Polarization — Malus's law, Brewster's angle derivation: Worked examples

This page hunts down every kind of question the parent note can throw at you. We first lay out a scenario matrix — a grid of all the case-classes — then work one full example per cell so you never meet a situation you haven't already seen.

Two tools do all the work here, and we earned both in the parent:

Malus's law $I = I_{0} cos^{2} θ$ — how much already-polarized light survives a polarizer tilted by $θ$ .
Brewster's law $tan θ_{B} = n_{2} / n_{1}$ — the incidence angle that makes reflected light perfectly polarized.

If any symbol below feels unfamiliar, re-read Unpolarized vs Polarized Light and Wave Nature of Light first — we assume you know that light is a wiggling transverse wave and that a polarizer keeps only the wiggle along its axis.

The scenario matrix

Every polarization problem is one (or a chain) of these cells:

#	Cell class	What's tricky about it	Example that hits it
A	Unpolarized → 1 polarizer	Use the $\frac{1}{2}$ rule, not $cos^{2}$	Ex 1
B	Polarized → polarizer at angle $θ$	Plain Malus, remember the square	Ex 2
C	Degenerate angles: $θ = 0^{\circ}$ and $θ = 9 0^{\circ}$	Full pass vs. total block (crossed)	Ex 3
D	Chain of $N$ polarizers, stepped angles	Apply Malus stage-by-stage; light reappears	Ex 4
E	"What angle gives a target fraction?" (inverse Malus)	Solve $cos^{2} θ = k$ , watch both roots	Ex 5
F	Brewster's angle, air → dense medium	$tan θ_{B} = n_{2} / n_{1}$ , check reflected ⟂ refracted	Ex 6
G	Brewster reversed: dense → rare medium	Ratio flips; $θ_{B}$ gets smaller	Ex 7
H	Real-world word problem (glare / sunglasses)	Translate physics words into $θ$ and axes	Ex 8
I	Exam twist: combine Brewster and Malus	Two tools in one chain	Ex 9

We now clear the whole grid.

Cell A — Unpolarized light, one polarizer

Worked example Ex 1 · The

$\frac{1}{2}$ rule (Cell A) Unpolarized sunlight of intensity $I_{0} = 800 W/m^{2}$ passes through a single polarizer. What comes out, and is it polarized?

Forecast: Guess before reading — will it be less than half, exactly half, or angle-dependent?

Recognise the input is unpolarized. Why this step? Unpolarized light is a random mix of every wiggle-angle $θ$ , so there is no single $θ$ to plug into $cos^{2} θ$ . We must average over all angles.
Use the averaged result $⟨ cos^{2} θ ⟩ = \frac{1}{2}$ : $I = \frac{1}{2} I_{0} = \frac{1}{2} (800) = 400 W/m^{2} .$ Why this step? The parent derived $\frac{1}{2 π} \int_{0}^{2 π} cos^{2} θ d θ = \frac{1}{2}$ — half the energy survives regardless of how you rotate this first filter.
State the output polarization. Why? Whatever survives is aligned with the transmission axis, so the beam is now fully polarized along that axis.

Verify: Units are $W/m^{2}$ in, $W/m^{2}$ out ✓. Rotating this single polarizer would not change the reading (still $400$ ) — the tell-tale signature of unpolarized input. If it did change, the input was already polarized.

Cell B — Polarized light, one polarizer

Worked example Ex 2 · Plain Malus with the square (Cell B)

Light already polarized vertically, intensity $I_{0} = 200 W/m^{2}$ , meets a polarizer whose axis is tilted $6 0^{\circ}$ from vertical. Find the transmitted intensity.

Forecast: At $6 0^{\circ}$ — closer to full pass or closer to blocked?

Identify $θ = 6 0^{\circ}$ = angle between the incoming polarization and the axis. Why this step? Malus's law needs the angle between the field and the axis, not the angle from anything else.
Apply Malus with the square: $I = I_{0} cos^{2} 6 0^{\circ} = 200 (\frac{1}{2})^{2} = 200 \cdot \frac{1}{4} = 50 W/m^{2} .$ Why the square? A detector reads intensity, and intensity $\propto E^{2}$ . Projection gives amplitude $E_{0} cos θ$ ; squaring gives $cos^{2} θ$ . Forgetting it would give the wrong $100$ .

Verify: $cos 6 0^{\circ} = 0.5$ , so $cos^{2} = 0.25$ ; $200 \times 0.25 = 50$ ✓. Sanity: $6 0^{\circ}$ is past $4 5^{\circ}$ , so more than half is lost — $50 < 100$ , consistent.

