HSC Exam Practice

Sample response. Polarisation is the restriction of the oscillations of a transverse wave to a single plane perpendicular to the direction of propagation. Only transverse waves can be polarised, because their oscillations are perpendicular to the direction of propagation, allowing a filter to select one perpendicular direction and absorb others. Longitudinal waves (such as sound in air) cannot be polarised because their oscillations are parallel to the direction of propagation — there is no perpendicular plane to restrict.

Marking notes. 1 mark for a correct definition of polarisation (restriction of oscillations to one plane, transverse waves only); 1 mark for stating transverse waves can be polarised and why (perpendicular oscillations); 1 mark for explaining why longitudinal waves cannot be polarised (parallel oscillations, no perpendicular plane to restrict).

Sample response. Malus’s Law: I = I₀ cos² θ, where I is the transmitted intensity (W/m²), I₀ is the incident intensity of already plane-polarised light (W/m²), and θ is the angle between the transmission axis of the analyser and the polarisation direction of the incident light. Calculation: I = 80 × cos² 60° = 80 × (0.5)² = 80 × 0.25 = 20 W/m².

Marking notes. 1 mark for correct statement of Malus’s Law (I = I₀ cos² θ); 1 mark for correctly defining I₀ and θ; 1 mark for correct substitution (cos² 60° = 0.25); 1 mark for correct final answer (20 W/m²) with unit.

Sample response. When sunlight reflects from the horizontal surface of a lake, the reflected light is partially (or, at Brewster’s angle, completely) plane-polarised with the electric field oscillating parallel to the surface — that is, horizontally. This horizontally polarised glare is the dominant cause of discomfort. Polarising sunglasses have lenses with a vertical transmission axis, which absorbs the horizontal component and blocks the horizontally polarised glare, while allowing vertically polarised or randomly oriented components of other light to pass through with reduced intensity.

Marking notes. 1 mark for identifying that reflected light from water is partially/completely plane-polarised parallel to the surface (horizontally polarised); 1 mark for correctly stating that the sunglass lens has a vertical transmission axis that absorbs horizontal polarisation; 1 mark for linking this to the reduction of glare by blocking the horizontally polarised reflected component.

Sample response. Application 1: LCD screens. Two crossed polarisers sandwich a layer of liquid crystals. Without voltage, liquid crystals rotate the polarisation of light so it can pass through the second polariser (pixel bright). When voltage is applied, the liquid crystals reorient and no longer rotate the polarisation, so the crossed second polariser blocks it (pixel dark). Property exploited: the ability of crossed polarisers to block light completely, and the control of polarisation rotation by liquid crystals. Application 2: 3D cinema. Two images are projected simultaneously, each with orthogonal polarisations (e.g. one horizontally, one vertically polarised). Each eye’s glasses contain a matching polarising filter, so each eye sees only its intended image, creating the 3D depth effect. Property exploited: the ability of two orthogonal polarisation states to be independently selected by matching filters.

Marking notes. 1 mark per application correctly named (2 marks); 1 mark per specific polarisation property correctly identified and linked to the application mechanism (2 marks). Accept photoelasticity (stress analysis using crossed polarisers and birefringent materials) or camera filters as alternatives.

Sample response. The claim is mostly correct but has one exception: if the third filter is inserted at 0° or 90° to either of the existing crossed filters, its effect is trivially zero or the same as removing it. Specifically, if the third filter is at 0° to the first polariser (same axis as P₁), it is also at 90° to the second (P₃), so no additional light passes. Similarly, if at 90° to P₁, it is at 0° to P₃, passing all light from P₁ but then blocked by P₃. At any angle strictly between 0° and 90° (exclusive), the intermediate filter does allow some light through, with a maximum of I₀/8 at 45°.

Marking notes. 1 mark for identifying the student’s claim is incorrect (or only partially correct) with a clear reason; 1 mark for correctly identifying 0° and/or 90° as the angles at which no additional light passes; 1 mark for explaining why (at 0° to P₁, the middle filter is aligned with P₁ but crossed with P₃; at 90° it is crossed with P₁ directly). Accept also the statement that only angles strictly between 0° and 90° (exclusive) produce non-zero transmitted intensity.

Sample response. At Brewster’s angle, the reflected beam is completely plane-polarised with its electric field oscillating parallel to the surface (the s-polarisation component only). The transmitted (refracted) beam contains both polarisation components but is partially polarised, with the p-component (perpendicular to the surface) dominating because the s-component has been entirely reflected. Brewster’s angle θ_B is related to the refractive indices by tan θ_B = n₂/n₁, where n₁ is the refractive index of the incident medium (usually air, n₁ = 1) and n₂ is the refractive index of the refracting medium (e.g. glass or water).

Marking notes. 1 mark for correctly stating that the reflected beam at Brewster’s angle is completely plane-polarised parallel to the surface; 1 mark for correctly describing the transmitted beam as partially polarised (or dominated by the p-polarisation component); 1 mark for correctly stating the Brewster’s angle formula tan θ_B = n₂/n₁ and defining symbols.

