Tip: Set δ = 90° and equal $E_x = E_y$ for circular polarisation. Change δ to see the ellipse rotate and squash.
Section 02
The Idea, Step by Step
Light is a wave that wiggles — but which way does it wiggle? Picture shaking a jump rope: you can flick it straight up and down, side to side, or swirl your hand so the rope traces a circle. The electric field of a light wave does exactly the same thing as the wave races toward you. Polarisation is simply the name for the direction (or pattern) of that wiggle in the plane facing you.
When the field stays along one fixed line, we call it linearly polarised. A polariser then acts like a picket fence: only the part of the wiggle lined up with the slots gets through. If your light arrives at an angle $\theta$ to the slots, only the slice along the slots survives, and its amplitude shrinks by $\cos\theta$. But brightness — the intensity our eyes and detectors measure — goes as amplitude squared, so the brightness you see is
Worked number: line two polarisers at $45°$. Since $\cos^2 45° = 0.5$, exactly half the light passes. Crank the angle to $90°$ (crossed polarisers) and $\cos^2 90° = 0$ — the view goes completely dark.
To capture every case at once, write the field as two perpendicular pieces that share a frequency but may be out of step by a phase $\delta$: a side-to-side part $E_x\cos(\omega t)$ and an up-and-down part $E_y\cos(\omega t + \delta)$. Over one cycle the tip of the combined arrow traces a shape, and that shape is the polarisation. If $\delta = 0$ the two pieces swing together and the tip rides a straight line (linear). If $\delta = 90°$ and $E_x = E_y$, the tip sweeps a perfect circle (circular). Everything in between is an ellipse. A quarter-wave plate — a crystal that delays one piece by exactly a quarter cycle ($\delta = 90°$) — is the everyday device that turns linear light into circular light, which is how 3D-movie glasses sort the left and right images.
Try this in the sim above
• In E-vector View, set $E_x = E_y$ and drag δ from $0$ up to $90°$: watch the straight line fatten into an ellipse, then round out into a circle.
• Switch to Malus Polariser and slide θ toward $90°$: the outgoing wave shrinks to nothing as $\cos^2\theta \to 0$ — Malus' law in motion.
• In Brewster Angle mode, raise $n_2$ and hunt for the incidence angle where the purple p-reflection fades out. That special angle is $\theta_B = \arctan(n_2)$ — the secret behind glare-cutting sunglasses.
Section 03
Equation Derivation
Light is a transverse electromagnetic wave. The electric field $\mathbf E$ oscillates perpendicular to the propagation direction. The pattern traced by the tip of $\mathbf E$ defines the polarisation state.
General Polarisation State
$$\mathbf E(z,t) = \hat{\mathbf x}\,E_x\cos(kz - \omega t) + \hat{\mathbf y}\,E_y\cos(kz - \omega t + \delta)$$
$$\boxed{\text{Polarisation} = \text{Pattern of }\mathbf E\text{ over one period}}$$
Symbol
Meaning
SI Unit
$\mathbf E$
Electric-field vector
V/m
$E_x, E_y$
Amplitudes along x, y
V/m
$\delta$
Phase difference between components
rad
$k$
Wavenumber = $2\pi/\lambda$
rad/m
$\omega$
Angular frequency = $2\pi f$
rad/s
$\theta$
Angle of polariser axis
rad
$I$
Intensity = $\tfrac12 c\epsilon_0 |\mathbf E|^2$
W/m²
$\theta_B$
Brewster angle = $\arctan(n_2/n_1)$
rad
$n_1, n_2$
Refractive indices of two media
—
Step 1 — The three states of polarisation
Linear — when $\delta = 0$ or $\pi$. The two components are in phase (or anti-phase), so $\mathbf E$ oscillates along a fixed line at angle $\arctan(E_y/E_x)$.
Circular — when $\delta = \pm\pi/2$ and $E_x = E_y$. The tip of $\mathbf E$ traces a circle. Right-handed if $\delta = +\pi/2$ (looking toward source); left-handed if $-\pi/2$.
Elliptical — the most general case. The tip traces an ellipse, characterised by ellipticity $\chi$ and azimuth $\psi$.
Step 2 — Malus' Law
A polariser passes only the component of $\mathbf E$ parallel to its transmission axis. If incident light is linearly polarised at angle $\theta$ from the axis, the transmitted field is
$$E_t = E_0\cos\theta$$
Since intensity $\propto |E|^2$:
$$\boxed{\;I = I_0\cos^2\theta\;}$$
For unpolarised light through one polariser, the average over all orientations gives $I_{\text{out}} = I_0/2$.
Step 3 — Brewster's Angle
When light hits an interface between two media, the Fresnel equations give different reflectivities for s- and p-polarisations. At one special angle, the p-reflection is exactly zero:
Geometrical meaning: at $\theta_B$, the reflected and refracted rays are perpendicular ($\theta_B + \theta_t = 90°$). Reflected light is then 100% s-polarised (perpendicular to the plane of incidence).
