Physics · Polarization

Malus’s law

Malus gives the intensity projection of already linearly polarized input at an ideal analyzer: $\cos\theta$ for field and $\cos^2\theta$ for intensity.

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Formula

Calculate

I = I_{0} cos^{2} θ

I,I₀ in W/m²; θ in degrees or radians with matching calculator mode. Axes are equivalent modulo 180°.

Variables & units

Variables & units
Symbol	Meaning	Unit
$I$	Output intensity after the chosen analyzer	W/m²
$I₀$	Already linearly polarized input intensity immediately before this analyzer	W/m²
$θ$	Angle between linear input direction and analyzer axis	rad or °

LaTeX

I=I_0\cos^2\theta

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Four intensity bars on one linear scale: 120, 60, 45 and 11.25. Three filter-axis glyphs show 0°, 30° and 90°; the first halves unpolarized input, then adjacent axis differences apply. — 120 W/m² of unpolarized input becomes 60, 45 and 11.25 W/m² after ideal filters at 0°, 30° and 90°. Adjacent axis differences are 30° and 60°. Beam direction remains unchanged.

Malus's law calculator

Find the intensity behind an ideal polarising filter from the input intensity and the angle, or the angle from both intensities.

Calculation model

Find

Linearly polarized input intensity I₀

Intensity of the already linearly polarised light in front of the analyser, e.g. 100 W/m².

Angle θ between the axes

Angle between the light’s polarisation direction and the transmission axis, 0° to 180°, e.g. 60°.

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

Working · rearrange for I, substitute, evaluate

1. Rearrange for I
$I=I_0\cos^2\theta$
2. Substitute values (in SI units)
- θ = 60 ° = 1.0472 rad
$I=100\cdot\cos^2\left(1.0471975512\right)$
3. Result
I = 25 W/m²

Result: Output intensity I

I = 25 W/m²

With I₀ = 100 W/m² and θ = 60 °, I = 25 W/m².

Quick tips

Values worth remembering for cos²θ: at 0° everything passes, at 30° three quarters, at 45° half, at 60° a quarter, at 90° nothing.
Unpolarised light always loses half at the first filter, whatever its orientation. The angle only matters from the second filter on.

Assumptions and limits

Already linearly polarized input immediately before one ideal analyzer. Axis angle 0–180°; inverse angle returns only the principal 0–90° value, not rotation direction. Same detection geometry; no real losses.

Derivation & background

The transmission axis projects electric-field amplitude; squaring gives $I=I_0\cos^2\theta$ . Afterward the transmitted field follows that axis. An ideal first stage halves unpolarized light; subsequent stages use neighboring axis differences. Ideal crossed filters give zero, real filters have leakage. The relation also applies to fully captured power with unchanged beam cross-section, not automatically to different detector apertures.

Exam blueprint

Validity range

An ideal linear analyzer with stable already linearly polarized input and unchanged collection. θ is relative axis angle. Unpolarized input requires first-stage half transmission; circular, elliptical or partially polarized input needs the appropriate state description. Real losses and leakage add further factors.

Derivation steps

Project a fixed linear field direction onto the analyzer axis, then square amplitude.

1I₀ is already linearly polarized input immediately before the analyzer.
2Define θ as angle between linear field direction and transmission axis.
3Projection onto the axis gives transmitted amplitude $E_\parallel=E_0\cos\theta$ .
4For the same medium and collection geometry, intensity is proportional to squared amplitude.
5Therefore $\frac{I}{I_0}=\cos^2\theta$ and $I=I_0\cos^2\theta$ ; negative projection never means negative intensity.
6A first ideal filter averages unpolarized input to half intensity. After each stage the new linear direction follows that filter’s axis.

Rearrangements

Angle from intensity ratio

\theta=\arccos\sqrt{\frac{I}{I_0}}

Require $I_0>0$ and $0\le I\le I_0$ . Restricting angle to 0°–90° selects the nonnegative cosine branch; intensity alone does not determine rotation sign.

Task variant

Already linearly polarized light reaches an ideal linear analyzer with 45° between polarization direction and analyzer axis. What intensity fraction remains?

Approach: apply Malus to intensity immediately before this stage.
Thus $\frac I{I_0}=\cos^245^\circ=\frac12$ .
Half is transmitted and the output is polarized along the analyzer axis.

Common mistakes

Applying $\cos\theta$ instead of $\cos^2\theta$ to intensity.

$\cos\theta$ projects field amplitude; intensity is proportional to its square.

Exam context

Analyzer orientations, polarizing filters and optical-wave intensity measurements.