Cell C — Degenerate angles: $0^{\circ}$ and $9 0^{\circ}$

Cell D — Chain of polarizers (light reappears)

Worked example Ex 4 · Three polarizers, stepped angles (Cell D)

Polarized light $I_{0} = 100 W/m^{2}$ enters along the axis of polarizer P1. P2 is at $3 0^{\circ}$ to P1; P3 is at $6 0^{\circ}$ to P1. Find the final intensity.

Forecast: Will the answer be tiny, or surprisingly large given the total turn is $6 0^{\circ}$ ?

Figure s01 — Ex 4: light steps through three polarizers P1, P2, P3 whose axes tilt $0^{\circ}, 3 0^{\circ}, 6 0^{\circ}$ ; intensity (W/m²) is printed above each beam segment.

Reading the figure: the beam travels left → right (yellow arrows, the number above each arrow is the intensity in $W/m^{2}$ ). Each shaded block is a polarizer; the dark line inside each block is its transmission axis, tilted $0^{\circ}$ , $3 0^{\circ}$ , $6 0^{\circ}$ from vertical for P1, P2, P3. Notice the axis rotates in equal $3 0^{\circ}$ steps from block to block — that stepping is the whole point.

P1 → P2, step angle $= 3 0^{\circ}$ (the tilt between P1's axis and P2's axis in the figure): $I_{2} = I_{0} cos^{2} 3 0^{\circ} = 100 (\frac{3}{2})^{2} = 100 \cdot \frac{3}{4} = 75 W/m^{2} .$ Why step-by-step? Each polarizer repolarizes the light along its own axis, so the next angle is measured from the previous axis, not from P1.
P2 → P3, step angle $= 6 0^{\circ} - 3 0^{\circ} = 3 0^{\circ}$ : $I_{3} = I_{2} cos^{2} 3 0^{\circ} = 75 \cdot \frac{3}{4} = 56.25 W/m^{2} .$ Why $3 0^{\circ}$ again? After P2 the light points along P2. P3 is $3 0^{\circ}$ beyond P2 (read the two axis lines in the figure), so the relevant tilt is $3 0^{\circ}$ .

Verify: If we (wrongly) removed P2, P1→P3 direct would give $100 cos^{2} 6 0^{\circ} = 25 W/m^{2}$ . Inserting the middle filter raised the output to $56.25 W/m^{2}$ , which is greater than $25 W/m^{2}$ — the reappearing-light surprise from the parent, because stepping the rotation wastes less than one big turn ✓.

Cell E — Inverse Malus (find the angle)

Worked example Ex 5 · Solve for the tilt (Cell E)

Polarized light must be dimmed to exactly $\frac{1}{4}$ of its intensity by one polarizer. Through what angle should the axis be set?

Forecast: One answer, or several?

Write the target as an equation: $\frac{I}{I _{0}} = cos^{2} θ = \frac{1}{4} .$ Why this step? "Fraction of intensity" is $cos^{2} θ$ ; the unknown is now $θ$ , so we invert Malus.
Take the square root, keeping both signs: $cos θ = \pm \frac{1}{2} .$ Why $\pm$ ? Squaring erased the sign; both $+ \frac{1}{2}$ and $- \frac{1}{2}$ satisfy the equation, and physical tilts can be on either side.
Read off all angles in $[0^{\circ}, 18 0^{\circ})$ : $θ = 6 0^{\circ} or θ = 12 0^{\circ} .$ Why two? $cos 6 0^{\circ} = + \frac{1}{2}$ and $cos 12 0^{\circ} = - \frac{1}{2}$ ; both give $cos^{2} θ = \frac{1}{4}$ . A polarizer doesn't care about the sign of the projection, only its square.

Verify: $cos^{2} 6 0^{\circ} = 0.25$ ✓ and $cos^{2} 12 0^{\circ} = (- 0.5)^{2} = 0.25$ ✓. Both dim to a quarter — the matrix cell "watch both roots" is satisfied.

Cell F — Brewster's angle, air → dense medium

Worked example Ex 6 · Brewster for diamond (Cell F)

Light in air ( $n_{1} = 1$ ) strikes diamond ( $n_{2} = 2.42$ ). Find Brewster's angle and confirm the reflected and refracted rays are perpendicular.

Forecast: Bigger or smaller than glass's $56. 3^{\circ}$ ?

Figure s02 — Ex 6: at Brewster's angle the reflected ray (coral) and refracted ray (mint) meet the normal at $θ_{B}$ and $θ_{r}$ , and the two rays are exactly $9 0^{\circ}$ apart.