For air-to-glass ($n_2 = 1.5$): $\theta_B = \arctan(1.5) \approx 56.3°$. This is the principle behind polarising sunglasses, photographic polarisers, and many laser components.
Step 4 — The Polarisation Ellipse
Eliminating time from $E_x = a\cos\phi$, $E_y = b\cos(\phi + \delta)$:
Reference: Hecht — Optics, 5th Ed., Pearson, 2017, Ch. 8: "Polarization"; Born & Wolf — Principles of Optics, 7th Ed., Cambridge UP, 1999, §1.4: "The Wave Equation in a Conducting Medium"; HRW 10th Ed., §33-7 to §33-9: "Polarization, Reflection, Brewster's Law".
Section 04
Frequently Asked Questions
A polariser passes only the field component along its transmission axis: $E_t = E_0\cos\theta$. But intensity (what detectors and our eyes measure) is proportional to the square of the field: $I \propto |E|^2$. Squaring $\cos\theta$ gives $\cos^2\theta$. This is why the intensity drops only to a quarter (not zero) when you cross polarisers at 60°: $\cos^2 60° = 0.25$, and at 90° it drops to zero.
Key: Field amplitude is $\cos\theta$; intensity is $\cos^2\theta$. The square comes from intensity ∝ |E|².
Polarised sunglasses block horizontally polarised light reflected from water/road, dramatically cutting glare (reflected light is partly polarised, with the polarisation depending on angle — the basis of the polarising sunglasses' selective absorption). LCD screens use polarisation: a backlight passes through a polariser, then liquid crystals rotate the polarisation pixel-by-pixel, then a second polariser converts that rotation into intensity. 3D movie glasses: each lens transmits a different circular polarisation, separating left/right images. Stress-induced birefringence in plastics (try a plastic ruler between two polarisers!) reveals stress patterns. Bee navigation uses sky polarisation. AM radio antennas are polarised; receiving them with a wrongly oriented antenna is an everyday Malus experiment.
Key: If you wear polarised sunglasses and tilt your head, you've just done a Malus experiment.
In E-vector mode you see the tip of the electric field $\mathbf E(t) = (E_x\cos\omega t, E_y\cos(\omega t + \delta))$ over one period. As you change the phase δ, the trajectory smoothly morphs from a line (δ=0) to an ellipse (intermediate δ) to a circle (δ=90° with equal amplitudes) and back. In Malus mode the simulation shows linearly polarised light hitting a polariser at angle θ; the transmitted intensity is shown alongside the live $\cos^2\theta$ curve. In Brewster mode you see incident, reflected, and refracted rays — at $\theta = \theta_B$ the reflected ray vanishes for p-polarisation. In wave plate mode, you can see how a quarter- or half-wave plate transforms the polarisation state.
Key: The shape traced by the E-vector tip is the polarisation. Trace it and you've named it.
Sound waves in fluids are longitudinal: the air molecules oscillate parallel to the direction of propagation. There's only one possible direction for the oscillation — there's nothing to choose. Polarisation requires transverse waves: oscillation perpendicular to propagation, which leaves a 2D plane of choices. Light is transverse (electric field perpendicular to direction of motion), so it can be polarised. In solids, sound can be transverse (shear waves) — these can be polarised, and shear waves through Earth's interior, for example, do not pass through the (liquid) outer core, which is how we know it's liquid.
Key: Polarisation = freedom of orientation in the perpendicular plane. Only transverse waves have it.
Send linearly polarised light at 45° through a birefringent crystal aligned with its fast axis along $x$ and slow along $y$. The light's two components $E_x = E_y = (E_0/\sqrt{2})\cos\omega t$ start in phase. The slow axis delays the y-component by a quarter period: $E_y \to (E_0/\sqrt{2})\cos(\omega t - \pi/2) = (E_0/\sqrt{2})\sin\omega t$. The output field is $\mathbf E = (E_0/\sqrt{2})(\cos\omega t, \sin\omega t)$ — exactly circular polarisation. Reverse direction (linear input perpendicular to fast axis) yields the opposite handedness.
Yes. In quantum optics, polarisation states correspond to photon spin angular momentum: right-circular polarised light = photons with $+\hbar$ spin along propagation; left-circular = $-\hbar$. Linear polarisation is an equal superposition of right and left. This deep relationship (Jones vectors ↔ qubit states) makes polarisation a perfect platform for quantum information: photons are easy to manipulate with polarisers and wave plates, and entangled-polarisation photon pairs are the workhorse of quantum cryptography (BB84, Ekert protocols) and Bell tests.
