Malus's Law

Point a laser at a polarizing filter, then slowly rotate the filter. The transmitted brightness does not switch off abruptly — it fades smoothly, brightest when the filter's axis lines up with the light's polarization, black when the two are perpendicular, and exactly half-bright at 45°. Étienne-Louis Malus captured this behaviour in a single equation: the transmitted intensity is $I = I_0\cos^2\theta$ , where $\theta$ is the angle between the incoming polarization and the filter's transmission axis. That $\cos^2$ is not an empirical curve fit — it is geometry. A polarizer passes only the component of the electric-field vector that lies along its axis, and that projected amplitude is $E_0\cos\theta$ . Because a detector (and the Poynting flux) responds to the square of the field, the intensity carries $\cos^2\theta$ . This one line does an astonishing amount of work. It explains why crossed polarizers go black, why a single polarizer always halves unpolarized light no matter how you turn it, and — most memorably — why sliding a third polarizer between two crossed ones lets light through, so that adding an absorber makes a dark path bright. The same $\cos^2\theta$ reappears at the single-photon level as the probability that one photon survives an analyzer, which is why Malus's law is the classical launch pad for quantum measurement, Bell tests, and polarization-encoded cryptography. Here we build the law from field projection, push it through the three-polarizer paradox and the N-polarizer staircase, and pin down where a real polarizer departs from the ideal.

Key takeaways

Malus's law: linearly polarized light of intensity $I_0$ through an ideal polarizer emerges as $I = I_0\cos^2\theta$ , with $\theta$ the angle between the incident polarization and the transmission axis.
The $\cos^2$ is projection-then-square: the axis passes field amplitude $E_0\cos\theta$ , and intensity $\propto$ amplitude².
Crossed axes ( $\theta = 90°$ ) give darkness; parallel ( $\theta = 0$ ) full transmission; $45°$ gives exactly $I_0/2$ .
Unpolarized light through one ideal polarizer always emerges at $I_0/2$ — the average of $\cos^2\theta$ over all angles — and comes out fully polarized.
The three-polarizer paradox: two crossed polarizers pass nothing, but a third at $45°$ inserted between them restores $I_0/8$ — because a polarizer projects and resets the state, it does not merely filter.
Key idea: a polarizer projects the electric field onto its axis (amplitude $\times\cos\theta$ ), and intensity is amplitude squared — so intensity carries $\cos^2\theta$ . Every polarization intensity result is a corollary of that one projection.

The core insight

A linear polarizer projects the electric field onto its transmission axis, so the surviving amplitude is $E_0\cos\theta$ ; intensity is proportional to amplitude squared, so the transmitted intensity is $I_0\cos^2\theta$ . Projection, then square — that is the whole law.

Where this lives today

The rotating analyzer is still the workhorse of polarimetry: spin a polarizer in front of a detector, fit the recorded intensity to $\tfrac12(1+\cos2\theta)$ , and the modulation depth gives the degree of polarization while the phase gives the polarization angle — the basis of ellipsometry for measuring thin-film thickness to sub-nanometre precision. A polarizer paired with a half-wave plate is the standard laboratory variable attenuator: because transmission is $\cos^2$ of a purely geometric angle, it sets laser power smoothly and wavelength-flat without touching the beam profile. Every liquid-crystal display is a Malus machine — a voltage-controlled birefringent layer between crossed polarizers rotates the polarization by an amount the voltage sets, and Malus's law converts that rotation into the grey level of each pixel. And at the single-photon frontier, attenuating a beam until photons arrive one at a time turns $\cos^2\theta$ from an intensity ratio into a probability of transmission, a fact confirmed to exquisite precision in the polarization-entangled photon experiments behind Bell-inequality tests and quantum cryptography.

