Brewster's Law & Polarization by Reflection/Scattering

The last article polarized light the brute-force way: shove it through a filter and throw away half. Nature has a subtler trick. Let unpolarized light glance off a sheet of glass or the surface of a lake, and the reflected beam comes back already polarized — no filter required. Tilt the surface to just the right angle and the reflected light is not merely partly polarized but completely polarized, every wave vibrating in the same plane. Polarization for free. That "right angle" is the Brewster angle, or polarizing angle, and the surprise is how simply it is set. When light strikes at the Brewster angle, the reflected ray and the refracted ray leave at exactly $90^\circ$ to each other. Feed that single geometric fact into Snell's law and out drops Brewster's law: $\tan\theta_p = n$ , where $n$ is the refractive index of the surface. Nothing but the tangent of one angle and the index of the material. For ordinary glass that angle is about $56^\circ$ from the normal; for water, about $53^\circ$ ; for diamond, about $68^\circ$ . The same physics — light polarized by the geometry of scattering rather than by a filter — paints the sky. Sunlight bouncing off the air molecules overhead is scattered, and light scattered through $90^\circ$ comes out fully polarized, which is why a patch of blue sky at right angles to the Sun is the most strongly polarized light in nature (and why that same Rayleigh scattering, stronger for short wavelengths as $1/\lambda^4$ , makes the sky blue in the first place). Bees navigate by it; photographers exploit it; polarizing sunglasses were invented to cut exactly the reflected glare this article explains. One law, $\tan\theta_p = n$ , ties reflection, refraction and the colour of the sky together.

Key takeaways

Unpolarized light reflecting off a dielectric (glass, water) is partially polarized; at one special angle it becomes 100% polarized, vibrating perpendicular to the plane of incidence.
That angle is the Brewster (polarizing) angle $\theta_p$ , and at it the reflected and refracted rays are exactly $90^\circ$ apart.
Brewster's law: $\tan\theta_p = n$ — the polarizing angle is set by the refractive index alone. Glass ( $n=1.5$ ): $56.3^\circ$ ; water ( $n=1.33$ ): $53.1^\circ$ ; diamond ( $n=2.42$ ): $67.5^\circ$ .
It is $\tan\theta_p = n$ , never $\sin\theta_p = n$ (that would demand a sine above 1 for glass) — a classic trap.
Polarization by scattering: light scattered by the air through $90^\circ$ is fully polarized, so the blue sky is polarized; the same Rayleigh scattering ( $\propto 1/\lambda^4$ ) makes it blue.
Key idea: reflection and scattering polarize light for free — at the Brewster angle $\tan\theta_p = n$ the reflected ray is fully polarized because it leaves at $90^\circ$ to the refracted ray.

The core insight

When unpolarized light strikes a transparent surface at the Brewster angle, the reflected and refracted rays are $90^\circ$ apart, which forces the reflected light to be completely polarized perpendicular to the plane of incidence; combining that geometry with Snell's law gives Brewster's law $\tan\theta_p = n$ , and the same $90^\circ$ -scattering rule polarizes the blue sky.

Where this lives today

Every pair of polarizing sunglasses is Brewster's law at work: they are tuned to kill the horizontally polarized glare that reflection off roads, water and car bonnets produces near the Brewster angle. Photographers rotate a polarizing filter to darken a blue sky (cutting its $90^\circ$ -scattered polarized light) and to erase reflections from glass and water, seeing through a shop window or into a pond. Laser engineers cut windows at the Brewster angle ("Brewster windows") so that light of the right polarization passes with zero reflection loss — essential inside gas lasers. Geologists and materials scientists identify minerals and map stress in transparent components using polarized light. And the polarization of scattered skylight, first explained here, is a real navigational compass: Viking sailors may have used "sunstones," and honeybees and many insects steer by the sky's polarization pattern to this day. One tangent, $\tan\theta_p = n$ , and one right angle in the sky, underpin all of it.

