Physics • Year 12 • Module 7 • Lesson 5
Polarisation of Light
Lock in the key vocabulary, Malus’s Law, and the distinction between unpolarised and plane-polarised light before tackling harder questions.
1. Term–definition match
The definitions below are shuffled. Write the matching term from this list in the right-hand column: polarisation, plane-polarised light, unpolarised light, polarising filter, analyser, Malus’s Law, transmission axis, Brewster’s angle, birefringence, transverse wave. 10 marks (1 each)
| # | Definition | Matching term |
|---|---|---|
| 1.1 | A wave in which the oscillations occur perpendicular to the direction of propagation; the only type of wave that can be polarised. | |
| 1.2 | The restriction of the oscillations of a transverse wave to a single plane perpendicular to the direction of propagation. | |
| 1.3 | Light in which the electric field oscillates in all directions perpendicular to the direction of propagation with no preferred orientation. | |
| 1.4 | Light in which the electric field oscillates in only one direction perpendicular to the direction of propagation. | |
| 1.5 | A material that transmits only one component of the electric field and absorbs the perpendicular component; produces plane-polarised light from unpolarised light. | |
| 1.6 | The direction along which a polarising filter transmits the electric field component of light. | |
| 1.7 | A second polarising filter placed after the first to detect or further reduce the intensity of already polarised light. | |
| 1.8 | The relationship I = I0 cos2 θ giving the transmitted intensity through an analyser at angle θ to the polarisation direction. | |
| 1.9 | The angle of incidence at which reflected light is completely plane-polarised, given by tan θB = n2/n1. | |
| 1.10 | A property of certain crystals (e.g. calcite) in which the refractive index differs for the two polarisation components, splitting unpolarised light into two polarised beams. |
2. True or false — with correction
Circle T or F for each statement. If the statement is false, write the corrected version on the line below it. 12 marks (1 T/F + 1 correction each)
2.1 Sound waves travelling through air can be polarised because they carry energy perpendicular to their direction of propagation. T / F
2.2 When unpolarised light passes through a single ideal polarising filter, the transmitted intensity is equal to half the incident intensity. T / F
2.3 Two polarising filters whose transmission axes are at 90° to each other transmit no light — this is called the “crossed polarisers” condition. T / F
2.4 Malus’s Law states that the transmitted intensity through an analyser is I = I0 sin2 θ, where θ is the angle between the transmission axes. T / F
2.5 Polarising sunglasses reduce glare from water because light reflected from non-metallic horizontal surfaces is partially polarised parallel to the surface (horizontally). T / F
2.6 Inserting a third polarising filter at 45° between two crossed polarisers causes the overall system to transmit less light than the crossed pair alone. T / F
3. Fill-in-the-blank paragraph
Use the word bank to complete the passage. Each word is used once. 8 marks (1 per blank)
Word bank:
analyser · cos2 θ · glare · half · longitudinal · plane-polarised · transverse · zero
Polarisation is a property of ___________ waves only; ___________ waves such as sound in air cannot be polarised because their oscillations are parallel to the direction of travel. When unpolarised light of intensity I0 passes through one ideal polarising filter, the transmitted intensity is ___________ of I0, and the emergent light is ___________. A second filter, called an ___________, is then placed in the path of the polarised beam. According to Malus’s Law, the transmitted intensity is proportional to ___________, where θ is the angle between the transmission axes. When the two axes are perpendicular (θ = 90°), the transmitted intensity is ___________, so no light emerges. This property is exploited in polarising sunglasses, which block horizontally polarised ___________ reflected from water and roads.
4. Function recall
Answer each question in 1–2 sentences using precise terms from the lesson. 8 marks (2 each)
4.1 Why is polarisation considered direct evidence that light is a transverse wave rather than a longitudinal wave?
4.2 State the equation for Malus’s Law, define each symbol, and state the two extreme cases (θ = 0° and θ = 90°).
4.3 Describe one application of polarisation technology in everyday life (other than sunglasses) and explain which aspect of polarisation it exploits.
