Physics • Year 12 • Module 7 • Lesson 5
Polarisation of Light
Apply Malus’s Law and your understanding of polarisation to numerical problems, graph interpretation, and real-world scenarios.
1. Malus’s Law calculations
Use I = I0 cos2 θ and I = I0/2 (for unpolarised light through one polariser) to answer each part. Show all working. 12 marks (2 each)
1.1 Plane-polarised light of intensity 60 W/m² passes through an analyser whose transmission axis is at 45° to the polarisation direction. Calculate the transmitted intensity.
1.2 Unpolarised light of intensity 120 W/m² passes through a single polarising filter. What is the transmitted intensity?
1.3 Plane-polarised light of intensity 80 W/m² passes through an analyser at 30°. Calculate the transmitted intensity. Give your answer to 1 decimal place.
1.4 Unpolarised light of intensity 200 W/m² passes through a polariser and then through an analyser at 60° to the polariser axis. Calculate the final transmitted intensity.
1.5 The transmitted intensity through an analyser is 20 W/m² when the incident plane-polarised intensity is 40 W/m². Find the angle between the analyser axis and the polarisation direction.
1.6 Three polarising filters are arranged in sequence. Filter 1 polarises unpolarised light of intensity 160 W/m². Filter 2 is at 30° to Filter 1. Filter 3 is at 30° to Filter 2. Calculate the intensity after each filter.
2. Interpret a graph — intensity vs analyser angle
A student rotated an analyser through 0° to 90° and measured the transmitted intensity for plane-polarised incident light of intensity I0 = 100 W/m². The graph below shows the results. 7 marks
Figure 1. Transmitted intensity vs analyser angle for plane-polarised light (I0 = 100 W/m²). Curve follows I = I0 cos2 θ. Illustrative data.
2.1 Describe the shape of the curve and explain, using Malus’s Law, why the transmitted intensity decreases from 100 W/m² at 0° to 0 W/m² at 90°. 2 marks
2.2 From the graph, read off the transmitted intensity at θ = 60° and verify your reading using Malus’s Law. 2 marks
2.3 A student claims the graph would look the same if the incident light were unpolarised rather than plane-polarised. Identify and correct the flaw in this reasoning. 3 marks
3. Compare polarised and unpolarised light across five features
Complete the two-column table below. For each feature, write a concise description contrasting the two types of light. 10 marks (1 per cell)
| Feature | Unpolarised light | Plane-polarised light |
|---|---|---|
| Electric field direction | ||
| Produced by | ||
| Intensity after one polariser | ||
| Behaviour between crossed polarisers | ||
| Application |
4. Predict and justify — the three-polariser scenario
Unpolarised light of intensity I0 passes through Polariser A (vertical axis). Two crossed polarisers A and C are placed 90° apart so that no light emerges. Polariser B is then inserted between A and C with its axis at 45° to A. 5 marks
4.1 Predict qualitatively (more, less, or same) how the transmitted intensity with B inserted compares to the crossed-polarisers-only case. Justify your prediction without yet doing the calculation. 2 marks
4.2 Calculate the transmitted intensity after each of the three polarisers in terms of I0. Show all steps. 3 marks
Q1 — Malus’s Law calculations
1.1 I = 60 × cos2 45° = 60 × 0.5 = 30 W/m².
1.2 I = I0/2 = 120/2 = 60 W/m².
1.3 I = 80 × cos2 30° = 80 × (√3/2)2 = 80 × 0.75 = 60.0 W/m².
1.4 After polariser: I1 = 200/2 = 100 W/m². After analyser at 60°: I = 100 × cos2 60° = 100 × 0.25 = 25 W/m².
1.5 I/I0 = 20/40 = 0.5. cos2 θ = 0.5, so cos θ = 1/√2, therefore θ = 45°.
1.6 After F1: I1 = 160/2 = 80 W/m². After F2 (30°): I2 = 80 × cos2 30° = 80 × 0.75 = 60 W/m². After F3 (30° to F2): I3 = 60 × cos2 30° = 60 × 0.75 = 45 W/m².
Q2.1 — Shape of curve
The curve decreases smoothly and non-linearly (following a cosine-squared shape) from 100 W/m² at 0° to 0 W/m² at 90°. By Malus’s Law, I = I0 cos2 θ: at 0° the analyser axis is aligned with the polarisation, so all light is transmitted; at 90° the axes are perpendicular (crossed), and cos2 90° = 0, so no light is transmitted.
Q2.2 — Reading at 60°
From the graph: approximately 25 W/m². Verification: I = 100 × cos2 60° = 100 × 0.25 = 25 W/m². Reading and calculation agree.
Q2.3 — Flaw with unpolarised claim
Flaw: if the incident light were unpolarised, the first filter encountered would act as a polariser, not just an analyser, and the starting intensity would be halved before Malus’s Law applies [1]. The curve shape described by the graph assumes plane-polarised light of a fixed intensity I0 arriving at the analyser. With unpolarised incident light, the whole curve would be scaled down by a factor of 2 (starting at 50 W/m² at 0°, not 100 W/m²) [1]. The student must distinguish the “polariser” (first filter, halving the intensity) from the “analyser” (second filter, to which Malus’s Law is applied) [1].
Q3 — Compare and contrast table
Electric field direction: Unpolarised: oscillates in all directions perpendicular to propagation with equal probability. Plane-polarised: oscillates in only one specific direction perpendicular to propagation.
Produced by: Unpolarised: natural sources (Sun, incandescent lamps) produce unpolarised light. Plane-polarised: produced by passing through a polarising filter, or by reflection at Brewster’s angle, or by scattering.
Intensity after one polariser: Unpolarised: I = I0/2 (half the incident intensity transmitted). Plane-polarised: I = I0 cos2 θ (depends on alignment angle θ).
Behaviour between crossed polarisers: Unpolarised: no light emerges; the first polariser transmits half, the second (crossed to first) blocks all of that. Plane-polarised: if the polarisation direction is aligned with the first, it passes through but is blocked by the crossed second; if already at 90° to the second, no light passes; if aligned with neither, Malus’s Law applies at the second filter.
Application: Unpolarised: standard illumination (light bulbs, sunlight for general lighting). Plane-polarised: polarising sunglasses, LCD displays, 3D cinema, photoelasticity.
Q4.1 — Qualitative prediction
The transmitted intensity with B inserted is greater than with only crossed polarisers (which transmit zero) [1]. Polariser B “re-polarises” the light at an intermediate angle (45°), so the light arriving at C is no longer at 90° to C’s axis; some light can now pass through C [1].
Q4.2 — Calculation of three-polariser chain
After A (first polariser from unpolarised): I1 = I0/2 [1]. After B (45° to A): I2 = (I0/2) × cos2 45° = (I0/2) × ½ = I0/4 [1]. After C (45° to B, since B is at 45° and C is at 90° from A): I3 = (I0/4) × cos2 45° = (I0/4) × ½ = I0/8 [1]. Final transmitted intensity = I0/8.