The physical picture

When light meets a boundary between air and a transparent medium, part reflects and part refracts into the medium. The reflection is not indifferent to polarization. Think of the light's electric field as having two independent parts: one vibrating in the plane of incidence (the plane containing the incident ray and the surface normal) and one vibrating perpendicular to it. The surface reflects the perpendicular component more efficiently than the in-plane component, so the reflected beam is always at least partially polarized, favouring the perpendicular direction. At most angles the effect is mild. But there is one incidence angle at which the in-plane component is not reflected at all — its reflection drops to zero — leaving the reflected beam 100% polarized perpendicular to the plane of incidence. The clean way to see when this happens is a geometric coincidence: it occurs precisely when the reflected ray and the refracted ray are at right angles to each other. At that instant the electric field of the would-be in-plane reflection would have to point along the direction the refracted ray travels, and light cannot have a field component along its own direction of motion (it is transverse) — so that reflection cannot exist. All that survives in reflection is the perpendicular component: full polarization.

Figure

Unpolarized light striking glass at the Brewster angle — the reflected ray is fully polarized perpendicular to the plane of incidence, and it leaves at 90 degrees to the refracted ray

Brewster's law

Put the geometry and Snell's law together. At the Brewster angle the reflected ray is at $90^\circ$ to the refracted ray. Since the reflected ray leaves at angle $\theta_p$ on the same side as the incident ray, and the refracted ray bends to angle $\theta_r$ , the right-angle condition works out to

\theta_p + \theta_r = 90^\circ. \qquad (1)

Snell's law at the surface reads $\sin\theta_p = n\sin\theta_r$ (going from air, index $1$ , into the medium, index $n$ ). Replace $\theta_r$ using (1), so $\theta_r = 90^\circ - \theta_p$ and $\sin\theta_r = \cos\theta_p$ :

\sin\theta_p = n\cos\theta_p \;\Rightarrow\; \boxed{\tan\theta_p = n} \qquad (2)

where: $\theta_p$ — the Brewster (polarizing) angle, measured from the normal; $n$ — the refractive index of the reflecting medium relative to the surrounding one. Equation (2) is Brewster's law, published by David Brewster in 1815. It is astonishingly economical: the polarizing angle depends only on the index, and knowing one gives the other. Because it is a tangent, $\theta_p$ can grow past $45^\circ$ without any trouble (unlike a sine, which is capped at $1$ ) — for glass $\tan\theta_p = 1.5$ gives $\theta_p \approx 56^\circ$ , comfortably above $45^\circ$ . The three canonical values every exam expects:

\theta_p^{\text{glass}} = \tan^{-1}(1.5) = 56.3^\circ, \quad \theta_p^{\text{water}} = \tan^{-1}(1.33) = 53.1^\circ, \quad \theta_p^{\text{diamond}} = \tan^{-1}(2.42) = 67.5^\circ. \qquad (3)

For glass the refracted angle is then $\theta_r = 90^\circ - 56.3^\circ = 33.7^\circ$ , and Snell's law checks out: $\sin 56.3^\circ / \sin 33.7^\circ = 0.832/0.555 = 1.50 = n$ . Everything is consistent, and the reflected beam at $56.3^\circ$ is completely polarized.

Polarization by scattering: the blue sky

The second free polarizer is the sky. Sunlight is unpolarized, but when it strikes an air molecule the molecule's electrons are set oscillating by the light's electric field and re-radiate (scatter) it in new directions — this is Rayleigh scattering. An oscillating charge is a tiny antenna, and an antenna radiates nothing along its own axis of oscillation. So if you look at scattered light coming from a direction at $90^\circ$ to the original sunbeam, only one of the two transverse vibration directions can reach you — the other would have had to radiate along its own axis, which it cannot. The $90^\circ$ -scattered light is therefore fully polarized.

\text{sunlight scattered through } 90^\circ \;\Rightarrow\; \text{completely polarized}. \qquad (4)

That is why a band of sky at right angles to the Sun is strongly polarized — you can see it darken and lighten by rotating a polarizing filter, and honeybees use exactly this polarization pattern as a compass. The very same Rayleigh scattering explains the colour: its strength grows as $1/\lambda^4$ , so short-wavelength blue light scatters far more than long-wavelength red. Comparing blue ( $\lambda \approx 450\ \mathrm{nm}$ ) with red ( $\lambda \approx 700\ \mathrm{nm}$ ), the ratio is $(700/450)^4 \approx 5.9$ — blue is scattered nearly six times as strongly, filling the sky with scattered blue while the direct Sun, robbed of blue, reddens at sunset. Colour and polarization, one mechanism.

