INTERACTIVE OPTICS LABORATORY
POLARIZATION OF LIGHT
Explore how light waves interact with polarizing filters, calculate transmitted amplitude and intensity, and discover the molecular physics behind polarized sunglasses.
Cambridge International AS & A Level Physics (9702) · for students, teachers and independent learners
What is light, and what does it mean for light to be polarized?
A transverse electromagnetic wave whose electric field has a definite oscillation direction.
How do polarizing filters alter amplitude and intensity?
Amplitude falls as |cos θ| and intensity as cos² θ — and unpolarized light loses half its intensity at the first filter.
How do the molecules inside a filter interact with the wave?
Aligned molecular structures absorb one electric-field component and transmit the perpendicular one.
SECTION 1
Calculators
Four calculators cover the amplitude projection, Malus's Law, whole filter stacks of up to six polarizers, and reverse algebra when the unknown is the angle. Every one of them shows its working line by line.
CALCULATOR 1
Electric-Field Amplitude After a Filter
Find the amplitude of a plane-polarized wave after an ideal polarizing filter. The filter passes the component of the electric field along its transmission axis.
FORMULA
A = A₀ |cos θ|
Signed component A₀ cos θ; magnitude uses |cos θ|
Any amplitude unit (V/m, or arbitrary)
Angle input unit
Measured from the horizontal
Measured from the horizontal
CALCULATION STEPS
- 1.A = A₀ |cos θ|
- 2.θ = |90° − 40°| = 50°
- 3.A/A₀ = |cos(50°)| = 0.6428
- 4.A = 0.6428 A₀ = 0.6428
Rotate the filter axis and watch the projected component change length.
CALCULATOR 2
Malus's Law — Transmitted Intensity
Find the intensity transmitted by an ideal polarizer from the angle between the incident polarization direction and the filter's transmission axis.
FORMULA
I = I₀ cos² θ
Ideal polarizer, linearly polarized incident light
Any intensity unit works
Intensity unit
Measured from the horizontal
Measured from the horizontal
CALCULATION STEPS
- 1.I = I₀ cos²θ
- 2.θ = |90° − 40°| = 50°
- 3.I = 100 × cos²(50°)
- 4.cos(50°) = 0.6428 ⇒ cos² = 0.4132
- 5.I = 100 × 0.4132 = 41.3176 W/m²
TRANSMITTED INTENSITY I/I₀ AGAINST θ (0°–180°)
- I / I₀ = cos²θ
The marker tracks your selected angle: currently θ = 50° ⇒ I/I₀ = 0.4132.
CALCULATOR 3
Multiple Polarizing Filters
Investigate light passing through up to six ideal polarizers. Each filter has its own transmission-axis angle, and the beam is re-polarized by every filter it passes.
FORMULA
Iₙ = Iₙ₋₁ cos²(θₙ − θₙ₋₁)
First polarizer on unpolarized light: I₁ = I₀ / 2
Initial light
Direction of the electric field, measured from the horizontal
FILTERS (2/6)
IN
100
OUT
0
OF ORIGINAL
0%
OUTPUT POLARIZATION
90°
CALCULATION CHAIN
| FILTER | AXIS | ANGLE TO PRECEDING POLARIZATION | INCIDENT | TRANSMITTED | % OF ORIGINAL |
|---|---|---|---|---|---|
| 1 | 0° | 90° | 100 | 0 | 0% |
| 2 | 90° | 90° | 0 | 0 | 0% |
LIGHT PATH THROUGH THE FILTER STACK
Linearly polarized at 90°
100
0%
0°
0%
90°
POLARIZED
0
Each filter's line shows its transmission axis, drawn at its angle measured from the horizontal. The bars show the intensity remaining at that point as text, so colour is never the only signal.
FULL CALCULATION CHAIN
- 1.Filter 1 (0°): θ = 90° ⇒ I1 = 100 × cos²(90°) = 0
- 2.Filter 2 (90°): θ = 90° ⇒ I2 = 0 × cos²(90°) = 0
- 3.Total: 0 — 0% of the original intensity
CALCULATOR 4
Solve for the Unknown
Choose the quantity to calculate and enter the values you know. Every algebraic rearrangement is shown, and physically impossible inputs are refused with an explanation.
FORMULA
I/I₀ = cos² θ ⇒ θ = arccos(√(I/I₀))
Returns the acute angle, 0°–90°
Quantity to calculate
Angle between the incident polarization direction and the transmission axis
TRANSMITTED INTENSITY I
25 W/m²
ALGEBRAIC REARRANGEMENT
- 1.I = I₀ cos²θ
- 2.I = 100 × cos²(60°)
- 3.I = 100 × 0.25
- 4.I = 25 W/m²
The amplitude falls to 0.5 A₀, but the intensity falls to 0.25 I₀ — intensity goes as amplitude squared.