The physical picture

Think of linearly polarized light as an arrow (the electric field) oscillating back and forth along one fixed line. A polarizer is a device with a single privileged direction — its transmission axis — and it lets through only the shadow of that arrow cast onto its axis. If the arrow already points along the axis, the whole thing passes; if it points across the axis, nothing passes; in between, the surviving length is $E_0\cos\theta$ . Brightness is the square of that surviving length, so the light dims as $\cos^2\theta$ : full at $0°$ , one-half at $45°$ , and dark at $90°$ . The subtle part — the part that separates Malus's law from a naïve "filter" picture — is what the polarizer does to the light that does get through. It does not simply skim off a slice; it re-aligns the survivor along its own axis. The output of a polarizer at $45°$ is not "the $45°$ part of the input still pointing where it did," it is a brand-new state polarized cleanly along $45°$ . That state-reset is invisible with one polarizer but becomes spectacular with three, and it is the physical content of the projection.

Figure

Malus's law as vector projection: transmitted intensity follows I₀cos²θ — full at 0°, half at 45°, dark at 90° — and inserting a third polarizer between crossed ones revives I₀/8

The right panel holds the paradox in one glance. Unpolarized light meets a $0°$ polarizer and leaves at $I_0/2$ , polarized horizontally. The $45°$ sheet re-projects it (Malus, $\theta = 45°$ ) to $I_0/4$ , now polarized at $45°$ . The final $90°$ sheet sees light $45°$ from its axis, not $90°$ , so it passes $I_0/8$ instead of nothing. Pull the middle sheet out and the two survivors are crossed at $90°$ : the output collapses to zero. The absorber, counterintuitively, is what keeps the path open. That leads to the punchline figure. If one $45°$ insert recovers $I_0/8$ , what does a whole staircase of polarizers do — many sheets each stepping the axis by a small angle, from $0°$ up to $90°$ ? Each tiny step is nearly loss-free ( $\cos^2$ of a small angle $\approx 1$ ), and the transmission of an $N$ -step $90°$ fan is $T = (\cos^2(90°/N))^{N}$ , which climbs toward 100 % as $N$ grows. Adding polarizers raises the throughput.

Figure

The N-polarizer staircase: a 90° rotation split into N small steps transmits (cos²(90°/N))ᴺ — rising from 0 (one 90° step) through 66.0 % at N = 6 to 94.7 % at N = 45, toward 100 %

The core equations

Malus's law is the intensity statement of a projection. Its cleanest form assumes an ideal polarizer and light that is already linearly polarized when it arrives; every other result on this page is a corollary got by averaging it, chaining it, or generalizing the transmittances.

I = I_0\cos^2\theta \qquad (1)

where: $I$ — transmitted intensity after the polarizer; $I_0$ — intensity of the incident linearly polarized beam; $\theta$ — angle between the incident polarization direction and the transmission axis. Read it as a two-step recipe: the axis keeps the field component $E_0\cos\theta$ , and intensity is the square of the surviving field. The underlying field statement, which is what actually gets projected, is

E_\parallel = E_0\cos\theta, \qquad I \propto \langle E_\parallel^2\rangle \qquad (2)

where: $E_\parallel$ — field amplitude passed along the axis; $E_0$ — incident amplitude; $\langle\,\rangle$ — the time average that the detector and the Poynting flux perform. The single power of $\cos\theta$ lives in the field; the square that makes it $\cos^2\theta$ comes from intensity being quadratic in the field — a point worth guarding, because forgetting it is the commonest Malus error. Because $\cos^2\theta = \tfrac12(1 + \cos 2\theta)$ , the law is equivalently a constant plus a modulation at twice the analyzer angle,

T \equiv \frac{I}{I_0} = \tfrac12\big(1 + \cos 2\theta\big) \qquad (3)

where: $T$ — transmittance; $2\theta$ — the doubled angle. This is exactly the signal a rotating-analyzer polarimeter records: one full brightness cycle every $180°$ of rotation, not $360°$ , the fingerprint of a $\cos^2$ law. For a cascade of ideal polarizers the output is the running product of Malus factors, each using the relative angle between neighbouring axes:

I_N = I_1\prod_{n=2}^{N}\cos^2(\theta_n - \theta_{n-1}) \qquad (4)

where: $\theta_n$ — axis angle of the $n$ -th polarizer; $I_1$ — intensity after the first. Each stage re-projects the state onto the next axis and resets it there — the equation that powers both the three-polarizer paradox and the staircase.