Formula cluster

Polarization and light intensity

Field projection determines transmitted intensity.

d·sin(α_k) = k·λ

Double-Slit Interference

n₁·sin(α₁) = n₂·sin(α₂)

Snell's Law of Refraction

ΦB = BA cos θ

Magnetic flux through a surface

Worked example

At 60°,25% of already linear input remains.72 W/m² at 37° gives 45.9229 W/m². Unpolarized 120 W/m² through 0°→30°→90° gives 60,45 and 11.25 W/m² after the stages.

Applications

Analyze analyzers, filter chains, polarization measurements and angular control; LCDs and real filters additionally need their actual polarization transformation and losses.

How do you calculate with Malus’s law?

Here is how to work through a typical Malus’s law (I = I₀ cos²θ) task step by step:

1
Task
Already linearly polarized light reaches an ideal linear analyzer with 45° between polarization direction and analyzer axis. What intensity fraction remains?
Solution path
1. Approach: apply Malus to intensity immediately before this stage.
2. Thus $\frac I{I_0}=\cos^245^\circ=\frac12$ .
3. Half is transmitted and the output is polarized along the analyzer axis.

Understand and apply the equation

Two crossed polarizers can almost extinguish light. Inserting a suitably rotated third filter between them makes the output brighter again. This creates no energy: each filter projects the remaining field onto a new axis. Malus’s law gives the resulting intensity fraction. Identify the input state first, then follow every stage separately.

Before calculating

Polarization describes the direction and time evolution of the electric field transverse to light propagation. A linearly polarized field oscillates along a fixed axis. A ray sketch alone does not show that axis. A polarizer selects one field component; using it to investigate an existing polarization makes it an analyzer.[1][2]

For the ideal basic calculation the analyzer’s input is already linearly polarized, and I₀ refers to that particular stage. Unpolarized light before the first ideal filter is first halved on average. You need cosine and the distinction between field amplitude and its square. Assume unchanged beam collection and no additional losses.[1]

I₀ belongs to the analyzer’s input stage

I=I_0\cos^2\theta

θ is the angle between the existing linear polarization direction and the analyzer transmission axis.[1]

I₀ is intensity immediately before the chosen analyzer, not automatically before the entire filter chain. Ideal parallel axes transmit I₀, crossed axes transmit zero. Axis orientation is defined modulo 180°:0° and 180° describe the same transmission direction. The intensity curve therefore repeats every 180°.[1]

Intensity is power per cross-sectional area in W/m². The same relation holds for total power if the unchanged beam cross-section is fully captured. If aperture or beam shape changes between stages, detector power ratios cannot be interpreted as $\frac{I}{I_0}$ without checking geometry.[1]

Intensity repeats every 180°

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Angle refers to existing linear input polarization. Do not apply the curve to unpolarized light without the initial averaging step.[1]

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From field projection to intensity

Let the analyzer axis be unit vector a. For a linearly polarized field with fixed direction, projection gives transmitted amplitude $E_\parallel=E_0\cos\theta$ . The ideal analyzer removes the perpendicular component. The outgoing field is linear along a, so a subsequent filter must compare its axis with a, not with the original direction before the chain.[2]

\begin{aligned}E_\parallel&=E_0\cos\theta\\I_0&=C|E_0|^2\\I&=C|E_\parallel|^2=C|E_0|^2\cos^2\theta\\\frac I{I_0}&=\cos^2\theta\end{aligned}

C is the same intensity factor before and after the ideal filter in the same medium. Squaring distinguishes intensity from amplitude.[2]

A negative projection represents field sign or phase, not negative intensity. Squaring makes intensity nonnegative. At 90° the projection vanishes, so a later ideal filter cannot restore energy already removed. An absorptive polarizer transfers rejected energy to its surroundings; other designs send it into a different output channel.[1]

The Brewster angle for polarization describes the incidence angle at which light reflected from a suitable interface is completely linearly polarized. Apply the analyzer intensity projection only once the linear input has been properly specified.[1]

Analyze a linearly polarized input

Analyzer at 37°

Given: I₀=72.0 W/m²; linear input; axis difference 37.0°.
Find: Find output intensity and rejected fraction.

Approach: An ideal analyser projects the electric-field amplitude onto its axis; intensity follows the square of that projection.
$\begin{aligned}E_{\parallel}&=E_0\cos\theta\\I&=I_0\cos^2\theta\end{aligned}$
Calculate and square the amplitude projection.
$\begin{aligned}&\cos37^\circ=0.798636\\&\cos^237^\circ=0.637819\end{aligned}$
Apply the intensity factor.
$I=72.0\cos^237^\circ=45.9229\,\mathrm{W\,m^{-2}}$

Result: I≈45.92 W/m²; 36.22% is removed from the transmitted channel.

Check: I lies between 0 and 72 W/m². Using cos 37° instead of its square would give 57.50 W/m², confusing amplitude and intensity.

