Polarisation and Malus's Law

Spec mapping: OCR H556 Module 4.4 — Waves: polarisation as restriction to a single plane; only transverse waves can be polarised; Malus's law $I = I_0 \cos^2\theta$I=I0​cos2θ; PAG 3 anchor (polarisation). (Refer to the official OCR H556 specification document for exact wording.)

Polarisation is arguably the subtlest of the basic wave phenomena. It is also one of the most powerful pieces of experimental evidence for the transverse nature of light, and it has genuine practical applications — from sunglasses and camera filters to LCD screens and 3D cinema. The OCR A-Level Physics A specification asks you to understand what polarisation is, why only transverse waves can be polarised, how polarising filters (like Polaroid) work, and — in the OCR context uniquely among the A-Level boards — to apply Malus's law:

$I = I_0 \cos^2 \theta$I=I0​cos2θ

This is the equation you must memorise and be able to deploy. The rest of this lesson establishes the physical content behind it.

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What Does It Mean for a Wave to Be Polarised?

In an unpolarised transverse wave, the direction of oscillation changes rapidly and randomly — there is no preferred plane. Natural light from the Sun or from a filament bulb is unpolarised: the electric field vector oscillates in all directions perpendicular to the direction of propagation, changing direction millions of times per second.

A polarised transverse wave is one in which the oscillation is restricted to a single plane containing the direction of propagation. That plane is called the plane of polarisation (OCR's convention is to define this as the plane containing the electric field vector and the direction of propagation).

Only transverse waves can be polarised, because only transverse waves have oscillations in multiple possible directions perpendicular to propagation. Longitudinal waves (like sound) oscillate only along the direction of travel — there is nothing left to restrict, so polarisation is meaningless for them.

This is the crucial experimental evidence that light is transverse: by 1808, Étienne-Louis Malus had observed that a beam of light reflected off glass at certain angles could be partially blocked by a second piece of glass oriented perpendicular to the first. Something about the reflected light made it asymmetric about the direction of travel — and that could only happen for a transverse wave.

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Polarising Filters (Polaroid)

A polarising filter (commonly called a Polaroid filter, after the commercial product invented by Edwin Land in 1929) contains long-chain conducting molecules all aligned in the same direction. These chains absorb the component of the electric field parallel to them, transmitting only the component perpendicular to them.

The direction along which the filter transmits is called its transmission axis.

flowchart LR
    U[Unpolarised light: E vibrates in all directions] --> F1[Polariser: transmission axis vertical]
    F1 --> P[Vertically polarised light: amplitude reduced by sqrt 2]
    P --> F2[Analyser at angle theta]
    F2 --> O[Transmitted light: amplitude A cos theta, intensity I cos squared theta]

What Happens to Unpolarised Light?

When unpolarised light of intensity $I_0$I0​ passes through a single polariser, the filter transmits only the component along its transmission axis. Since the incoming light is random in orientation, averaging $\cos^2\theta$cos2θ uniformly over all directions gives $\langle \cos^2\theta \rangle = 1/2$⟨cos2θ⟩=1/2, so only half the intensity is transmitted:

$I_1 = \frac{I_0}{2}$I1​=2I0​​

This is the starting point of any polarisation problem.

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Malus's Law

Once light has been polarised by a first filter (the polariser), you can analyse it with a second filter (the analyser). If the analyser's transmission axis makes an angle $\theta$θ with the plane of polarisation of the incoming light, the transmitted intensity is given by Malus's law:

$I = I_0 \cos^2 \theta$I=I0​cos2θ

where $I_0$I0​ is the intensity of the already polarised light incident on the analyser, $I$I is the intensity of the light transmitted through the analyser, and $\theta$θ is the angle between the analyser's transmission axis and the plane of polarisation of the incoming polarised light.

Why the $\cos^2$cos2 Factor?

The analyser transmits the component of the electric field along its transmission axis. If the incoming polarised light has electric field amplitude $E_0$E0​ at angle $\theta$θ to the transmission axis, then the transmitted amplitude is:

$E = E_0 \cos \theta$E=E0​cosθ

But intensity is proportional to the square of the amplitude:

$I \propto E^2 \quad \Rightarrow \quad I = I_0 \cos^2 \theta$I∝E2⇒I=I0​cos2θ

The squaring is why $\cos\theta$cosθ appears in amplitude equations but $\cos^2\theta$cos2θ appears in intensity equations. Be vigilant about this distinction. It is the single most common slip in OCR Malus questions.

Exam Tip: OCR is one of the few A-Level Physics specifications that names Malus's law explicitly and asks you to apply it. Be ready for a worked calculation in which you must identify $\theta$θ correctly and compute $I = I_0 \cos^2 \theta$I=I0​cos2θ.