4.4 Light reflected from a lake surface at Brewster’s angle is described as completely plane-polarised. In what direction is the electric field oscillating in this reflected light?
5. Build a concept map
Draw labelled arrows between the six terms below to show how they connect. Each arrow must carry a linking phrase (e.g. “is governed by”, “produces”, “is a type of”). Aim for at least 6 labelled arrows. 6 marks (1 per valid labelled arrow)
Supplied terms: polarisation · transverse wave · Malus’s Law · polarising filter · intensity · analyser angle.
Q1 — Term–definition match
1.1 transverse wave • 1.2 polarisation • 1.3 unpolarised light • 1.4 plane-polarised light • 1.5 polarising filter • 1.6 transmission axis • 1.7 analyser • 1.8 Malus’s Law • 1.9 Brewster’s angle • 1.10 birefringence.
Q2 — True / false with correction
2.1 False. Sound waves travelling through air are longitudinal — oscillations are parallel to the direction of propagation. Because there is no perpendicular plane to restrict, sound waves cannot be polarised.
2.2 True. An ideal polariser absorbs one component and transmits the other, so I = I0/2 for unpolarised incident light.
2.3 True. At θ = 90°, Malus’s Law gives I = I0 cos2 90° = 0. No light is transmitted through crossed polarisers.
2.4 False. Malus’s Law uses cosine squared, not sine squared: I = I0 cos2 θ. Maximum transmission occurs at θ = 0°, where cos2 0° = 1.
2.5 True. Reflected light from non-metallic horizontal surfaces (water, roads, glass) is partially polarised with the electric field oscillating parallel to the surface (horizontally). Polarising sunglasses with a vertical transmission axis block this component.
2.6 False. Inserting a third polariser at 45° between two crossed polarisers increases the transmitted light from zero to I0/8. The crossed pair alone transmits zero; with the middle polariser, some light gets through.
Q3 — Cloze paragraph
In order: transverse / longitudinal / half / plane-polarised / analyser / cos2 θ / zero / glare.
Q4.1 — Polarisation as evidence of transverse nature
Polarisation is direct evidence that light is a transverse wave because only transverse waves have oscillations perpendicular to the direction of propagation. A polarising filter can select one perpendicular orientation and block others. Longitudinal waves (like sound) oscillate parallel to propagation, so there is no perpendicular plane to restrict — they cannot be polarised at all.
Q4.2 — Malus’s Law and extreme cases
I = I0 cos2 θ, where I is the transmitted intensity, I0 is the incident (already polarised) intensity, and θ is the angle between the transmission axis of the analyser and the polarisation direction. At θ = 0°: cos2 0° = 1, so I = I0 (maximum transmission). At θ = 90°: cos2 90° = 0, so I = 0 (no transmission — crossed polarisers).
Q4.3 — Application of polarisation
Accept any one of: LCD screens — use two crossed polarisers with a liquid crystal layer that rotates polarisation when voltage is applied, allowing pixels to switch between bright and dark. 3D cinema — two images are projected with orthogonal polarisations; each eye’s matching filter blocks the other image. Photoelasticity — stressed transparent materials rotate polarisation, revealing stress patterns as coloured fringes between crossed polarisers. Award full marks for any valid application with a correct mechanism.
Q4.4 — Direction of reflected polarised light
At Brewster’s angle, the reflected light is completely plane-polarised with the electric field oscillating parallel to the reflecting surface (horizontally, if the surface is horizontal). This is because the component with an electric field perpendicular to the surface is transmitted into the medium at this special angle rather than being reflected.
Q5 — Sample concept map
Correct maps should include arrows such as:
- transverse wave — can undergo → polarisation
- polarising filter — produces → polarisation
- Malus’s Law — predicts → intensity
- analyser angle — determines (via Malus’s Law) → intensity
- polarising filter — reduces → intensity
- polarisation — is quantified by → Malus’s Law
Award 1 mark per valid labelled arrow (minimum 6, maximum 6 marked).