A first ideal filter halves unpolarized light

Statistically unpolarized light has no preferred transverse field direction. An ideal linear filter transmits half of mean intensity. This is an average over polarization directions, not Malus with one fixed angle describing the complete input. The output then has the filter’s linear direction and subsequent stages follow Malus.[1]

\langle\cos^2\phi\rangle=\frac1{2\pi}\int_0^{2\pi}\cos^2\phi\,d\phi=\frac12

Averaging gives the ideal half transmission. Partially polarized or elliptical input needs additional information.[1]

Circularly polarized light also gives half intensity through an ideal linear analyzer, but is not unpolarized: its components have a definite phase relation. One intensity value after a linear filter cannot distinguish these states. Complete polarization analysis needs additional optical settings, such as a retardation plate.[2]

Insert a third filter between crossed axes

Three ideal filters at 0°,30° and 90°

Given: Unpolarized Iin=120 W/m²; order 0°→30°→90°; no additional losses.
Find: Find intensity after each stage.

Approach: The first ideal filter halves unpolarized input intensity. Subsequent stages use Malus with each adjacent axis difference.
$\begin{aligned}I_1&=\frac{I_{\mathrm{in}}}{2}\\I_2&=I_1\cos^2\theta_{12}\\I_3&=I_2\cos^2\theta_{23}\end{aligned}$
The first stage halves intensity and gives 0° linear polarization.
$I_1=\frac{120}{2}=60\,\mathrm{W\,m^{-2}}$
Between 0° and 30°, the axis difference is 30°.
$I_2=60\cos^230^\circ=45\,\mathrm{W\,m^{-2}}$
The third stage receives 30° polarization; its relative angle is 60°.
$I_3=45\cos^260^\circ=11.25\,\mathrm{W\,m^{-2}}$

Result: After the stages: 60, 45 and 11.25 W/m². Without the middle filter, the outer ideal filters are crossed and output is zero.

Check: Output energy is below every preceding stage. The added filter creates no energy; it changes the surviving field’s direction before the final projection. Applying 120 cos²90° directly would misdescribe this chain.

Infer angles and uncertainty from intensity ratios

\begin{aligned}&\theta=\arccos\sqrt{\frac{I}{I_0}}\quad(0\le\theta\le90^\circ)\\&\frac{dI}{d\theta}=-I_0\sin2\theta\end{aligned}

Angle inference requires a chosen interval; other equivalent or mirrored axis orientations produce the same intensity. Differentiate with θ in radians.[1]

For $\frac{I}{I_0}=0.36$ , choosing 0° to 90° gives 53.1301°. Intensity alone does not identify clockwise versus counterclockwise rotation. With I₀=60 W/m² and 45°, I=30 W/m². A±1° angle shift gives 31.047 or 28.953 W/m², approximately±3.49% around this operating point.[1]

Near crossed axes, ideal absolute output is small but relative error becomes sensitive. Scattering or finite extinction can dominate the dark signal. Measure background and normalize against parallel transmission before inferring a precise angle from a near-zero detector reading.[1]

With $I_0=0$ , the ideal model also gives $I=0$ ; no angle can be inferred. Likewise, a measurement $I=0$ at exactly crossed axes does not determine input intensity: every nonnegative input is fully blocked.[1]

Explore further: real polarizers

I=I_0\bigl(T_\parallel\cos^2\theta+T_\perp\sin^2\theta\bigr)

A simple real model for linear input; principal-axis transmissions refer to the specified spectrum and geometry.[1]

For Tparallel=0.90 and $T_\perp=0.002$ , crossed output is 0.002 I₀ rather than zero and parallel output 0.90 I₀ rather than I₀. Their ratio 450 is the extinction ratio as defined here. Product specifications may use reciprocal conventions or different measurement conditions, so read the numerical definition explicitly.[1]

The ideal relation assumes fixed linear input polarization, stable intensity, an appropriate linear analyzer and unchanged collection. Depolarization, elliptical states, angle-sensitive coatings or a saturated detector require more than a cosine-squared factor. Malus describes directional analysis, not every loss in a complete optical system.[2]

Leakage through a nonideal filter

Given: Linear input I₀=80.0 W/m²; $T_\parallel=0.900$ ; $T_\perp=0.00200$ ; unchanged collection.
Find: Find intensity at 0°, 45° and 90° and the extinction ratio defined here.