Special Angles

$\theta$θ	$\cos^2 \theta$cos2θ	$I / I_0$I/I0​	Interpretation
$0°$0°	$1$1	$1$1	Full transmission — analyser aligned with polariser
$30°$30°	$0.75$0.75	$3/4$3/4	Mostly transmitted
$45°$45°	$0.5$0.5	$1/2$1/2	Half transmitted
$60°$60°	$0.25$0.25	$1/4$1/4	One quarter transmitted
$90°$90°	$0$0	$0$0	Crossed — no light transmitted

The last row is worth dwelling on: when the analyser is oriented perpendicular to the polariser (crossed polarisers), no light gets through at all. This is the definitive test for whether light is polarised — if rotating the second filter through $90°$90° extinguishes the light, the light was polarised. Conversely, if rotating the second filter changes the brightness but does not extinguish it, the incident light was unpolarised.

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Worked Example 1 — Single Polariser

Q. Unpolarised light of intensity $200$200 W m$^{-2}$−2 is incident on a polarising filter. Calculate the transmitted intensity.

A. For unpolarised incident light, a single polariser transmits half the intensity:

$I = \frac{I_0}{2} = \frac{200}{2} = 100 \text{ W m}^{-2}$I=2I0​​=2200​=100 W m−2

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Worked Example 2 — Two Polarisers with an Angle Between Them

Q. Unpolarised light of intensity $800$800 W m$^{-2}$−2 passes first through a polariser, then through an analyser whose transmission axis is at $30°$30° to that of the polariser. Calculate the final transmitted intensity.

A. After the first filter the light is polarised with intensity:

$I_1 = \frac{I_0}{2} = \frac{800}{2} = 400 \text{ W m}^{-2}$I1​=2I0​​=2800​=400 W m−2

This light is now polarised. Applying Malus's law for the second filter:

$I_2 = I_1 \cos^2(30°) = 400 \times \left(\frac{\sqrt{3}}{2}\right)^2 = 400 \times 0.75 = 300 \text{ W m}^{-2}$I2​=I1​cos2(30°)=400×(23​​)2=400×0.75=300 W m−2

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Worked Example 3 — Finding the Angle

Q. Polarised light of intensity $60$60 W m$^{-2}$−2 passes through an analyser and emerges with intensity $15$15 W m$^{-2}$−2. Calculate the angle between the light's plane of polarisation and the analyser's transmission axis.

A. Using $I = I_0 \cos^2 \theta$I=I0​cos2θ:

$\cos^2 \theta = \frac{I}{I_0} = \frac{15}{60} = 0.25$cos2θ=I0​I​=6015​=0.25 $\cos \theta = 0.5 \quad \Rightarrow \quad \theta = 60°$cosθ=0.5⇒θ=60°

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Worked Example 4 — Three Polarisers

Q. Unpolarised light of intensity $I_0$I0​ passes through three polarisers in succession. The second is at $45°$45° to the first, and the third is at $90°$90° to the first. Calculate the final transmitted intensity.

A.

• After polariser 1: $I_1 = I_0/2$I1​=I0​/2.
• After polariser 2 (angle $45°$45° to polariser 1): $I_2 = I_1 \cos^2(45°) = (I_0/2)(1/2) = I_0/4$I2​=I1​cos2(45°)=(I0​/2)(1/2)=I0​/4.
• After polariser 3 (the angle between polariser 2 and polariser 3 is $90° - 45° = 45°$90°−45°=45°, since polariser 2 is at $45°$45° and polariser 3 is at $90°$90°): $I_3 = I_2 \cos^2(45°) = (I_0/4)(1/2) = I_0/8$I3​=I2​cos2(45°)=(I0​/4)(1/2)=I0​/8.

This is a famous counter-intuitive result: inserting a filter increases the transmitted intensity (from $0$0 with just polarisers 1 and 3 crossed, to $I_0/8$I0​/8 when filter 2 is inserted between them). Each polariser projects the light onto a new axis, gradually "rotating" the plane of polarisation. The physical content: a polariser is not a passive blocker but an active projector.

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Worked Example 5 — Successive Rotation

Q. Polarised light of intensity $40$40 W m$^{-2}$−2 passes through an analyser whose transmission axis is rotated by $20°$20° from the light's plane of polarisation. The transmitted light then passes through a second analyser, whose transmission axis is rotated by another $20°$20° in the same sense. Calculate the final transmitted intensity.

A. Each analyser projects through the same angle of $20°$20°:

After analyser 1: $I_1 = 40 \times \cos^2 20° = 40 \times 0.883 = 35.3$I1​=40×cos220°=40×0.883=35.3 W m$^{-2}$−2.