Figure

Sunlight scattered by an air molecule emerges fully polarized when viewed at 90 degrees to the incoming beam

Derivation ABrewster's law from the right-angle condition and Snell's law

The goal is to derive $\tan\theta_p = n$ from two ingredients: the geometric fact that at the polarizing angle the reflected and refracted rays are perpendicular, and Snell's law of refraction. The strategy is to write the right-angle condition as a relation between $\theta_p$ and $\theta_r$ , substitute it into Snell's law, and simplify with a single trig identity. Only algebra is needed.

Step 1. State the right-angle condition. At $\theta_p$ the reflected ray (at angle $\theta_p$ from the normal) and the refracted ray (at $\theta_r$ ) are $90^\circ$ apart. Following the geometry of the normal and the surface, this reduces to $\theta_p + \theta_r = 90^\circ$ , so $\theta_r = 90^\circ - \theta_p$ .

Step 2. Write Snell's law from air into the medium: $\sin\theta_p = n\sin\theta_r$ , with $n$ the medium's index (air taken as $1$ ).

Step 3. Substitute $\theta_r = 90^\circ - \theta_p$ . Since $\sin(90^\circ - \theta_p) = \cos\theta_p$ , Snell's law becomes $\sin\theta_p = n\cos\theta_p$ .

Step 4. Divide both sides by $\cos\theta_p$ : $\tan\theta_p = n$ . Read off the values: glass $\tan^{-1}(1.5) = 56.3^\circ$ , water $\tan^{-1}(1.33) = 53.1^\circ$ , diamond $\tan^{-1}(2.42) = 67.5^\circ$ . For glass, $\theta_r = 33.7^\circ$ and the Snell ratio $\sin 56.3^\circ/\sin 33.7^\circ = 1.50$ confirms $n$ . Verified: $\tan\theta_p = n$ for glass, water and diamond gives $56.3^\circ$ , $53.1^\circ$ , $67.5^\circ$ ; the glass refracted angle $33.7^\circ$ and the Snell cross-check $1.50$ are recomputed in verify_hs2-brewster-polarization.py.

Go deeper · Derivation B · 1 minWhy 90-degree scattered light is fully polarized

Derivation BWhy 90-degree scattered light is fully polarized

The goal is to explain, without any new formula, why light scattered through a right angle comes out completely polarized — the fact behind the polarized blue sky. The strategy is to model the scattering molecule as an oscillating electric dipole driven by the incoming light, and use the one property of dipole radiation you need: a dipole emits no radiation along its own axis of oscillation.

Step 1. Set up the driving. Unpolarized sunlight travelling horizontally has its electric field vibrating in all directions in the vertical plane across the beam. It drives the molecule's electrons to oscillate in those same transverse directions — a superposition of a vertical oscillation and a horizontal one (both perpendicular to the beam).

Step 2. Look from directly above, at $90^\circ$ to the beam. The molecule's vertical oscillation points straight at you (along your line of sight); a dipole radiates nothing along its own axis, so that component sends you no light. The molecule's horizontal oscillation is across your line of sight and radiates toward you fully.

Step 3. Conclude. Only the horizontal vibration reaches the $90^\circ$ observer, so the scattered light is polarized in that single direction — completely polarized. At other scattering angles both components partly reach you, so the light is only partially polarized; head-on (forward scatter) it stays unpolarized.

Step 4. Attach the payoff. This makes a $90^\circ$ patch of sky the most polarized light in nature, exploited by bees, photographers and navigators; and because Rayleigh scattering $\propto 1/\lambda^4$ , blue scatters about six times more than red, colouring the sky. Verified: the $90^\circ$ full-polarization condition and the blue/red scattering ratio $(700/450)^4 \approx 5.9$ are recomputed in verify_hs2-brewster-polarization.py.

Limiting cases

case	result	meaning
incidence at $\theta_p$	reflected 100% polarized	the polarizing angle; in-plane reflection vanishes
any other angle	reflected partially polarized	perpendicular component still favoured, but not fully
$n = 1$ (no index step)	$\theta_p = 45^\circ$	$\tan^{-1}1 = 45^\circ$ ; a limiting formal case, no real surface
larger index $n$	larger $\theta_p$	diamond ( $2.42$ ) polarizes at $67.5^\circ$ , glass ( $1.5$ ) at $56.3^\circ$
scattering at $90^\circ$	fully polarized	dipole radiates nothing along its own axis
scattering head-on ( $0^\circ$ )	unpolarized	both transverse components reach the observer