Equivalent orientations: 60° and 120° give the same result because cos²θ is symmetric.
GRAPHING TOOLS
Transmission Curves
Compare how amplitude and intensity respond to the relative angle. The gold marker follows the angle you choose; every graph uses mathematically correct axes, labels and units, and updates instantly.
Moves the marker on all three curves below
θ
60°
AMPLITUDE A/A₀
50%
INTENSITY I/I₀
25%
INTENSITY REMOVED
75%
1 · RELATIVE AMPLITUDE A/A₀ AGAINST FILTER ANGLE
- A / A₀ = |cos θ|
2 · RELATIVE INTENSITY I/I₀ AGAINST FILTER ANGLE
- I / I₀ = cos²θ
3 · COMPARISON OF AMPLITUDE AND INTENSITY TRANSMISSION
- Amplitude A / A₀
- Intensity I / I₀
The intensity curve sits below the amplitude curve everywhere except at θ = 0° and θ = 90°, because intensity is the amplitude ratio squared. At 45° the gap is widest: 70.7% of the amplitude survives but only 50% of the intensity.
4 · TRANSMITTED INTENSITY AFTER EACH FILTER
A separate demonstration chain: unpolarized light enters at 100 units and passes three ideal filters.
FINAL
12.5%
OF ORIGINAL INTENSITY
12.50 units
INTENSITY AFTER EACH STAGE
INTERACTIVE LABORATORY
Interactive Polarizing Shades Laboratory
Rotate the lenses of a pair of polarizing sunglasses and look at a sunlit scene of water, road and glass. The glare is partially polarized; the lenses are polarizers. Every brightness change on screen is computed from the same equations the calculators use.
Simulation mode
Exact ideal Malus's Law behaviour: glare fully polarized, filters perfect, no tint.
SUNGLASSES
0°–180°
0°–180°
Rotating the pair rotates both lenses together
EXTRA POLARIZERS
Only the sunglasses are in the light path. Add another polarizer in front of the lenses to try the crossed-polarizer experiment on the scene itself.
SCENE & INCOMING LIGHT
0° = horizontally polarized reflected glare
GLARE BLOCKED (LEFT)
100%
GLARE BLOCKED (RIGHT)
100%
LENS RELATIONSHIP
Aligned with the glare
Drag either lens ring to rotate its transmission axis (touch, stylus or mouse), or use the sliders for precise values. The axis line is drawn at its angle measured from the horizontal: 0° is horizontal, 90° is vertical.
Axis separation between the lenses: 0°
The lenses are aligned: maximum transmission relative to the light entering the second lens.
Aligned axes give maximum transmission relative to the light entering the second polarizer. Crossed axes give zero. At 45° an ideal second polarizer transmits exactly half of the polarized intensity incident upon it.
LEFT LENS CALCULATION
LEFT LENS
- 1.Filter stack: the left lens at 90°
- 2.Glare (100% polarized): θ = |90° − 0°| = 90°
- 3.Glare transmitted = 0 of 1 — 100% removed
- 4.Skylight (unpolarized) transmitted = 0.275 of 0.55 — half is lost in the first polarizer
- 5.Ideal mode: no tint and perfect transmission, exactly as Malus's Law describes
- 6.Total scene brightness = 0.275 of 1.55 ⇒ 17.74%
RIGHT LENS CALCULATION
RIGHT LENS
- 1.Filter stack: the right lens at 90°
- 2.Glare (100% polarized): θ = |90° − 0°| = 90°
- 3.Glare transmitted = 0 of 1 — 100% removed
- 4.Skylight (unpolarized) transmitted = 0.275 of 0.55 — half is lost in the first polarizer
- 5.Ideal mode: no tint and perfect transmission, exactly as Malus's Law describes
- 6.Total scene brightness = 0.275 of 1.55 ⇒ 17.74%
What Is Light?
Before you can polarize light, you need to know what is oscillating and in which direction. The animation below is a plane electromagnetic wave — pause it, change the wavelength, and rotate the polarization to see how the fields move.
0° = electric field vertical on screen; 90° = electric field into the screen (depth axis)
A = 0.75 of the maximum drawn amplitude
Drawn wavelength = 370 px on the diagram
Speed affects the animation only, never the physics
Light is an electromagnetic wave
Light is electromagnetic radiation: a traveling wave carrying oscillating electric and magnetic fields. In a plane wave in a vacuum the electric field oscillates perpendicular to the direction of propagation, the magnetic field oscillates perpendicular to the electric field, and both are perpendicular to the direction of travel. The fields travel together at the speed of light in vacuum, c = 3.00 × 10⁸ m s⁻¹.
Light is a transverse wave
Because both fields oscillate perpendicular to the direction of travel, light is a transverse wave. This is exactly what makes polarization possible: the oscillation has a definite direction within the plane perpendicular to the beam.