Approach: weight both orthogonal input components by their own power transmissions.
$\begin{aligned}I(\theta)&=I_0\left(T_\parallel\cos^2\theta+T_\perp\sin^2\theta\right)\\\mathcal E&=\frac{T_\parallel}{T_\perp}\end{aligned}$
Substitute the three angles separately. At 45° both squared projections are equal.
$\begin{aligned}I(0^\circ)&=80\cdot0.900=72.0\,\mathrm{W\,m^{-2}}\\I(45^\circ)&=80\frac{0.900+0.00200}{2}=36.08\,\mathrm{W\,m^{-2}}\\I(90^\circ)&=80\cdot0.00200=0.160\,\mathrm{W\,m^{-2}}\\\mathcal E&=\frac{0.900}{0.00200}=450\end{aligned}$

Result: Crossed orientation transmits 0.160 W/m² here. Parallel-to-crossed transmission ratio is 450; the filter is not a perfectly blocking polarizer.

Check: All outputs are below 80 W/m²; the 45° value is exactly halfway between parallel and crossed output. Setting $T_\parallel=1$ and $T_\perp=0$ recovers Malus’s law.

The wave relation for propagation speed, wavelength and frequency connects $c=\lambda f$ . That relation alone does not determine filter transmission: the polarization state and field-amplitude projection are still needed.[3][1]

Five calculations keeping filter stages distinct

Calculate · Foundation

Linear I₀=90 W/m²; analyzer 32°. Calculate I.

Show hint

Intensity contains the squared field factor.

Learning goal: Calculate transmitted intensity for specified linear polarization.

Show solution and calculation

Approach: the input is already linearly polarized.
Apply Malus directly and square the field projection factor.
This gives I≈64.7267 W/m².

Check: intensity ratio is about 0.719186, between zero and one. A passive ideal filter yields neither negative intensity nor more than 90 W/m².

I=90\cos^232^\circ\,\mathrm{W\,m^{-2}}

Transfer · Application

Unpolarized 80 W/m² reaches two ideal filters 60° apart. Find output.

Show hint

The first stage halves; the second uses Malus.

Learning goal: Halve unpolarized input first, then use relative axis orientation.

Show solution and calculation

Approach: the first stage halves unpolarized intensity on average.
The second then receives the linear polarization selected by the first.
Intensities are 40 W/m² and then 10 W/m².

Check: directly using $80\cos^260^\circ=20$ would omit the first halving. Each correctly treated stage reduces transmitted intensity here.

I_2=\frac{80}2\cos^260^\circ=10\,\mathrm{W\,m^{-2}}

Calculate · Application

Reduce linear input to 36%. Which ideal analyzer angle between 0° and 90° is required?

Show hint

Take the square root of the intensity ratio first.

Learning goal: Find an analyser angle from a target intensity fraction.

Show solution and calculation

Approach: the chosen interval 0° to 90° fixes the nonnegative cosine branch.
Take the square root of the intensity ratio before inverse cosine.
This gives θ≈53.1301°.

Check: $\cos^2\theta=0.36$ . The negative angle would give the same intensity but is outside the chosen interval.

\begin{aligned}\cos\theta&=\sqrt{0.36}=0.60\\\theta&=\arccos0.60\approx53.1301^\circ\end{aligned}

Interpret · Application

Light is already linearly polarized at 0° with intensity 64 W/m². An ideal analyser at 90° transmits nothing. An ideal polarizer at 45° is now inserted before it. Calculate intensity after both elements and explain why an extra element permits more output light here. Absorption beyond ideal polarization selection is excluded.

Show hint

Apply Malus’s law sequentially. After the first element polarization is at 45°, so the second relative angle is also 45°.

Learning goal: Explain the effect of inserting another polarizer between crossed axes.

Show solution and calculation

Approach: treat each polarizer relative to the polarization of its own input.
After 45°, intensity is 32 W/m²; after the 90° analyser it is 16 W/m².
The intermediate polarizer changes the transmitted light’s polarization, so the analyser no longer receives a perpendicular input.

Check: output remains below input intensity. Increased transmission does not create energy.

\begin{aligned}I_1&=I_0\cos^245^\circ=32\,\mathrm{W\,m^{-2}}\\I_2&=I_1\cos^2(90^\circ-45^\circ)=16\,\mathrm{W\,m^{-2}}\\I_{\mathrm{without}}&=I_0\cos^290^\circ=0\end{aligned}

[1][2]

Error analysis · Foundation

Unpolarized light of intensity 50 W/m² reaches a single ideal linear polarizer. Someone chooses a laboratory axis and uses Malus’s law to predict zero transmission at rotation angle 90°. Find actual average output intensity and explain which premise of that single-angle calculation is missing.

Show hint

Unpolarized light has no fixed input polarization axis. Averaging cos² over uniformly distributed directions gives one half.

Learning goal: Avoid assigning unpolarized light an arbitrary direction when applying Malus’s law.

Show solution and calculation

Approach: average the unpolarized input over polarization directions.
A single ideal polarizer transmits 25 W/m², independently of absolute laboratory orientation.
An angle to a laboratory axis is not an angle to an existing input polarization axis.