Reading the figure: the horizontal line is the air–diamond surface; the dashed vertical line is the normal (perpendicular to the surface). The yellow arrow coming down from the upper left is the incident ray, the coral arrow going up to the upper right is the reflected ray, and the mint arrow going down into the diamond is the refracted ray. The angle each ray makes with the normal is marked: $θ_{B}$ for incident and reflected (they are equal by the law of reflection), $θ_{r}$ for refracted. The label on the right shows the $9 0^{\circ}$ between the reflected and refracted arrows.

Apply Brewster's law: $tan θ_{B} = \frac{n _{2}}{n _{1}} = \frac{2.42}{1} = 2.42.$ Why $tan$ , not $sin$ ? The parent's derivation combined Snell's law with $θ_{r} = 9 0^{\circ} - θ_{B}$ ; dividing by $cos θ_{B}$ turned the ratio into a tangent. ("Brewster Tangles.")
Invert with arctan (the angle whose tangent is $2.42$ ): $θ_{B} = arctan (2.42) = 67. 5^{\circ} .$
Where does $θ_{r} = 9 0^{\circ} - θ_{B}$ come from? (mini-derivation, read off the figure). Look at the point where the beam hits the surface. Three things fan out above and below along the normal on the right-hand side: the reflected ray sits $θ_{B}$ from the normal (upper right), the refracted ray sits $θ_{r}$ from the normal (lower right), and between the reflected and refracted rays is the $9 0^{\circ}$ Brewster corner. These three angles are stacked around the straight normal line, which spans $18 0^{\circ}$ from straight up to straight down. So they must add to a straight angle: $θ_{B} + 9 0^{\circ} + θ_{r} = 18 0^{\circ} \Rightarrow θ_{r} = 9 0^{\circ} - θ_{B} = 9 0^{\circ} - 67. 5^{\circ} = 22. 5^{\circ} .$ Why does this matter? This is the geometric fact that turns Snell into Brewster; without it we could not eliminate $θ_{r}$ .

Verify (plug into Snell, watching rounding): $n_{1} sin θ_{B} = 1 \cdot sin 67. 5^{\circ} = 0.9239$ ; $n_{2} sin θ_{r} = 2.42 \cdot sin 22. 5^{\circ} = 2.42 \times 0.3827 = 0.9261$ . These differ only in the third decimal ( $0.9239$ vs $0.9261$ , a $0.2%$ gap) — that is pure rounding from writing $θ_{B}$ as $67. 5^{\circ}$ instead of $arctan 2.42 = 67.52 3^{\circ}$ . Using the un-rounded angle they match exactly ✓. Denser medium ⇒ larger $θ_{B}$ than glass's $56. 3^{\circ}$ , as forecast.

Cell G — Brewster reversed: dense → rare medium

Worked example Ex 7 · Light leaving glass into air (Cell G)

Light travelling inside glass ( $n_{1} = 1.5$ ) hits the glass–air surface ( $n_{2} = 1.0$ ). Find Brewster's angle for this internal reflection.

Forecast: Larger or smaller than the air→glass value of $56. 3^{\circ}$ ?

Keep the ratio in the right order: $tan θ_{B} = \frac{n _{2}}{n _{1}} = \frac{1.0}{1.5} = 0.667.$ Why the flip? $n_{2}$ is the medium light goes into (air), $n_{1}$ is where it starts (glass). Getting this backwards is the classic error flagged in the parent's mistakes.
Invert: $θ_{B} = arctan (0.667) = 33. 7^{\circ} .$

Verify: Notice $33. 7^{\circ} + 56. 3^{\circ} = 9 0^{\circ}$ : the two Brewster angles for a given interface pair are complementary. Why? $tan θ_{B}^{in} = 1/ tan θ_{B}^{out}$ , and angles whose tangents are reciprocals sum to $9 0^{\circ}$ ✓. Reversed direction gives the smaller angle, as forecast.

Cell H — Real-world word problem (glare)

Worked example Ex 8 · Fisherman's sunglasses (Cell H)

Sunlight reflects off a still lake and reaches your eyes as glare. (a) At what viewing angle from the vertical is the glare fully polarized? (b) In which direction is that glare polarized, and how should the sunglasses' axis be oriented to kill it? Water $n = 1.33$ . (The plane of incidence was defined at the top of this page.)

Forecast: Vertical or horizontal transmission axis for the glasses?