Polarization is defined by the electric field
The polarization direction of linearly polarized light is conventionally taken to be the direction in which the electric field oscillates. When a problem says light is 'vertically polarized', the electric field vector oscillates in the vertical plane and the magnetic field oscillates horizontally.
Unpolarized light has no preferred direction
In unpolarized light the transverse electric-field orientation varies rapidly and randomly, with no preferred direction in the plane perpendicular to propagation — for example light from a filament lamp or the Sun.
Polarization survives reflection and scattering
Reflection from a smooth surface and scattering by air molecules both introduce partial polarization, which is why glare from water, roads, and glass is partially horizontally polarized.
Photons and polarization
Light also behaves as particles: energy is exchanged with matter in discrete quanta called photons, each carrying energy E = hf, where h is the Planck constant and f is the frequency of the radiation. Ordinary linear polarizers primarily select polarization states rather than photon energies: an ideal polarizer transmits a fraction cos² θ of the incident intensity at essentially the same frequency. A polarizer is not a sieve that photons squeeze between molecular gaps — the material's electrons respond differently to electric fields in different directions, so light in one polarization is absorbed while light in the perpendicular polarization passes.
Unpolarized & Polarized Light
The AICE calculators on this page deal with unpolarized light and linearly polarized light. Circular and elliptical polarization are introduced here for completeness — and to make clear when Malus's Law does and does not apply.
Unpolarized light
The transverse electric-field direction varies rapidly and randomly. No direction in the plane perpendicular to travel is preferred. An ideal polarizer transmits half the intensity, because the average of cos² θ over uniformly distributed orientations is 1/2.
Central to AICE problems: I₁ = I₀/2 through the first ideal polarizer.
Linearly polarized light
The electric field oscillates along one fixed line in the transverse plane. Malus's Law applies directly: I = I₀ cos² θ, and the transmitted amplitude is A = A₀ |cos θ|.
The standard case for every calculation on this page.
Circularly polarized light
The electric-field vector has constant magnitude and rotates steadily as the wave travels, tracing a circle. It can be described as two orthogonal components of equal amplitude 90° out of phase.
Introductory only: the simple A₀ cos θ projection does not describe an ideal linear polarizer acting on circular light — such light passes an ideal polarizer unchanged in intensity at any axis angle.
Elliptically polarized light
The general case: the electric-field vector traces an ellipse, so its magnitude and direction both change as the wave travels. Linear and circular polarization are special cases of it.
Introductory only: without knowing the ellipse's orientation and axes, Malus's Law cannot be applied directly.
WHY THE FIRST POLARIZER HALVES UNPOLARIZED LIGHT
Average of cos² θ over uniformly distributed polarization orientations = 1/2
I = I₀ / 2
For unpolarized light the field directions are uniformly distributed over 0°–180°, and cos² θ averages to exactly one half. So an ideal polarizer always transmits 50% of unpolarized light — regardless of the polarizer's orientation, and even as the polarizer is rotated.
How Do Polarizing Filters Work at the Molecular Level?
Polarization is an electromagnetic interaction, not a geometric one. Here is what the electric field is doing inside the material, and why one direction is absorbed while the other passes through.
Molecular structure
Many common sheet polarizers are made from aligned polymer structures containing light-absorbing molecules or conductive molecular segments. A classic example is stretched polyvinyl alcohol (PVA) film treated with iodine: stretching aligns the polymer chains, and the iodine-related structures produce anisotropic optical absorption — strong absorption in one direction and weak absorption in the perpendicular direction.
The electric field of the incoming light interacts with charged particles, particularly electrons, inside the material. The electronic response depends on the direction of the electric field relative to the aligned molecular structures, so the film absorbs one electric-field component far more strongly than the other.
Interaction with the electric field
- ▸The electric-field component parallel to the principal absorbing molecular alignment is strongly absorbed.
- ▸The perpendicular component is preferentially transmitted.
- ▸The transmitted light is therefore polarized along the filter's transmission axis, which is perpendicular to the principal absorbing direction in this idealized model.
- ▸Absorbed optical energy is not destroyed — it is transferred to the material (typically appearing as internal energy, including heat).
Important correction
A polarizing film is not a mechanical screen with microscopic gaps that waves squeeze through. If it were, the transmitted intensity would not depend so precisely on angle, and the effect would not scale with electric-field direction.
Polarization happens because the material's electrons respond differently to electric fields in different directions. It is an electromagnetic interaction between the field and the material's charges, not a geometric filter.
INTERACTIVE MOLECULAR DIAGRAM
Red = absorbed component (parallel to the absorbing direction) · green = transmitted component · the transmission axis is perpendicular to the principal absorbing direction.