Check: only a second polarizer could use its angle relative to the first to modify the now linearly polarized output through Malus’s law.

\begin{aligned}\left\langle\cos^2\theta\right\rangle&=\frac12\\I_{\mathrm{out}}&=\frac12I_{\mathrm{unpol}}=25\,\mathrm{W\,m^{-2}}\end{aligned}

[1]

What to take away

Before calculating, identify the intensity immediately before this filter and its polarization. Linear input uses the squared field projection. An ideal first filter halves unpolarized intensity on average; later stages use adjacent axis differences. An extra filter can change direction but cannot create energy. Real residual light needs a leakage or background model.[1][2]

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

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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Change history

October 2, 2026Linked polarization and wave context, added inverse zero cases and placed one decimal value wholly inside its math delimiters; preserved examples and intensity curve.
October 1, 2026Explained input stage and field projection; added a periodicity plot and real-transmission example; clarified filter chains and exercise checks.
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. Malus’s law. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/malus-gesetz-polarisation

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Frequently asked questions about Malus’s law

What does Malus’s law describe physically?+

Malus describes projection of already linearly polarized light onto an ideal linear analyzer axis. Field amplitude contains $\cos\theta$ ; intensity contains $\cos^2\theta$ . θ is the difference between input direction and analyzer axis, with I₀ immediately before that stage. At 60°,25% of polarized input remains. Ideal parallel axes transmit all, crossed axes none. After the analyzer, surviving light has its polarization direction. Unpolarized input has no single fixed direction and a first ideal stage halves mean intensity. Include that halving before a subsequent Malus calculation whenever the task specifies intensity before the complete filter chain.

How can I check a calculation involving Malus’s law?+

First identify input polarization and the stage referenced by I₀. Linear 72 W/m² input at 37° gives 45.9229 W/m². The ratio 0.637819 lies between zero and one; passive ideal output cannot exceed input. For a chain, use differences between neighboring axes. Unpolarized 120 W/m² through 0°,30°,90° gives 60, then 45, then 11.25 W/m². The third filter receives 30° polarization, making its relative angle 60°. Check those three intermediate values and decreasing energy. Applying one 90° factor to the entire three-filter chain would ignore the middle filter’s change of polarization direction. No energy is created by inserting that additional filter.

What assumptions and limits apply to Malus’s law?+

The ideal relation requires stable known linear input, an ideal linear analyzer and unchanged beam collection. Real filters lose some wanted polarization and leak unwanted polarization when crossed. A simple extension uses different principal-axis transmissions. Partially polarized, circular or elliptical light needs a suitable state description rather than one fixed angle. Circular and unpolarized light both transmit half through an ideal linear filter but are different states. Detector saturation, stray light or changing aperture can also distort a measured intensity curve. These deviations should not be hidden by arbitrary angular corrections to an otherwise ideal cosine-squared model.

How do I check units and scale for Malus’s law?+

I and I₀ share an intensity unit, such as W/m², making their ratio dimensionless. A fully captured unchanged beam cross-section permits the same calculation with power. Use degrees or radians consistently with calculator mode. Axes are equivalent modulo 180° and intensity repeats accordingly. Differentiation with respect to angle, however, uses radians. For 60 W/m² at 45°, output is 30 W/m²;±1° gives 31.047 or 28.953 W/m², plausibly close to half input. Near crossed axes, relative error becomes sensitive and leakage may dominate, so percent-based interpretations require careful background and parallel-transmission normalization there.

When should I combine Malus’s law with another relation?+

In a filter chain, combine Malus with first-stage half transmission for ideal unpolarized input. Each subsequent stage sets the new linear direction. For reflected light at Brewster incidence, first calculate Fresnel power reflection and resulting s polarization; Malus alone does not describe the surface. Real analyzers add principal-axis transmissions, and detector measurements may add background. A known intensity ratio can infer an angle within a chosen interval:36% gives 53.1301° between 0° and 90°. Intensity does not determine rotation sign. Distinguishing circular from unpolarized light requires additional optical settings beyond a single linear analyzer measurement.

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Related formulas

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

Brewster angle for polarization

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d\sin\theta_m=m\lambda

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\sin\theta_c=\frac{n_2}{n_1}\quad(n_1>n_2)

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\theta_{\min}\approx1.22\frac{\lambda}{D}

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Malus’s law

Malus gives the intensity projection of already linearly polarized input at an ideal analyzer:

\cos\theta

for field and

\cos^2\theta

for intensity.

AdvancedUpper secondaryUniversity/collegeBachelor foundations

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Symbol

Meaning

Unit

I

Output intensity after the chosen analyzer

W/m²

I₀

Already linearly polarized input intensity immediately before this analyzer

W/m²

θ

Angle between linear input direction and analyzer axis

rad or °

Derivation & background

Exam blueprint

Validity range

Derivation steps

Project a fixed linear field direction onto the analyzer axis, then square amplitude.