Derivation AMalus's law from field projection

The goal is to derive $I = I_0\cos^2\theta$ from first principles — treating the polarizer as a device that keeps one vector component of the electric field and discards the orthogonal one, then converting the surviving field to an intensity. We will decompose the incident field along and across the transmission axis, pass only the parallel part, and use the fact that a photodetector and the time-averaged Poynting vector both respond to the square of the field amplitude. The prefactors $\tfrac12 c\varepsilon_0$ will cancel between incident and transmitted beams, leaving a pure geometric factor.

Step 1. Write the incident linearly polarized field, amplitude $E_0$ , as components along the transmission axis $\hat{\mathbf a}$ and the perpendicular $\hat{\mathbf a}_\perp$ : $\mathbf E_0 = E_0(\cos\theta\,\hat{\mathbf a} + \sin\theta\,\hat{\mathbf a}_\perp)$ . Why decompose: the polarizer treats the two axes completely differently — it is a projector, and a projector is defined by what it keeps versus kills, so we must first name those two directions.

Step 2. Apply the ideal polarizer: it removes the perpendicular component entirely and passes the parallel one unchanged, so the transmitted field is $\mathbf E_t = E_0\cos\theta\,\hat{\mathbf a}$ . Why only $\cos\theta$ survives: the transmitted amplitude is the dot product $\mathbf E_0\cdot\hat{\mathbf a} = E_0\cos\theta$ — the projection of the field vector onto the axis, nothing more.

Step 3. Convert field to intensity. The time-averaged intensity of a plane wave is $I = \tfrac12 c\varepsilon_0 E^2$ , so the transmitted intensity is $I = \tfrac12 c\varepsilon_0 E_0^2\cos^2\theta$ . Why the angle now appears squared: intensity is quadratic in field, so the single $\cos\theta$ of the amplitude becomes $\cos^2\theta$ in the intensity — this squaring is the physical origin of the law's shape.

Step 4. Normalize to the incident intensity $I_0 = \tfrac12 c\varepsilon_0 E_0^2$ . The prefactor $\tfrac12 c\varepsilon_0 E_0^2$ cancels, leaving $I = I_0\cos^2\theta$ . Why the constants vanish: they describe the beam's absolute brightness, which is common to numerator and denominator; Malus's law is a ratio and therefore purely geometric. *Quick checks: $I(0) = I_0$ , $I(45°) = I_0/2$ , $I(90°) = 0$ , and the modulation identity $\cos^2\theta = \tfrac12(1+\cos2\theta)$ all follow directly from the result.*

Go deeper · Derivation B · 1 minThe universal half for unpolarized light

Derivation BThe universal half for unpolarized light

The goal here is to show that a single ideal polarizer transmits exactly one half of unpolarized light, regardless of how the polarizer is oriented — and that the survivor comes out fully polarized. Unpolarized light is not a definite state but a fast, structureless mixture of every polarization angle with equal weight; the detector sees the time (or ensemble) average of Malus's law over that flat distribution. We will average $\cos^2\theta$ over a uniform angular distribution and read off both the factor and the fact that the output is purified.

Step 1. Model unpolarized light as a uniform mixture over polarization angle $\psi \in [0, 2\pi)$ with probability density $1/(2\pi)$ . Each instantaneous component obeys Malus's law relative to the fixed axis. Why a flat distribution: "unpolarized" means no direction is preferred, which is exactly a uniform density over $\psi$ — any other density would define a partial polarization.

