to the p component.

Learning goal: Interpret zero reflection for pure p input.

Show solution and calculation

Approach: input contains only p polarization, whose ideal reflection coefficient is zero at Brewster incidence.
No power reflects; 8 mW crosses this single lossless interface.

Check: reflected and transmitted powers sum to 8 mW. No s-reflected beam can appear because the input contains no s component.

P_r=R_pP_i=0

Interpret · Challenge

For n₁=n₂=1.40, tanθB=1 formally gives 45°. Is this the only reflection-free angle?

Show hint

Check the complete Fresnel coefficients.

Learning goal: Recognize the index-matched degenerate case.

Show solution and calculation

Approach: inspect the full Fresnel coefficients rather than just evaluating tangent.
Matching index and permeability give an optically matched interface, so both reflection components vanish at all nongrazing incidence angles.
The formal 45° value is not an isolated Brewster zero.
Claiming that angle alone is special is therefore incorrect.

R_s=R_p=0\quad(n_1=n_2)

Interpret · Challenge

An unpolarized 8 mW beam hits a single ideal nonmagnetic lossless interface from n₁=1 to n₂=√3 at Brewster incidence. Find reflected polarization and power using the article’s s-amplitude coefficient.

Show hint

Unpolarized means equal mean s and p power; only p has zero reflection.

Learning goal: Separate the p-reflection zero from total reflected power.

Show solution and calculation

Approach: θB=60° and θ₂=30°.
For s, $r_s=-\frac12$ , hence $R_s=\frac14$ .
Incident s power is 4 mW, giving 1 mW reflected.
The reflected beam is purely s-polarized.

Check: 7 mW crosses this single lossless interface, not all 8 mW. Squaring removes the amplitude sign for reflectance.

\begin{aligned}r_s&=\frac{1\cos60^\circ-\sqrt3\cos30^\circ}{1\cos60^\circ+\sqrt3\cos30^\circ}=-\frac12\\P_r&=\frac{8\,\mathrm{mW}}2\left|r_s\right|^2=1\,\mathrm{mW}\end{aligned}

[1][2]

Error analysis · Application

An absorbing surface has a complex refractive index with real part 2. A solution computes $\arctan 2\approx63.435^\circ$ and claims exactly zero p-reflection there. Is that zero established? Describe the required calculation instead.

Show hint

A complex index cannot simply be replaced by its real part.

Learning goal: Avoid applying the real Brewster formula to absorbing media.

Show solution and calculation

Approach: the real tangent relation assumes suitable lossless dielectric media.
The real part alone determines neither location nor depth of a p-reflection minimum.
The full complex index, incident medium and appropriate Fresnel conditions are needed.

Check: only a genuinely zero complex reflection coefficient would give Rp=0. A commonly nonzero reflection minimum is not an established Brewster zero.

\begin{aligned}R_p&=|r_p|^2\\R_p=0&\ \Longleftrightarrow\ r_p=0\end{aligned}

[3][2]

What to take away

Sources for this explanation

Quanta · Updated October 2, 2026

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An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 2, 2026Added relevant connections; preserved existing derivations, calculations, exercises, figures and model limits.
October 1, 2026Explained p/s polarization before calculation; expanded Brewster derivation and power checks; added index measurement and an optional Fresnel plot.
October 1, 2026Added prerequisites and model assumptions; clarified derivation steps and arithmetic checks; included an additional original example, diagram and summary.

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Quanta. Brewster angle for polarization. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/brewster-winkel

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Frequently asked questions about Brewster angle for polarization

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Retain Brewster angle for polarization for exams

Create a curated FSRS exam set for tan θB = n₂/n₁: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential

Brewster angle

Variables & units

Brewster angle calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Polarization

Worked example

Applications

How do you calculate with Brewster angle for polarization?

Before calculating

p and s refer to the plane of incidence

A Fresnel amplitude zero and perpendicular rays

Align an external Brewster window

Air against a specified glass model

Only the p curve reaches zero

How much unpolarized light is reflected?