1I₀ is already linearly polarized input immediately before the analyzer.
2Define θ as angle between linear field direction and transmission axis.
3Projection onto the axis gives transmitted amplitude $E_\parallel=E_0\cos\theta$ .
4For the same medium and collection geometry, intensity is proportional to squared amplitude.
5Therefore $\frac{I}{I_0}=\cos^2\theta$ and $I=I_0\cos^2\theta$ ; negative projection never means negative intensity.
6A first ideal filter averages unpolarized input to half intensity. After each stage the new linear direction follows that filter’s axis.

How do you calculate with Malus’s law?

Here is how to work through a typical Malus’s law (I = I₀ cos²θ) task step by step:

Task

Already linearly polarized light reaches an ideal linear analyzer with 45° between polarization direction and analyzer axis. What intensity fraction remains?

Solution path

Approach: apply Malus to intensity immediately before this stage.
Thus $\frac I{I_0}=\cos^245^\circ=\frac12$ .
Half is transmitted and the output is polarized along the analyzer axis.

Understand and apply the equation

Before calculating

I₀ belongs to the analyzer’s input stage

I=I_0\cos^2\theta

θ is the angle between the existing linear polarization direction and the analyzer transmission axis.[1]

Intensity repeats every 180°

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Angle refers to existing linear input polarization. Do not apply the curve to unpolarized light without the initial averaging step.[1]

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From field projection to intensity

\begin{aligned}E_\parallel&=E_0\cos\theta\\I_0&=C|E_0|^2\\I&=C|E_\parallel|^2=C|E_0|^2\cos^2\theta\\\frac I{I_0}&=\cos^2\theta\end{aligned}

C is the same intensity factor before and after the ideal filter in the same medium. Squaring distinguishes intensity from amplitude.[2]

Analyze a linearly polarized input

Analyzer at 37°

Given: I₀=72.0 W/m²; linear input; axis difference 37.0°.
Find: Find output intensity and rejected fraction.

Approach: An ideal analyser projects the electric-field amplitude onto its axis; intensity follows the square of that projection.
$\begin{aligned}E_{\parallel}&=E_0\cos\theta\\I&=I_0\cos^2\theta\end{aligned}$
Calculate and square the amplitude projection.
$\begin{aligned}&\cos37^\circ=0.798636\\&\cos^237^\circ=0.637819\end{aligned}$
Apply the intensity factor.
$I=72.0\cos^237^\circ=45.9229\,\mathrm{W\,m^{-2}}$

Result: I≈45.92 W/m²; 36.22% is removed from the transmitted channel.

Check: I lies between 0 and 72 W/m². Using cos 37° instead of its square would give 57.50 W/m², confusing amplitude and intensity.

A first ideal filter halves unpolarized light

\langle\cos^2\phi\rangle=\frac1{2\pi}\int_0^{2\pi}\cos^2\phi\,d\phi=\frac12

Averaging gives the ideal half transmission. Partially polarized or elliptical input needs additional information.[1]

Insert a third filter between crossed axes

Three ideal filters at 0°,30° and 90°

Given: Unpolarized Iin=120 W/m²; order 0°→30°→90°; no additional losses.
Find: Find intensity after each stage.

Approach: The first ideal filter halves unpolarized input intensity. Subsequent stages use Malus with each adjacent axis difference.
$\begin{aligned}I_1&=\frac{I_{\mathrm{in}}}{2}\\I_2&=I_1\cos^2\theta_{12}\\I_3&=I_2\cos^2\theta_{23}\end{aligned}$
The first stage halves intensity and gives 0° linear polarization.
$I_1=\frac{120}{2}=60\,\mathrm{W\,m^{-2}}$
Between 0° and 30°, the axis difference is 30°.
$I_2=60\cos^230^\circ=45\,\mathrm{W\,m^{-2}}$
The third stage receives 30° polarization; its relative angle is 60°.
$I_3=45\cos^260^\circ=11.25\,\mathrm{W\,m^{-2}}$

Result: After the stages: 60, 45 and 11.25 W/m². Without the middle filter, the outer ideal filters are crossed and output is zero.

Infer angles and uncertainty from intensity ratios

\begin{aligned}&\theta=\arccos\sqrt{\frac{I}{I_0}}\quad(0\le\theta\le90^\circ)\\&\frac{dI}{d\theta}=-I_0\sin2\theta\end{aligned}

Angle inference requires a chosen interval; other equivalent or mirrored axis orientations produce the same intensity. Differentiate with θ in radians.[1]

Explore further: real polarizers

I=I_0\bigl(T_\parallel\cos^2\theta+T_\perp\sin^2\theta\bigr)

A simple real model for linear input; principal-axis transmissions refer to the specified spectrum and geometry.[1]

Leakage through a nonideal filter

Given: Linear input I₀=80.0 W/m²; $T_\parallel=0.900$ ; $T_\perp=0.00200$ ; unchanged collection.
Find: Find intensity at 0°, 45° and 90° and the extinction ratio defined here.