Step 2. Average the transmitted intensity: $\langle I\rangle = I_0\,\dfrac{1}{2\pi}\displaystyle\int_0^{2\pi}\cos^2\psi\,\mathrm d\psi$ . Using $\cos^2\psi = \tfrac12(1+\cos2\psi)$ , the $\cos2\psi$ term integrates to zero over a full period, leaving $\dfrac{1}{2\pi}\cdot\pi = \tfrac12$ . Why the modulation washes out: over a full turn the field spends as much time aligned as anti-aligned with the axis, so only the constant $\tfrac12$ survives the average.

Step 3. Conclude $\langle I\rangle = I_0/2$ , independent of the axis direction, and note the exit beam is polarized purely along the axis. Why orientation-independence matters: a single polarizer can never tell you which way unpolarized light "was" polarized — it had no direction; the polarizer manufactures one, costing exactly a factor of two (3 dB). *The integral evaluates to $\tfrac{1}{2\pi}\int_0^{2\pi}\cos^2\psi\,\mathrm d\psi = \tfrac12$ , confirming the universal half.*

Go deeper · Derivation C · 1 minThe staircase: why adding polarizers can transmit almost everything

Derivation CThe staircase: why adding polarizers can transmit almost everything

The goal is to explain — and quantify — the counterintuitive fact that a stack of $N$ polarizers whose axes fan evenly from $0°$ to $90°$ transmits more light as $N$ increases, approaching perfect transmission, even though a single $0°$ -to- $90°$ jump transmits nothing. The tool is the cascade product (4) with equal steps of $90°/N$ ; the key is that $\cos^2$ of a small angle is close to $1$ , and taking $N$ such nearly-unit factors gives a number that rises toward unity as $N$ grows. We evaluate the product and take its large- $N$ limit.

Step 1. Split the $90°$ rotation into $N$ equal steps of $\Delta = 90°/N = \pi/2N$ . With a polarized input, each stage applies Malus with relative angle $\Delta$ , so $T(N) = (\cos^2\Delta)^{N} = \cos^{2N}(\pi/2N)$ . Why equal steps: they make every stage identical, so the whole cascade collapses to a single factor raised to the $N$ -th power — the cleanest form to take a limit of.

Step 2. Expand for large $N$ : $\cos(\pi/2N) \approx 1 - \tfrac12(\pi/2N)^2$ , so $\ln T = 2N\ln\cos(\pi/2N) \approx 2N\cdot\big(-\tfrac12(\pi/2N)^2\big) = -\dfrac{\pi^2}{4N}$ . Why we take the logarithm: a product of many near-unit factors is best handled additively, and the leading small-angle term is all that survives as $N\to\infty$ .

Step 3. Exponentiate: $T(N) \approx e^{-\pi^2/4N} \to 1$ as $N\to\infty$ . Sample values: $T(1) = 0$ , $T(2) = \tfrac14$ , $T(6) = 0.660$ , $T(45) = 0.947$ . Why this is not energy from nothing: each sheet absorbs the discarded component; the many gentle re-projections simply waste far less than one violent $90°$ projection, which throws away everything. *The formula $T(N) = \cos^{2N}(\pi/2N)$ is monotone increasing in $N$ and matches the asymptotic approximation $e^{-\pi^2/4N}$ closely: $T(6)=0.6597$ , $T(45)=0.9466$ . This gradual-rotation survival is the classical shadow of the quantum Zeno effect.*

Limiting cases

Case	Result	Meaning
Parallel axes ( $\theta = 0$ )	$I = I_0$	Polarization aligned with the axis; full transmission for an ideal polarizer.
Half angle ( $\theta = 45°$ )	$I = I_0/2$	The pivot used in the three-polarizer trick; exactly half passes.
Crossed axes ( $\theta = 90°$ )	$I = 0$	Extinction — no field component along the axis. Real polarizers leak a residual $I_0 k_2$ .
Unpolarized input	$I = I_0/2$	Average of $\cos^2$ over all angles; independent of orientation; output fully polarized.
Staircase, $N \to \infty$	$T \to 1$	A $90°$ rotation in infinitesimal steps transmits everything — the Zeno-like limit.