One interface with 20 mW incident power

Reverse incidence, total reflection and angle conventions

Infer index from a reflection minimum

An index ratio with angular bounds

When the simple tangent formula is insufficient

Five exercises checking polarization and limits

n1=1.00 toward n2=1.33: calculate θB and θ2.

A p-polarized 8 mW beam hits an ideal lossless air-glass interface exactly at θB. What is reflected?

For n1=n2=1.40, tanθB=1 formally gives 45°. Is this the only reflection-free angle?

An unpolarized 8 mW beam hits a single ideal nonmagnetic lossless interface from n1=1 to n2=√3 at Brewster incidence. Find reflected polarization and power using the article’s s-amplitude coefficient.

An absorbing surface has a complex refractive index with real part 2. A solution computes arctan⁡2≈63.435∘\arctan 2\approx63.435^\circarctan2≈63.435∘ and claims exactly zero p-reflection there. Is that zero established? Describe the required calculation instead.

What to take away

Sources for this explanation

Frequently asked questions about Brewster angle for polarization

Subject references and further reading

More formulas: Optics

Retain Brewster angle for polarization for exams

Brewster angle

Variables & units

Brewster angle calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Polarization

Worked example

Applications

How do you calculate with Brewster angle for polarization?

Before calculating

p and s refer to the plane of incidence

A Fresnel amplitude zero and perpendicular rays

Align an external Brewster window

Air against a specified glass model

Only the p curve reaches zero

How much unpolarized light is reflected?

One interface with 20 mW incident power

Reverse incidence, total reflection and angle conventions

Infer index from a reflection minimum

An index ratio with angular bounds

When the simple tangent formula is insufficient

Five exercises checking polarization and limits

n1=1.00 toward n2=1.33: calculate θB and θ2.

A p-polarized 8 mW beam hits an ideal lossless air-glass interface exactly at θB. What is reflected?

For n1=n2=1.40, tanθB=1 formally gives 45°. Is this the only reflection-free angle?

An unpolarized 8 mW beam hits a single ideal nonmagnetic lossless interface from n1=1 to n2=√3 at Brewster incidence. Find reflected polarization and power using the article’s s-amplitude coefficient.

An absorbing surface has a complex refractive index with real part 2. A solution computes arctan⁡2≈63.435∘\arctan 2\approx63.435^\circarctan2≈63.435∘ and claims exactly zero p-reflection there. Is that zero established? Describe the required calculation instead.

What to take away

Sources for this explanation

Frequently asked questions about Brewster angle for polarization

Subject references and further reading

More formulas: Optics

Retain Brewster angle for polarization for exams

n₁=1.00 toward n₂=1.33: calculate θB and θ₂.

For n₁=n₂=1.40, tanθB=1 formally gives 45°. Is this the only reflection-free angle?

An unpolarized 8 mW beam hits a single ideal nonmagnetic lossless interface from n₁=1 to n₂=√3 at Brewster incidence. Find reflected polarization and power using the article’s s-amplitude coefficient.

An absorbing surface has a complex refractive index with real part 2. A solution computes $\arctan 2\approx63.435^\circ$ and claims exactly zero p-reflection there. Is that zero established? Describe the required calculation instead.

n₁=1.00 toward n₂=1.33: calculate θB and θ₂.

For n₁=n₂=1.40, tanθB=1 formally gives 45°. Is this the only reflection-free angle?

An unpolarized 8 mW beam hits a single ideal nonmagnetic lossless interface from n₁=1 to n₂=√3 at Brewster incidence. Find reflected polarization and power using the article’s s-amplitude coefficient.

An absorbing surface has a complex refractive index with real part 2. A solution computes $\arctan 2\approx63.435^\circ$ and claims exactly zero p-reflection there. Is that zero established? Describe the required calculation instead.