Approach: weight both orthogonal input components by their own power transmissions.
$\begin{aligned}I(\theta)&=I_0\left(T_\parallel\cos^2\theta+T_\perp\sin^2\theta\right)\\\mathcal E&=\frac{T_\parallel}{T_\perp}\end{aligned}$
Substitute the three angles separately. At 45° both squared projections are equal.
$\begin{aligned}I(0^\circ)&=80\cdot0.900=72.0\,\mathrm{W\,m^{-2}}\\I(45^\circ)&=80\frac{0.900+0.00200}{2}=36.08\,\mathrm{W\,m^{-2}}\\I(90^\circ)&=80\cdot0.00200=0.160\,\mathrm{W\,m^{-2}}\\\mathcal E&=\frac{0.900}{0.00200}=450\end{aligned}$

Result: Crossed orientation transmits 0.160 W/m² here. Parallel-to-crossed transmission ratio is 450; the filter is not a perfectly blocking polarizer.

Check: All outputs are below 80 W/m²; the 45° value is exactly halfway between parallel and crossed output. Setting $T_\parallel=1$ and $T_\perp=0$ recovers Malus’s law.

Five calculations keeping filter stages distinct

Calculate · Foundation

Linear I₀=90 W/m²; analyzer 32°. Calculate I.

Show hint

Intensity contains the squared field factor.

Learning goal: Calculate transmitted intensity for specified linear polarization.

Show solution and calculation

Approach: the input is already linearly polarized.
Apply Malus directly and square the field projection factor.
This gives I≈64.7267 W/m².

Check: intensity ratio is about 0.719186, between zero and one. A passive ideal filter yields neither negative intensity nor more than 90 W/m².

I=90\cos^232^\circ\,\mathrm{W\,m^{-2}}

Transfer · Application

Unpolarized 80 W/m² reaches two ideal filters 60° apart. Find output.

Show hint

The first stage halves; the second uses Malus.

Learning goal: Halve unpolarized input first, then use relative axis orientation.

Show solution and calculation

Approach: the first stage halves unpolarized intensity on average.
The second then receives the linear polarization selected by the first.
Intensities are 40 W/m² and then 10 W/m².

Check: directly using $80\cos^260^\circ=20$ would omit the first halving. Each correctly treated stage reduces transmitted intensity here.

I_2=\frac{80}2\cos^260^\circ=10\,\mathrm{W\,m^{-2}}

Calculate · Application

Reduce linear input to 36%. Which ideal analyzer angle between 0° and 90° is required?

Show hint

Take the square root of the intensity ratio first.

Learning goal: Find an analyser angle from a target intensity fraction.

Show solution and calculation

Approach: the chosen interval 0° to 90° fixes the nonnegative cosine branch.
Take the square root of the intensity ratio before inverse cosine.
This gives θ≈53.1301°.

Check: $\cos^2\theta=0.36$ . The negative angle would give the same intensity but is outside the chosen interval.

\begin{aligned}\cos\theta&=\sqrt{0.36}=0.60\\\theta&=\arccos0.60\approx53.1301^\circ\end{aligned}

Interpret · Application

Light is already linearly polarized at 0° with intensity 64 W/m². An ideal analyser at 90° transmits nothing. An ideal polarizer at 45° is now inserted before it. Calculate intensity after both elements and explain why an extra element permits more output light here. Absorption beyond ideal polarization selection is excluded.

Show hint

Apply Malus’s law sequentially. After the first element polarization is at 45°, so the second relative angle is also 45°.

Learning goal: Explain the effect of inserting another polarizer between crossed axes.

Show solution and calculation

Approach: treat each polarizer relative to the polarization of its own input.
After 45°, intensity is 32 W/m²; after the 90° analyser it is 16 W/m².
The intermediate polarizer changes the transmitted light’s polarization, so the analyser no longer receives a perpendicular input.

Check: output remains below input intensity. Increased transmission does not create energy.

\begin{aligned}I_1&=I_0\cos^245^\circ=32\,\mathrm{W\,m^{-2}}\\I_2&=I_1\cos^2(90^\circ-45^\circ)=16\,\mathrm{W\,m^{-2}}\\I_{\mathrm{without}}&=I_0\cos^290^\circ=0\end{aligned}

[1][2]

Error analysis · Foundation

Unpolarized light of intensity 50 W/m² reaches a single ideal linear polarizer. Someone chooses a laboratory axis and uses Malus’s law to predict zero transmission at rotation angle 90°. Find actual average output intensity and explain which premise of that single-angle calculation is missing.