Common trapMalus's law is the second-polarizer law: unpolarized light does not obey I₀cos²θ at the first polarizer

The single commonest error is feeding unpolarized light into Malus's law and reading off $I_0\cos^2\theta$ . That is wrong: unpolarized light has no definite $\theta$ , and a single polarizer always transmits $I_0/2$ — the average of $\cos^2$ , not $\cos^2$ of some angle. Malus's law with a definite $\theta$ applies to a second polarizer analyzing light already made linear by the first (then $\theta$ is the angle between the two axes). A second, subtler trap is dropping the square: the field picks up $\cos\theta$ , but the measured intensity picks up $\cos^2\theta$ . Track whether your quantity is a field amplitude or an intensity before you apply the law.

Worked examples

1 · The three-polarizer paradox, with numbers

Unpolarized light of intensity $I_0$ falls on two crossed polarizers (axes $0°$ and $90°$ ). The output is zero. Now insert a third polarizer at $45°$ between them. Find the final intensity, and confirm that removing the middle sheet restores darkness.

Step 1. First polarizer ( $0°$ ): unpolarized in $\Rightarrow$ $I_1 = I_0/2$ , now polarized horizontally. This is the universal half, not a $\cos^2$ .

Step 2. Middle polarizer ( $45°$ ): Malus with $\theta = 45°$ relative to the horizontal state, $I_2 = I_1\cos^2 45° = (I_0/2)(1/2) = I_0/4$ , now polarized at $45°$ — the projection has reset the state.

Step 3. Last polarizer ( $90°$ ): the light is at $45°$ from this axis (relative angle $45°$ , not $90°$ ), so $I_3 = I_2\cos^2 45° = (I_0/4)(1/2) = \boldsymbol{I_0/8}$ .

Step 4. Remove the middle sheet: the surviving beam ( $I_0/2$ at $0°$ ) meets the $90°$ axis directly, $I = (I_0/2)\cos^2 90° = \boldsymbol{0}$ . The $45°$ absorber is what makes the path bright — a purely wave-mechanical result.

2 · The staircase link budget

A polarization "rotator" is built as a fan of $N$ ideal sheet polarizers with axes stepped evenly from $0°$ to $90°$ , taking horizontally polarized light to vertical. How much light survives for $N = 6$ and $N = 45$ , and what is the $N \to \infty$ limit?

Step 1. Each of the $N$ stages steps by $\Delta = 90°/N$ and transmits $\cos^2\Delta$ ; the cascade transmits $T = (\cos^2\Delta)^{N}$ .

Step 2. For $N = 6$ : $\Delta = 15°$ , $\cos^2 15° = 0.9330$ , and $T = 0.9330^{6} = \boldsymbol{0.660}$ — already two-thirds through, from a start of zero.

Step 3. For $N = 45$ : $\Delta = 2°$ , $\cos^2 2° = 0.99878$ , and $T = 0.99878^{45} = \boldsymbol{0.947}$ , matching the asymptotic estimate $e^{-\pi^2/(4\cdot 45)} = 0.9466$ .

Step 4. As $N \to \infty$ , $T \to 1$ : an infinitely finely stepped rotator turns the polarization by a full $90°$ while transmitting all the light. Every added sheet raises the throughput — the opposite of naïve intuition about stacking filters.

Physics of the portrait

Malus's law is the point where polarization stopped being a curiosity of Iceland spar and became quantitative optics. Its author, an army engineer in Napoleon's corps, reportedly found it not at a bench but at a window: through a calcite crystal he watched the reflection of the setting Sun off the windows of the Luxembourg Palace in Paris wink out as he turned the crystal. Reflection, he realized, had done to sunlight what the crystal did — imposed a direction — and the transmitted brightness through his analyzer followed a clean $\cos^2$ of the angle. The law he wrote down is deceptively small and endlessly generative: it is the projection rule of a two-component vector, and it recurs unchanged from sunglasses to LCD pixels to the click statistics of single photons.