After analyser 2: the light leaving analyser 1 is polarised along analyser 1's axis. The angle between this and analyser 2's transmission axis is again $20°$20°, so:

$I_2 = 35.3 \times \cos^2 20° = 35.3 \times 0.883 = 31.2 \text{ W m}^{-2}.$I2​=35.3×cos220°=35.3×0.883=31.2 W m−2.

Total $\cos^4 20°$cos420° factor: $\cos^4(20°) = (0.940)^4 \approx 0.780$cos4(20°)=(0.940)4≈0.780. In general, $N$N polarisers each separated by $\theta$θ from the previous one transmit a fraction $\cos^{2N}\theta$cos2Nθ of the originally polarised beam — provided the rotations accumulate in the same sense. For small $\theta$θ this approaches $(1 - \theta^2)^N \approx e^{-N\theta^2}$(1−θ2)N≈e−Nθ2, which is the adiabatic-rotation limit familiar from quantum mechanics.

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Sources of Polarised Light

Besides polarising filters, there are several natural ways to produce polarised light.

1. Reflection (Partial Polarisation)

Light reflected off a non-metallic surface (water, glass, a wet road) is partially polarised parallel to the reflecting surface. At a specific angle — Brewster's angle — the reflected light is completely polarised.

This is why polarising sunglasses work so well for reducing glare: their transmission axis is vertical, so they block the horizontally polarised light reflected off horizontal surfaces like water or roads.

Exam Tip: OCR will ask why polarising sunglasses reduce glare more effectively than ordinary tinted sunglasses. Your answer should mention that light reflected off horizontal surfaces is partially polarised horizontally, and that vertically transmitting sunglasses block this reflected light while still allowing direct (unpolarised) scene light through.

2. Scattering

Sunlight scattered off molecules in the atmosphere (Rayleigh scattering) is partially polarised. If you look at a blue sky with a polarising filter and rotate it, the sky will alternately brighten and darken. Bees and other insects can use this atmospheric polarisation pattern to navigate.

3. Transmission Through Polaroid Filters

Already covered above. This is the most direct laboratory method of producing polarised light and is how polarisation is demonstrated in school.

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Applications of Polarisation

• Polaroid sunglasses — block horizontally polarised reflected glare. The transmission axis is vertical, so horizontally polarised reflections (from water, road surfaces, car bonnets) are absorbed while the unpolarised scene light is transmitted at half intensity. The result is improved contrast and reduced eye strain — the reason fly-fishers, drivers and skiers wear them.
• Photographic polarising filters — darken skies, reduce reflections from water and glass. A circular polariser screwed onto a camera lens can completely eliminate sky reflections from a lake surface, revealing fish and underwater rocks otherwise invisible. The same filter darkens the blue sky (which is partially polarised by Rayleigh scattering) by transmitting only the un-scattered component.
• Liquid crystal displays (LCDs) — sandwich liquid crystals between crossed polarisers. Without an applied voltage, the liquid crystal twists the polarisation by $90°$90°, so light passes through both polarisers (the pixel appears bright). With a voltage applied, the twist is undone, and the second polariser blocks the light (the pixel appears dark). Every pixel of every laptop, phone and TV LCD screen in your environment depends on this physics.
• 3D cinema — left and right eye images are projected through perpendicular (linear) or oppositely-handed (circular) polarisers; matching glasses deliver each image to the correct eye. The brain then fuses the two images into a stereoscopic perception of depth.
• Photoelastic stress analysis — transparent materials under mechanical stress become birefringent (their refractive index becomes direction-dependent). Viewing a stressed transparent component through crossed polarisers reveals colourful fringe patterns whose density and orientation map the local stress field. Engineers use this technique to find stress concentrations in scale-model components before metal versions are built.
• Microwave-grid polarisers — a grid of parallel conducting wires acts as a polariser for microwaves: the oscillating electric field component along the wires drives currents and is absorbed; the perpendicular component passes through. This is the classroom demonstration of Malus's law in a wavelength regime where the polariser-grid is mechanically obvious.
• Communications — modern wireless communication systems often use polarisation diversity: transmitting two independent signals on the same frequency but with orthogonal polarisations doubles the information capacity of a given bandwidth. Mobile networks, satellite TV downlinks and Wi-Fi MIMO all exploit this principle.

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Demonstrating Polarisation with Microwaves (PAG 3 anchor)

A standard A-Level laboratory demonstration uses microwaves, listed in OCR's Practical Activity Group 3 (Waves) as the canonical polarisation experiment. A microwave transmitter emits vertically polarised waves (the antenna sets the orientation). If a grid of parallel wires is placed in front of the receiver:

• Wires vertical (parallel to $\vec{E}$E): the wave is absorbed because the oscillating electric field drives currents in the wires, which dissipate the energy. Minimum signal received.
• Wires horizontal (perpendicular to $\vec{E}$E): the wave passes through. Maximum signal received.