Visual density only — the physics does not change
ABSORBED AMPLITUDE
100%
TRANSMITTED AMPLITUDE
0%
TRANSMITTED INTENSITY
0%
TRANSMISSION AXIS
0°
component along the absorbing direction = A₀ |cos 0°| = 1 A₀ → strongly absorbed
component along the transmission axis = A₀ |sin 0°| = 0 A₀ → transmitted
I / I₀ = cos²90° = sin²0° = 0
Not all polarizers absorb
Absorption is only one mechanism. Reflection, transmission, and beam separation are physically different routes to the same outcome — light with one polarization left behind, light with the other continuing on.
Dichroic (absorbing) polarizers
AbsorptionAligned polymer or crystal structures absorb one electric-field component and transmit the perpendicular component. Stretched PVA with iodine is the common sheet polarizer; tourmaline and some dichroic crystals absorb in the same basic way.
Wire-grid polarizers
Reflection / re-emissionA grid of fine parallel conducting wires reflects or re-radiates the field component parallel to the wires and transmits the perpendicular component. The grid spacing must be much smaller than the wavelength, so wire grids work well for microwaves and mid-infrared light and need sub-micron manufacture for visible light.
Polarizing beam splitters
Beam separationA birefringent crystal or multilayer coating separates an unpolarized beam into two orthogonally polarized beams that travel in different directions. Both components survive, so the device separates rather than absorbs.
Polarization by reflection
Selective reflectionReflection at a surface polarizes light to a degree that depends on the angle of incidence and the refractive indices. At Brewster's angle the reflected beam is fully polarized parallel to the surface, which is why glare from horizontal surfaces is partially horizontally polarized.
Worked AICE-Style Examples
Six fully worked problems covering the structures that appear most often in Cambridge International AS & A Level Physics (9702) polarization questions. Try each one before revealing the solution.
All wording and diagrams are original teaching material. No copyrighted examination questions are reproduced.
GIVEN
- •Vertically polarized light (polarization direction 90° from horizontal).
- •Polarizer transmission axis 40° above the horizontal.
REQUIRED
The transmitted amplitude ratio A/A₀.
FORMULA
A = A₀ |cos θ|, where θ is the angle between the incident polarization direction and the transmission axis.
Substitution: θ = |90° − 40°| = 50°
WORKING
- 1.A/A₀ = |cos 50°|
- 2.A/A₀ = 0.6428
- 3.A = 0.6428 A₀
ANSWER
A/A₀ ≈ 0.643
Common Student Mistakes
Every one of these errors costs marks in examinations. Read them alongside the worked examples.
MISTAKE 1
Using the angle from the horizontal directly
The θ in Malus's Law is the angle between the incident polarization direction and the filter's transmission axis — not the filter's angle measured from the horizontal. 40° from horizontal with vertically polarized light gives θ = 50°, not 40°.
MISTAKE 2
Confusing amplitude with intensity
The amplitude follows the cosine projection, A = A₀ |cos θ|. The intensity follows the cosine squared, I = I₀ cos² θ. At 45° the amplitude is 70.7% of A₀ while the intensity is only 50% of I₀.
MISTAKE 3
Forgetting the half-intensity rule for unpolarized light
An ideal polarizer transmits exactly half the intensity of unpolarized light, whatever its orientation (I₁ = I₀/2). This factor applies once, at the first polarizer only.
MISTAKE 4
Assuming crossed polarizers always block light
Crossed polarizers do transmit nothing — until a further polarizer is placed between them. Inserting one at 45° lets 25% through, because it re-orients the polarization of the light reaching the final filter. No energy is created: the final intensity is still below the original.
MISTAKE 5
Treating a polarizer as a mechanical slit
A polarizing film has no microscopic gaps for waves to squeeze through. Polarization arises because the material's electrons respond differently to electric fields in different directions; one component is absorbed, and the perpendicular component is transmitted.
MISTAKE 6
Applying Malus's Law to every polarization state
Malus's Law assumes linearly polarized incident light and an ideal polarizer. For unpolarized light you must use the average (½). For circularly and elliptically polarized light the simple projection does not apply. Real filters also absorb some of the transmitted component, so measured transmission is a little below the ideal value.
ORIGINAL PRACTICE CHALLENGES
Test Yourself
30 original questions across three difficulty levels — multiple choice and numerical response. Every answer is verified, and every wrong answer is explained, not just marked.
QUESTION
1 / 30
ANSWERED
0 / 30
CORRECT
0
SCORE
—
Vertically polarized light passes through an ideal polarizer whose transmission axis is 40° from the horizontal. Find the transmitted amplitude ratio A/A₀.
Conventions used throughout: angles are measured from the horizontal unless stated otherwise; θ is always the angle between the incident polarization direction and the filter's transmission axis; amplitude is reported as a magnitude via |cos θ|; and ideal polarizers are assumed in the calculators, so measured real-world transmission will be slightly lower. The laboratory states clearly when it switches to its approximate realistic model.