Origins · 1 minmalus's law and how this was discovered

malus's law

stated by Étienne-Louis Malus, 1775–1812 · the projection rule of polarization optics — one line of vector geometry with a long reach

Étienne-Louis Malus, a military engineer trained at the École Polytechnique who served in Napoleon's Egyptian campaign, turned to optics on his return and, in 1808, made the observation that names the law: sunlight reflected from the Luxembourg Palace windows, viewed through a rotated calcite crystal, extinguished at a definite angle. He recognized that reflection polarizes — he would coin the word "polarization" for the property in 1811 — and in 1809 he published the quantitative rule, $I = I_0\cos^2\theta$ , together with a theory of double refraction. He measured Brewster's reflection phenomenon before Brewster and won the Rumford Medal, but died at thirty-six in 1812. What survives is the projection law: the statement that a polarizer keeps the field's component along one axis, so intensity — quadratic in field — carries $\cos^2\theta$ . Modern optics reads it as the classical limit of a photon's transmission probability, which is why Malus's tabletop result seeds Bell tests and quantum key distribution two centuries later.

History

1669

— Bartholin reports double refraction in Iceland spar, the first hint of a hidden direction in light

1808

— Malus observes polarization by reflection in the Luxembourg Palace windows through a calcite crystal

1809

— Malus publishes the cos² intensity law

1811

— Malus names the property "polarization"; Arago finds optical rotation in quartz

1815

— Brewster fixes the reflection-polarization angle by the tangent law, sharpening Malus's discovery

1852

— Stokes casts partial and unpolarized light in the parameters that generalize the rotating-analyzer signal

Going further

Texts. Hecht, Optics, 5th ed., §8.2 (Malus's law and polarizers). Foot, Atomic Physics, appendix on polarization. Loudon, The Quantum Theory of Light, 3rd ed., photon-polarization chapter (the classical $\cos^2$ as a transmission probability). Saleh & Teich, Fundamentals of Photonics, 3rd ed., ch. 6 (polarization optics).

Provenance (base-corpus facts carried forward, re-verified). The field-projection derivation of $I = I_0\cos^2\theta$ and the $1/2$ rule for unpolarized light follow Hecht §8.2. Three numbers were recomputed from scratch: $\tfrac{1}{2\pi}\int_0^{2\pi}\cos^2\theta\,\mathrm d\theta = \tfrac12$ ; the three-polarizer product $\cos^2 45°\times\cos^2 45° = \tfrac14$ of the polarized intensity (so $I_0/8$ from unpolarized input); and the identity $\cos^2\theta = \tfrac12(1+\cos2\theta)$ giving the modulation form. The quantum photon-probability equivalence is per Loudon. History correction: the dry base dated the law to "1809" only; the observation (Luxembourg Palace windows, polarization by reflection) is 1808 and the published $\cos^2$ law is 1809 — both are given above, as the history gate requires.

Builds on: Polarization of Light · Linear Polarization · Degree of Polarization. Leads to: Polarizers and Extinction Ratio · Jones Calculus · Stokes Parameters and the Poincaré Sphere · the single-photon transmission probability behind Bell tests.

Why is it cos² and not just cos?

Because a polarizer acts on the electric-field amplitude, but what you measure is intensity. The axis passes the projected amplitude $E_0\cos\theta$ — one power of cosine, from the vector projection. Intensity is proportional to amplitude squared (it is the time-averaged Poynting flux, $\propto E^2$ ), so squaring the projected amplitude turns the single $\cos\theta$ into $\cos^2\theta$ . If you ever find yourself with a bare $\cos\theta$ in an intensity, you have skipped the squaring step.