Show hint

Unpolarized light has no fixed input polarization axis. Averaging cos² over uniformly distributed directions gives one half.

Learning goal: Avoid assigning unpolarized light an arbitrary direction when applying Malus’s law.

Show solution and calculation

Approach: average the unpolarized input over polarization directions.
A single ideal polarizer transmits 25 W/m², independently of absolute laboratory orientation.
An angle to a laboratory axis is not an angle to an existing input polarization axis.

Check: only a second polarizer could use its angle relative to the first to modify the now linearly polarized output through Malus’s law.

\begin{aligned}\left\langle\cos^2\theta\right\rangle&=\frac12\\I_{\mathrm{out}}&=\frac12I_{\mathrm{unpol}}=25\,\mathrm{W\,m^{-2}}\end{aligned}

[1]

What to take away

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

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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, 2026Linked polarization and wave context, added inverse zero cases and placed one decimal value wholly inside its math delimiters; preserved examples and intensity curve.
October 1, 2026Explained input stage and field projection; added a periodicity plot and real-transmission example; clarified filter chains and exercise checks.
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. Malus’s law. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/malus-gesetz-polarisation

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Frequently asked questions about Malus’s law

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Malus’s law

Variables & units

Malus's law calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Polarization and light intensity

Worked example

Applications

How do you calculate with Malus’s law?

Before calculating

I0 belongs to the analyzer’s input stage

Intensity repeats every 180°

From field projection to intensity

Analyze a linearly polarized input

Analyzer at 37°

A first ideal filter halves unpolarized light

Insert a third filter between crossed axes

Three ideal filters at 0°,30° and 90°

Infer angles and uncertainty from intensity ratios

Explore further: real polarizers

Leakage through a nonideal filter

Five calculations keeping filter stages distinct

Linear I0=90 W/m2; analyzer 32°. Calculate I.

Unpolarized 80 W/m2 reaches two ideal filters 60° apart. Find output.

Reduce linear input to 36%. Which ideal analyzer angle between 0° and 90° is required?

Unpolarized light of intensity 50 W/m2 reaches a single ideal linear polarizer. Someone chooses a laboratory axis and uses Malus’s law to predict zero transmission at rotation angle 90°. Find actual average output intensity and explain which premise of that single-angle calculation is missing.

What to take away

Sources for this explanation

Frequently asked questions about Malus’s law

Subject references and further reading

Related formulas

More formulas: Optics

Retain Malus’s law for exams

Malus’s law

Variables & units

Malus's law calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Polarization and light intensity

Worked example

Applications

How do you calculate with Malus’s law?

Before calculating

I0 belongs to the analyzer’s input stage

Intensity repeats every 180°

From field projection to intensity

Analyze a linearly polarized input

Analyzer at 37°

A first ideal filter halves unpolarized light

Insert a third filter between crossed axes

Three ideal filters at 0°,30° and 90°

Infer angles and uncertainty from intensity ratios

Explore further: real polarizers

Leakage through a nonideal filter

Five calculations keeping filter stages distinct

Linear I0=90 W/m2; analyzer 32°. Calculate I.

Unpolarized 80 W/m2 reaches two ideal filters 60° apart. Find output.

Reduce linear input to 36%. Which ideal analyzer angle between 0° and 90° is required?

Unpolarized light of intensity 50 W/m2 reaches a single ideal linear polarizer. Someone chooses a laboratory axis and uses Malus’s law to predict zero transmission at rotation angle 90°. Find actual average output intensity and explain which premise of that single-angle calculation is missing.

What to take away

Sources for this explanation

Frequently asked questions about Malus’s law

Subject references and further reading

Related formulas

More formulas: Optics

Retain Malus’s law for exams

I₀ belongs to the analyzer’s input stage

Linear I₀=90 W/m²; analyzer 32°. Calculate I.

Unpolarized 80 W/m² reaches two ideal filters 60° apart. Find output.

Unpolarized light of intensity 50 W/m² reaches a single ideal linear polarizer. Someone chooses a laboratory axis and uses Malus’s law to predict zero transmission at rotation angle 90°. Find actual average output intensity and explain which premise of that single-angle calculation is missing.

I₀ belongs to the analyzer’s input stage

Linear I₀=90 W/m²; analyzer 32°. Calculate I.

Unpolarized 80 W/m² reaches two ideal filters 60° apart. Find output.

Unpolarized light of intensity 50 W/m² reaches a single ideal linear polarizer. Someone chooses a laboratory axis and uses Malus’s law to predict zero transmission at rotation angle 90°. Find actual average output intensity and explain which premise of that single-angle calculation is missing.