HSC Exam Practice

Sample response. Constructive interference occurs when two waves arrive at a point in phase, causing their amplitudes to add together and produce a bright fringe. The path difference condition is: path difference = nλ, where n = 0, 1, 2, … Destructive interference occurs when two waves arrive out of phase by half a wavelength, causing their amplitudes to cancel and produce a dark fringe. The path difference condition is: path difference = (n + ½)λ, where n = 0, 1, 2, …

Marking criteria. 1 mark: defines constructive interference as waves arriving in phase with amplitudes adding (bright fringe). 1 mark: states path difference = nλ. 1 mark: defines destructive interference as waves arriving out of phase by ½λ with amplitudes cancelling (dark fringe). 1 mark: states path difference = (n + ½)λ.

Sample response. Even though both sodium lamps emit the same wavelength (~589 nm), they are incoherent: each lamp is an independent source of light, with atoms emitting photons randomly and with randomly changing phase. The phase relationship between the two lamps is not constant — it fluctuates many millions of times per second. As a result, any interference fringes that form instantaneously shift position so rapidly that the human eye perceives only a uniform average brightness, with no visible fringes.

Marking criteria. 1 mark: identifies that the two lamps are not coherent. 1 mark: explains that their phase relationship fluctuates randomly over time. 1 mark: explains that rapidly shifting fringes average out to uniform brightness, making no stable pattern visible.

Sample response. (a) Doubling λ: Δx = λL/d, so Δx doubles (λ is in the numerator). (b) Halving d: d is in the denominator, so halving d doubles Δx. (c) Decreasing L to one-third: L is in the numerator, so Δx decreases to one-third of the original value.

Marking criteria. 1 mark per correct answer with justification from the formula (3 marks total). Accept any phrasing that correctly identifies the direction and magnitude of change and links it to the relevant variable in Δx = λL/d.

Sample response (a). Δx = λL/d = (633×10⁻⁹ m × 2.50 m) / (0.30×10⁻³ m) = 5.275×10⁻³ m ≈ 5.3 mm.

Sample response (b). Distance to 5th fringe = 5 × Δx = 5 × 5.275 mm = 26.4 mm ≈ 2.6 cm from the central maximum.

Marking criteria. Part (a): 1 mark for correct substitution with unit conversions; 1 mark for correct answer (5.3 mm or equivalent). Part (b): 1 mark for correct method (5 × Δx); 1 mark for correct answer (26.4 mm or 2.64 cm).

Sample response. The central fringe (n = 0) would appear white, because all wavelengths in white light have zero path difference at the centre and all interfere constructively simultaneously. Moving away from the centre, each wavelength forms its own set of fringes with spacing Δx = λL/d. Because longer wavelengths produce larger fringe spacings, the first-order fringes form a spectrum: violet/blue on the inner edge, red on the outer edge. At higher orders, the spectra from adjacent orders overlap and the pattern washes out to white again.

Marking criteria. 1 mark: central fringe is white (all λ constructively interfere at path difference = 0). 1 mark: outer fringes show spectral colours. 1 mark: correct colour order — violet/blue inner, red outermost (red has largest λ so largest Δx). Do not award the third mark if the order is reversed.

Sample response. The single slit acts as a point source of coherent wavefronts. Light diffracts through the narrow single slit, spreading as a single expanding wavefront. This wavefront reaches both double slits simultaneously, ensuring that the two double slits receive light from the same coherent source. Because both double slits are illuminated by the same single wavefront, they maintain the same frequency and a constant phase difference, satisfying the coherence condition required for a stable interference pattern. Without the single slit, light arriving at the double slits from an extended incoherent source (like the sun or a lamp) would illuminate the slits incoherently, preventing a clear, stable pattern.

Marking criteria. 1 mark: single slit creates a coherent point source by diffraction. 1 mark: same wavefront illuminates both double slits, ensuring constant phase relationship. 1 mark: coherence is required for a stable interference pattern; without it, fringes wash out.