Common trapIt is tan θp = n, not sin θp = n

The single most common Brewster error is to write Snell-style $\sin\theta_p = n$ or to confuse Brewster's law with the critical angle for total internal reflection ( $\sin\theta_c = 1/n$ ). Brewster's law is a tangent: $\tan\theta_p = n$ . The tell is that $\sin\theta_p = n$ is impossible for any ordinary medium — it would demand $\sin\theta_p = 1.5 > 1$ for glass, which no angle satisfies. A tangent has no such ceiling, so $\theta_p$ can and does exceed $45^\circ$ (glass $56.3^\circ$ , diamond $67.5^\circ$ ). Two more cautions: the reflected beam is fully polarized only exactly at $\theta_p$ (elsewhere merely partial), and it is polarized perpendicular to the plane of incidence, not in it. Brewster ( $\tan$ ), critical angle ( $1/n$ sine), and Snell ( $n\sin$ ) are three different laws — keep them apart.

Worked examples

Board level — Brewster angles for common media (CBSE and NEET pattern)

Find the polarizing angle for (a) glass of refractive index $1.5$ , (b) water of index $1.33$ , and (c) diamond of index $2.42$ . For the glass, also find the angle of refraction and confirm Snell's law.

Step 1. Brewster's law $\tan\theta_p = n$ . Glass: $\theta_p = \tan^{-1}(1.5) = \boxed{56.3^\circ}$ .

Step 2. Water: $\theta_p = \tan^{-1}(1.33) = \boxed{53.1^\circ}$ . Diamond: $\theta_p = \tan^{-1}(2.42) = \boxed{67.5^\circ}$ . Higher index, larger angle.

Step 3. Glass refraction: $\theta_r = 90^\circ - \theta_p = 90^\circ - 56.3^\circ = \boxed{33.7^\circ}$ .

Step 4. Snell check: $\sin 56.3^\circ / \sin 33.7^\circ = 0.832/0.555 = 1.50$ , which equals $n$ ✓ — reflected and refracted rays are $90^\circ$ apart as required.

Board level — index from the polarizing angle (ICSE and JEE pattern)

The light reflected from a certain liquid surface is found to be completely polarized when the angle of incidence is $53.1^\circ$ . Identify the liquid's refractive index and name a likely liquid.

Step 1. At complete polarization the incidence angle is the Brewster angle, so $n = \tan\theta_p = \tan 53.1^\circ$ .

Step 2. Compute: $\tan 53.1^\circ = \boxed{1.33}$ .

Step 3. A refractive index of $1.33$ is that of water — the likely liquid.

Step 4. Cross-check with the right-angle rule: $\theta_r = 90^\circ - 53.1^\circ = 36.9^\circ$ , and $\sin 53.1^\circ/\sin 36.9^\circ = 0.800/0.600 = 1.33$ ✓.

Competitive level — diamond and the scattering angle (JEE-Advanced and NEET pattern)

(a) At what angle must light strike a diamond ( $n = 2.42$ ) for the reflection to be fully polarized, and what is the refracted angle there? (b) At what scattering angle in the atmosphere is sunlight completely polarized, and roughly how much more strongly is blue light ( $450\ \mathrm{nm}$ ) scattered than red ( $700\ \mathrm{nm}$ )?

Step 1. Diamond Brewster angle: $\theta_p = \tan^{-1}(2.42) = \boxed{67.5^\circ}$ .

Step 2. Refracted angle: $\theta_r = 90^\circ - 67.5^\circ = \boxed{22.5^\circ}$ . Snell check: $\sin 67.5^\circ/\sin 22.5^\circ = 0.924/0.383 = 2.41 \approx n$ ✓ (rounding).

Step 3. Scattering: sunlight is completely polarized when scattered through $\boxed{90^\circ}$ — look at right angles to the Sun.

Step 4. Rayleigh scattering $\propto 1/\lambda^4$ , so the blue-to-red ratio is $(700/450)^4 = \boxed{5.9}$ — blue is scattered nearly six times as strongly, which is why the sky is blue and its polarization strongest overhead at sunset. Timing note: about 60 seconds. Brewster problems are one-liners once you spot "fully polarized reflection" — it means $\tan\theta_p = n$ . The only trap is reaching for $\sin$ instead of $\tan$ , or confusing the polarizing angle with the critical angle.