Yes — the famous "three-polariser paradox". Cross two polarisers (axes 0° and 90°): $\cos^2 90° = 0$ → no light. Now insert a third polariser at 45° between them. After the first: $I_0/2$ (unpolarised → polarised). After the 45° polariser: $\cos^2 45° \cdot (I_0/2) = I_0/4$. After the third polariser at 90° (now the previous axis is 45°): $\cos^2 45° \cdot (I_0/4) = I_0/8$. So adding a polariser between two crossed ones gives output $I_0/8$ instead of zero! This works because each polariser projects the field onto its axis — successive projections need not align.
Key: Three polarisers can pass more light than two — successive projection is not multiplication of factors.
Best Resource: Hecht — Optics, 5th Ed., Ch. 8 (gold-standard textbook); HyperPhysics — "Polarisation of Light", hyperphysics.phy-astr.gsu.edu/hbase/phyopt/polclas.html; Khan Academy — "Polarisation"; MIT OCW 8.03 — Optics, Lectures 22–24.
Section 05
Common Misconceptions
❌ Polarised light contains "less" light than unpolarised light.
✅ Both have the same intensity; only the orientation distribution differs.
Unpolarised light is a mixture of all polarisation orientations with equal probability — it's not "smaller" than polarised light, just statistically scrambled. A laser pointer can be 100% polarised yet bright; a 100 W incandescent bulb is unpolarised yet brighter. What changes is the symmetry: unpolarised light passing through a polariser loses half its intensity (only half the orientations survive); already-polarised light at the same angle as the polariser loses none.
📖 Hecht — Optics, 5th Ed., §8.2: "Natural Light".
❌ Light polarisation tells you the direction of the photon's motion.
✅ Polarisation is perpendicular to the direction of motion.
For an electromagnetic wave travelling in direction $\hat z$, the electric field $\mathbf E$ lies in the $xy$-plane (transverse). Polarisation describes the orientation of $\mathbf E$ within that perpendicular plane. The propagation direction is the same regardless of polarisation. Confusing the two leads to wrong intuition about polarisers and Brewster's angle. The propagation direction is given by the wavevector $\mathbf k$; polarisation refers only to $\mathbf E$.
📖 Born & Wolf — Principles of Optics, 7th Ed., §1.4.
❌ Crossed polarisers always block all light.
✅ Only when there is nothing optically active between them.
Two ideal polarisers at 90° (crossed) transmit no light if nothing is between them. But many transparent materials are birefringent or optically active — they rotate or modify the polarisation state of light passing through. Sugar solutions rotate linear polarisation (used to measure concentration). Stressed plastics show colourful patterns between crossed polarisers because birefringence is wavelength-dependent. Liquid-crystal cells in LCDs deliberately rotate polarisation by an electrically controlled amount, allowing or blocking transmission — that's how every screen pixel works.
❌ Brewster's angle works for any pair of refractive indices, including air-to-air.
✅ It requires a refractive-index difference between the two media.
Brewster's angle $\theta_B = \arctan(n_2/n_1)$ requires $n_1 \ne n_2$. If the two media have the same refractive index, light doesn't reflect at all (no interface in the optical sense), and there is no Brewster angle to speak of. The bigger the index contrast, the more pronounced the Brewster effect: glass-to-air gives a clear effect ($\theta_B \approx 56°$), while two adjacent layers of slightly different glass give only a weak effect.
📖 Hecht — Optics, 5th Ed., §4.6: "Reflection".
❌ Circular polarisation requires a special "rotating" wave that doesn't exist for normal light.
✅ It's just the sum of two perpendicular linear waves with a 90° phase difference.
There is no exotic mechanism — circular polarisation is mathematically equivalent to two perpendicular linear oscillations 90° out of phase. Any laser plus a quarter-wave plate produces it instantly. "Circular" describes the path traced by the tip of $\mathbf E$, not a physically rotating wave. The same holds for elliptical polarisation: it's just the most general superposition $E_x\cos\omega t + E_y\cos(\omega t + \delta)$ of two perpendicular linear waves with arbitrary phase and amplitude.
❌ Polarisation only matters in physics labs — it has no everyday relevance.
✅ Every smartphone, sunglass, and 3D movie depends on it.
Every LCD screen — phone, TV, computer — works by rotating polarisation pixel-by-pixel. Every pair of polarised sunglasses cuts glare via Brewster-aware blocking of horizontally polarised reflections. Every 3D movie uses circular polarisation (left-circular for one eye, right-circular for the other). Every camera with a polarising filter uses it to deepen sky colours and remove water reflections. Every fibre-optic cable preserves or scrambles polarisation deliberately for signal-integrity reasons. Polarisation is the foundation of an industry worth hundreds of billions.
📖 Yu & Yang — Polarization Devices and Their Applications, SPIE, 2014.
Misconception research: Mestre et al. — "Optics misconceptions in introductory physics", AJP 80, 421 (2012); Galili & Hazan — "Learners' knowledge in optics: interpretation, structure and analysis", Int. J. Sci. Educ. 22, 57 (2000); Driver et al. — Making Sense of Secondary Science.