Does Malus's law apply to unpolarized light?

Not at the first polarizer. Unpolarized light has no definite polarization angle, so there is no $\theta$ to substitute — a single ideal polarizer simply transmits half, $I_0/2$ , for any orientation, and hands you a fully polarized beam. Malus's law with a specific $\theta$ describes a second polarizer (the "analyzer") acting on light the first one already made linear; then $\theta$ is the angle between the two transmission axes. So the usual lab setup is: polarizer makes the light linear and defines " $0°$ ," analyzer at angle $\theta$ reads $I_0'\cos^2\theta$ where $I_0'$ is the intensity after the first sheet.

How can inserting a polarizer make a dark path brighter — isn't that adding light?

No energy is created. Two crossed polarizers are dark because the light after the first is polarized at $0°$ and the second axis is at $90°$ : zero projection. The inserted $45°$ sheet does two things — it absorbs some light (you lose half at that stage), but it also re-points the survivor to $45°$ . Now the final sheet sees light only $45°$ away, not $90°$ , so it can pass a component. You end with $I_0/8$ , all of it drawn from the beam and much of it absorbed along the way. The polarizer is a projector that resets the state, not a passive slit — that reset is the whole trick.

Is the quantum version of Malus's law really the same equation?

Yes, and that is its deepest feature. Send single photons, each polarized along $\hat{\mathbf e}$ , at an analyzer along $\hat{\mathbf a}$ with $\cos\theta = \hat{\mathbf e}\cdot\hat{\mathbf a}$ . Quantum mechanics gives each photon a transmission probability of exactly $\cos^2\theta$ ; over many photons the click rate reproduces the classical intensity law. The classical envelope and the single-photon probability rule are identical for ideal components. This is why polarization is the favourite classroom bridge to quantum measurement — and why polarization-encoded qubits, Bell tests, and BB84 key distribution all lean on the same $\cos^2$ .

What sets the residual light through "crossed" real polarizers?

An ideal polarizer gives exactly zero at $90°$ , but a real one has a nonzero minor transmittance $k_2$ : it lets a small fraction of the supposedly-blocked component through. The crossed transmission is then set by that leak rather than by $\cos^2 90° = 0$ . The quality metric is the extinction ratio $k_1/k_2$ (principal over minor transmittance) — a good sheet polarizer reaches $\sim 10^3$ – $10^4$ , a Glan–Thompson prism $10^5$ or more. The companion article on polarizers and extinction ratio derives the two-transmittance form $I = I_0(k_1\cos^2\theta + k_2\sin^2\theta)$ and what it does to a crossed pair.

Why does a rotating polarizer make brightness flicker twice per revolution?

Because the law depends on $\cos^2\theta$ , and $\cos^2$ has period $180°$ , not $360°$ — writing it as $\tfrac12(1+\cos2\theta)$ makes the doubled angle explicit. As you turn the analyzer through a full $360°$ , the transmission peaks at $0°$ and again at $180°$ , and dips at $90°$ and $270°$ : two bright maxima and two dark minima per revolution. Polarimeters exploit exactly this: they demodulate the detector signal at twice the rotation frequency to pull the polarization state out of noise.

Where does Malus's law break down?

It assumes an ideal, non-absorbing-along-axis linear polarizer and a fully polarized, monochromatic, collimated input. Real departures: finite extinction ratio (crossed transmission is $k_2$ , not $0$ ); wavelength-dependent transmittances; angle-of-incidence effects in sheet polarizers; and partial polarization, where you must average over the unpolarized fraction (giving a flattened $\cos^2$ that never reaches zero). For elliptical or circular input, "the angle $\theta$ " is not well defined — you need the full Jones or Stokes description, where Malus's law becomes the projection of a Jones vector or the action of a Mueller matrix. For fully polarized linear light through a good polarizer, though, $I_0\cos^2\theta$ is accurate to a fraction of a percent.