Exam traps

Writing $\sin\theta_p = n$ instead of $\tan\theta_p = n$ . Brewster's law is a tangent. For glass, $\sin\theta_p = 1.5$ is impossible (sine cannot exceed $1$ ), which is the built-in warning that you have the wrong function. The tangent lets $\theta_p$ exceed $45^\circ$ , as it must.

Confusing Brewster's angle with the critical angle. The critical angle for total internal reflection is $\sin\theta_c = 1/n$ and applies to light going from dense to rare medium. Brewster's angle is $\tan\theta_p = n$ for reflection off the surface. Different phenomena, different formulas — a favourite mix-up.

Getting the polarization direction wrong. The reflected beam at the Brewster angle is polarized perpendicular to the plane of incidence (parallel to the surface). This is why polarizing sunglasses have their axis vertical: horizontal glare reflected off roads and water is perpendicular-polarized (horizontal), and a vertical axis blocks it.

Thinking reflection is fully polarized at every angle. Only at exactly $\theta_p$ is the reflected light $100\%$ polarized. At other angles it is partially polarized — the perpendicular component is favoured but the in-plane component is not fully suppressed.

Getting the scattering geometry wrong. Scattered sunlight is fully polarized at $90^\circ$ to the incoming beam, not head-on. Forward-scattered light stays unpolarized; the right-angle view is the polarized one.

Board scope. Brewster's law and polarization by reflection and scattering are examined by CBSE, ICSE, JEE-Main, JEE-Advanced and NEET. Like the previous article, they are a structural zero on IB (SL and HL) and on A-Level 9702, and are not part of the current AP Physics 2 core — neither IB nor A-Level 9702 covers polarization. This is India-facing content; IB and A-Level students can skip it.

Origins · 1 minDavid Brewster and how this was discovered

David Brewster

1781–1868 · Scotland

Sir David Brewster was a Scottish physicist — licensed to preach but too nervous ever to take the pulpit — and prolific inventor who devoted his life to the study of light — especially its polarization. In 1815 he established the law that carries his name: that unpolarized light reflected from a transparent surface is completely polarized when the angle of incidence has a tangent equal to the medium's refractive index, $\tan\theta_p = n$ . He showed that at this angle the reflected and refracted rays are perpendicular, tying polarization directly to Snell's law. Brewster catalogued the polarizing angles of dozens of materials and studied the colours of crystals and stressed glass, founding the field of optical mineralogy. He is also, delightfully, the inventor of the kaleidoscope (1816), which he patented, and an improver of the stereoscope. A founder of the British Association for the Advancement of Science, he wrote a celebrated biography of Newton.

History

1669

— Erasmus Bartholin observes double refraction in Iceland spar, an early polarization phenomenon.

1808

— Étienne-Louis Malus discovers polarization by reflection, later (1811) coining the term polarization.

1815

— David Brewster states his law, tan θp = n, for the angle of complete polarization on reflection.

1871

— Rayleigh (J. W. Strutt) explains the blue and polarized sky by scattering that grows as one over wavelength to the fourth power.

1929

— Edwin Land patents an inexpensive sheet polarizer (later branded Polaroid), making polarizing filters and glare-cutting sunglasses everyday objects.

Going further

Texts. NCERT Physics Class 12, Ch. 10 ("Wave Optics"), §10.7 on polarization by reflection and Brewster's law · H. C. Verma, Concepts of Physics Vol. 1, "Light Waves," Brewster and scattering · Halliday, Resnick and Walker, "Polarization by reflection."

Provenance. Standard results. Recompute: $\tan\theta_p = n$ gives $56.3^\circ$ (glass, $1.5$ ), $53.1^\circ$ (water, $1.33$ ), $67.5^\circ$ (diamond, $2.42$ ); glass $\theta_r = 33.7^\circ$ , Snell cross-check $\sin 56.3^\circ/\sin 33.7^\circ = 1.50$ ; diamond $\theta_r = 22.5^\circ$ ; index from $\theta_p = 53.1^\circ$ is $\tan 53.1^\circ = 1.33$ (water); scattering full polarization at $90^\circ$ ; Rayleigh blue/red ratio $(700/450)^4 = 5.9$ .

Prerequisites. Polarization & Malus's Law — the transverse-wave and filter physics this article builds on. Leads to. the university seams Brewster's Angle and Polarization of Light.