Sample response (a) — predicted intensities. After P₁ (from unpolarised): I₁ = 320/2 = 160 W/m². After P₂ (20° to P₁): I₂ = 160 × cos² 20° = 160 × 0.8830 = 141.3 W/m². After P₃: I₃ = 141.3 × 0.8830 = 124.7 W/m². After P₄: I₄ = 124.7 × 0.8830 = 110.1 W/m². After P₅: I₅ = 110.1 × 0.8830 = 97.2 W/m². [1 mark for P₁ explicitly; 1 mark for P₂ explicitly with correct cos² 20°; 1 mark for remaining filters correctly propagated.]

Sample response (b) — systematic trend. The recorded values are consistently slightly lower than the predicted values by a roughly constant fraction at each stage (e.g. ~1 W/m² lower at each filter after P₂). This systematic trend suggests a consistent source of energy loss at each filter beyond the ideal Malus’s Law prediction, such as partial absorption within the filter material (imperfect transmission even for the aligned component), surface reflections at each filter interface, or a slight misalignment of filter axes [1 for identifying systematic pattern; 1 for a valid physical explanation; 1 for using specific data values to support the claim].

Sample response (c) — P₁ + P_X at 80°. After P₁: I₁ = 160 W/m². After P_X (80° to P₁): I = 160 × cos² 80° = 160 × 0.0302 = 4.8 W/m², compared to ~97 W/m² with the full five-filter chain [1]. The two configurations give very different intensities even though P₅ ends up at 80° cumulative angle from P₁, because in the five-filter chain each filter only needs to rotate the polarisation direction by 20° (giving cos² 20° = 0.883), whereas P_X is applied directly at 80° to the polarised beam (cos² 80° = 0.030). The intermediate filters progressively “re-polarise” the beam so that the angle presented to each subsequent filter is small, preventing the severe cos² 80° suppression that occurs in the two-filter case [1].

Sample response. Polarisation is widely regarded as the most direct evidence that light is a transverse wave, and this claim is well supported. The ability to polarise light means that there must exist perpendicular directions to the propagation axis in which the oscillation can be selectively restricted. Only transverse waves have oscillations in a plane perpendicular to propagation; if light were a longitudinal wave, like sound, every oscillation would be along the direction of travel and there would be no directional component to select with a filter. The direct observational test — passing sunlight through two polarising filters rotated to 90° and observing zero transmission — is reproducible, quantitatively described by Malus’s Law, and has no equivalent explanation in a longitudinal wave model. This is why polarisation is considered uniquely direct evidence for the transverse nature. By contrast, diffraction and interference (e.g. Young’s double-slit experiment) demonstrate that light has wave properties but do not distinguish transverse from longitudinal waves: sound waves also diffract and interfere. Therefore these phenomena are evidence for the wave nature of light in general, but not specifically for its transverse character. Refraction (Snell’s Law) similarly applies to both wave types and does not indicate transversality. To challenge the polarisation claim, one would need to show that a longitudinal wave model could also produce the observed directional blocking behaviour — which is physically impossible, since longitudinal waves have no perpendicular oscillation component. A specific historical example is the Malus experiment (1808), where Étienne-Louis Malus discovered that light reflected from glass at certain angles was polarised, establishing the directional nature of light oscillations; this was directly incompatible with Newton’s corpuscular (particle) theory and confirmed the transverse wave model of Young and Fresnel. A second example is 3D cinema: the ability to project two simultaneously visible images with orthogonal polarisations, each selectable by the corresponding polarised lens, only works because light has two independent transverse oscillation components — a fact with no longitudinal-wave analogue. In conclusion, polarisation does provide the most direct evidence for the transverse nature of light, because it uniquely identifies the existence of perpendicular oscillation directions; diffraction, interference, and refraction cannot make this specific distinction.

Marking criteria (7 marks). 1 = correctly explains what polarisation reveals about the transverse nature of light (perpendicular oscillation direction, unique to transverse waves). 1 = correctly contrasts with longitudinal waves (no perpendicular component, cannot be polarised). 1 = correctly analyses the evidence that would challenge the claim (would need to show a longitudinal model that produces directional blocking — impossible). 1 = correctly assesses diffraction and/or interference as evidence for wave nature generally but not transverse nature specifically (applies to longitudinal waves too). 1 = correctly assesses refraction as not specific to transverse waves. 1 = first named specific experiment or example correctly cited and explained (Malus/Brewster’s angle experiment, Young’s double-slit showing wave nature, 3D cinema, or photoelasticity). 1 = evaluative judgement: explicitly concludes that polarisation is uniquely direct evidence for transverse nature and integrates comparison with other phenomena. Award marks for any coherent evaluative structure that addresses the claim, supports it with specific evidence, considers alternatives fairly, and reaches an explicit conclusion.