Glare is fully polarized at Brewster's angle: $tan θ_{B} = \frac{n _{water}}{n _{air}} = \frac{1.33}{1} = 1.33, θ_{B} = arctan (1.33) = 53. 1^{\circ} .$ Why? Reflection off a surface is fully polarized only at Brewster's angle — the physics word "fully polarized glare" is code for " $θ_{B}$ ." Answer to (a): $53. 1^{\circ}$ from the vertical (i.e. from the normal to the water).
Direction of the glare's polarization: the reflected light is polarized perpendicular to the plane of incidence (the "s" direction defined above), which lies parallel to the water surface = horizontal. Why? The parent's electron-oscillation argument kills the in-plane ("p") part at $θ_{B}$ , so only the surface-parallel ("s") wiggle survives in the reflection.
Orient the sunglasses to block it: set the transmission axis vertical, perpendicular to the horizontal glare. Why? Malus with $θ = 9 0^{\circ}$ gives $I = I_{0} cos^{2} 9 0^{\circ} = 0$ — the horizontal glare is completely absorbed. Answer to (b): the glare is horizontal, so the sunglasses' axis must be vertical to kill it.

Verify: $arctan (1.33) = 53. 1^{\circ}$ ✓; blocking condition uses $cos^{2} 9 0^{\circ} = 0$ ✓. This is exactly why polarized sunglasses have vertical axes — matching the parent's Feynman note.

Cell I — Exam twist: Brewster feeds Malus

Worked example Ex 9 · Two tools in one chain (Cell I)

Unpolarized light of intensity $I_{0} = 400 W/m^{2}$ reflects off glass ( $n = 1.5$ ) at Brewster's angle, becoming fully horizontally polarized. It then passes through a polarizer whose axis is $4 0^{\circ}$ from horizontal. Find the final intensity. (For a real reflection only part of the light reflects; here assume the reflected beam that emerges carries intensity $I_{0}$ and is fully polarized — a common exam idealisation.)

Forecast: Which law governs the last step — the $\frac{1}{2}$ rule or $cos^{2}$ ?

Brewster step fixes the polarization state, not the number here. Why this step? We're told the reflected beam is fully horizontally polarized with intensity $I_{0} = 400$ . At $θ_{B} = arctan (1.5) = 56. 3^{\circ}$ the reflection is 100% polarized — that is what makes the next step a $cos^{2}$ problem, not a $\frac{1}{2}$ problem.
Now the light is polarized, so use Malus (not $\frac{1}{2}$ ): angle between the horizontal polarization and the polarizer axis is $θ = 4 0^{\circ}$ : $I = I_{0} cos^{2} 4 0^{\circ} = 400 cos^{2} 4 0^{\circ} .$ Why not halve? The $\frac{1}{2}$ rule applies only to unpolarized input; after Brewster reflection the beam is already polarized, so $cos^{2} θ$ is correct.
Compute: $cos 4 0^{\circ} = 0.766$ , so $cos^{2} 4 0^{\circ} = 0.587$ : $I = 400 \times 0.587 = 234.7 W/m^{2} .$

Verify: $θ_{B} = arctan 1.5 = 56. 3^{\circ}$ ✓; $400 \times cos^{2} 4 0^{\circ} \approx 234.7 W/m^{2}$ ✓; result lies in $[0, 400]$ ✓. If we had wrongly applied the $\frac{1}{2}$ rule first, we'd get $\frac{1}{2} (400) cos^{2} 4 0^{\circ} = 117.4$ — half the correct value, the exact trap in the matrix.

Recall

Recall Which rule for the

first filter — $\frac{1}{2}$ or $cos^{2}$ ? If the input is unpolarized → $\frac{1}{2}$ . If it is already polarized → $cos^{2} θ$ . ::: Unpolarized → $\frac{1}{2} I_{0}$ ; polarized → $I_{0} cos^{2} θ$ . Why does inverse-Malus give two angles for one fraction? ::: Because $cos^{2} θ = k$ ⇒ $cos θ = \pm k$ , giving a supplementary pair like $6 0^{\circ}$ and $12 0^{\circ}$ . Why is the glass↔air Brewster pair complementary? ::: The tangents are reciprocals ( $n_{2} / n_{1}$ vs $n_{1} / n_{2}$ ), and reciprocal-tangent angles sum to $9 0^{\circ}$ .

Connections

Snell's Law and Refraction — every Brewster verification plugs back into Snell.
Electromagnetic Waves — the $I \propto E^{2}$ that forces the square in Malus.
Reflection and Refraction at Interfaces — the reflected ⟂ refracted geometry.
Unpolarized vs Polarized Light — decides $\frac{1}{2}$ vs $cos^{2}$ at each stage.