Go deeper (UG). The full Fresnel equations — which give the reflected fraction of each polarization at every angle, and from which the Brewster zero drops out — are developed in Brewster's Angle; the general treatment of polarization states is in Polarization of Light. Enrichment only; the geometry above is all this article needs.

Why is Brewster's law a tangent and not a sine like Snell's law?

Because it comes from combining Snell's law with the right-angle condition. Snell's law is $\sin\theta_p = n\sin\theta_r$ , a sine relation. But at the Brewster angle the refracted ray is $90^\circ - \theta_p$ , and $\sin(90^\circ - \theta_p) = \cos\theta_p$ . Substituting turns Snell's law into $\sin\theta_p = n\cos\theta_p$ , and dividing by $\cos\theta_p$ gives $\tan\theta_p = n$ . The tangent appears only after you feed in the geometry. A practical bonus: a tangent has no upper limit, so the polarizing angle can exceed $45^\circ$ — which it must, since $\tan^{-1}(1.5) = 56^\circ$ for glass. If you ever write $\sin\theta_p = n$ , the impossible $\sin > 1$ for glass is your signal to switch to the tangent.

How is the Brewster angle different from the critical angle?

They describe opposite situations. The critical angle governs total internal reflection: light travelling from a dense medium toward a rarer one is completely reflected when it strikes beyond $\theta_c$ , where $\sin\theta_c = 1/n$ . The Brewster angle governs polarization on reflection: unpolarized light from the rarer side reflects off the surface fully polarized at $\theta_p$ , where $\tan\theta_p = n$ . They involve different directions of travel, different formulas ( $1/n$ sine versus $n$ tangent), and different physics (total reflection versus polarization). For glass, $\theta_c \approx 42^\circ$ but $\theta_p \approx 56^\circ$ — not even close, so if your two answers come out equal you have mixed the laws.

Why is the sky polarized, and where is it strongest?

Because sunlight is scattered by air molecules, and light scattered through $90^\circ$ comes out fully polarized. A scattering molecule acts like a tiny antenna set vibrating by the sunlight; an antenna radiates nothing along its own vibration axis, so a viewer at right angles to the Sun receives only one vibration direction — polarized light. The polarization is therefore strongest in the band of sky $90^\circ$ from the Sun, and weakest looking straight toward or away from it. You can see this with any polarizing filter: rotate it while looking at that band of sky and it darkens and brightens. Bees, ants and migrating birds read this pattern as a compass even when the Sun itself is hidden.

Is the reflected light at the Brewster angle bright?

No — it is completely polarized but weak. Only a small fraction of the incident light reflects off glass (a few per cent), and at the Brewster angle the in-plane component reflects zero, so the reflected beam is even dimmer than at other angles. Most of the light refracts on into the medium. What makes the reflected beam special is its quality (100% polarized), not its brightness. The transmitted (refracted) beam, by contrast, is only slightly polarized, because it carries most of both components. This is why a single Brewster reflection is a good polarizer of a weak beam but a poor one if you need most of the light.

Do polarizing sunglasses use the Brewster effect?

They exploit it. Glare bouncing off horizontal surfaces — wet roads, water, car bonnets — hits near the Brewster angle and comes back strongly polarized in the horizontal direction. Polarizing sunglasses set their transmission axis vertical, so by Malus's law that horizontal glare meets an angle near $90^\circ$ and is heavily cut, while the rest of the scene passes. That is also why the glare flares back if you tilt your head sideways: you rotate your glasses' axis toward the glare's horizontal polarization. The Brewster effect creates the polarized glare; Malus's law (previous article) describes how the glasses remove it — the two articles are two halves of one everyday device.

Does Brewster's law work for light going from glass into air too?

Yes, with the index taken the right way round. Brewster's law uses the ratio of indices across the boundary: $\tan\theta_p = n_2/n_1$ , where the light travels from medium $1$ into medium $2$ . Going from air ( $n_1 = 1$ ) into glass ( $n_2 = 1.5$ ) gives $\tan\theta_p = 1.5$ , $\theta_p = 56.3^\circ$ . Going from inside glass out to air gives $\tan\theta_p = 1/1.5 = 0.667$ , $\theta_p = 33.7^\circ$ — which is exactly the refracted angle from the air side, as the $90^\circ$ geometry demands. The two Brewster angles for the same interface are complementary, adding to $90^\circ$ . For this course, unless told otherwise, take light going from air into the medium and use $\tan\theta_p = n$ .