Sample response (a). λ = Δx × d / L = (4.26×10⁻³ m × 0.20×10⁻³ m) / 1.80 m = 4.73×10⁻⁷ m = 473 nm. This is in the blue region of the visible spectrum (450–495 nm).

Marking: 1 mark for correctly rearranging the formula; 1 mark for correct unit conversions and substitution; 1 mark for final answer 473 nm with correct colour identification (blue). Accept 470–476 nm.

Sample response (b). In the liquid (n = 1.47): λ_liquid = λ_air/n = 473×10⁻⁹/1.47 = 3.218×10⁻⁷ m = 321.8 nm. New fringe spacing: Δx′ = λ_liquid × L / d = (3.218×10⁻⁷ × 1.80) / (0.20×10⁻³) = 2.896×10⁻³ m ≈ 2.90 mm.

Marking: 1 mark for correctly applying λ_liquid = λ_air/n; 1 mark for correct substitution into Δx formula; 1 mark for final answer ~2.90 mm. Accept 2.88–2.92 mm.

Sample response (c). Light travels more slowly in a denser medium, so its wavelength decreases while its frequency remains unchanged (frequency depends on the source, not the medium). Since Δx = λL/d and λ is in the numerator, a shorter wavelength in the liquid produces a smaller fringe spacing. The fringe spacing decreased from 4.26 mm to ~2.90 mm because the effective wavelength decreased by a factor of n = 1.47.

Marking: 1 mark for stating λ decreases in a denser medium (while frequency is unchanged). 1 mark for linking reduced λ to reduced Δx via the formula.

Sample response (d). Assumption: L (slit-to-screen distance) and d (slit separation) remain unchanged when the apparatus is submerged. This is justified because the physical dimensions of the apparatus (ruler distance, slit slide) do not change significantly in a liquid — the liquid does not affect mechanical dimensions, only the optical wavelength.

Marking: 1 mark for identifying L and/or d unchanged, with brief justification. Accept: assumption that refractive index is uniform throughout the liquid.

Sample response. Young’s double slit experiment (1801) was highly significant because it provided the first quantitative, reproducible experimental evidence that light exhibits wave behaviour through interference — a phenomenon that could not be explained by Newton’s corpuscular (particle) theory, which had dominated physics for over a century. Newton’s model predicted that particles passing through two slits would produce two bright patches on a screen. Young observed instead a series of alternating bright and dark fringes, which is the definitive signature of wave superposition: bright fringes arise from constructive interference (path difference = nλ) and dark fringes from destructive interference (path difference = (n + ½)λ). No particle model can account for cancellation of intensity at dark fringes. Furthermore, Young’s formula Δx = λL/d allowed him to quantitatively predict fringe positions with high accuracy, giving the wave model strong predictive power. The observation that covering one slit destroys the pattern further confirmed that two coherent sources are required for interference, precisely as wave theory demands. However, the experiment’s decisiveness has limits: the full acceptance of the wave model required additional evidence (notably Fresnel’s mathematical wave theory and Foucault’s 1850 measurement showing light travels more slowly in water — as predicted by waves, not particles). Young’s result was also initially resisted by supporters of Newton’s authority. In summary, Young’s experiment was decisive in demonstrating that light can interfere and must have wave properties, but the broader acceptance of the wave model depended on a cumulative body of evidence from multiple independent experiments.

Marking criteria (5 marks). 1 = clearly explains what the particle model predicted and why Young’s observed pattern was inconsistent with it. 1 = correctly explains how the wave model (superposition, constructive/destructive interference) accounts for the alternating fringes. 1 = discusses a specific piece of additional evidence or limitation that shows why the experiment alone was not entirely decisive (e.g. Foucault, Fresnel, initial resistance, or wave-particle duality later). 1 = uses quantitative reasoning or specific reference to the formula Δx = λL/d or path difference conditions. 1 = reaches an explicit evaluative judgement about the significance and decisiveness of the experiment, rather than merely describing it.