Physics • Year 12 • Module 7 • Lesson 3
Interference — Young’s Double Slit
Apply the fringe-spacing equation, interpret experimental data, and reason about how changing apparatus variables affects the interference pattern.
1. Interpret experimental data — comparing four double slit setups
A student performed four double slit experiments, varying one parameter at a time. The table below shows the apparatus settings and measured fringe spacing. Some values are missing. 10 marks
| Exp. | λ (nm) | d (mm) | L (m) | Δx (mm) | Notes |
|---|---|---|---|---|---|
| A | 550 | 0.20 | 1.5 | Green laser; calculate Δx | |
| B | 450 | 0.20 | 1.5 | Blue laser; same apparatus as A | |
| C | 550 | 0.40 | 1.5 | d doubled from Exp. A | |
| D | 550 | 0.20 | 3.0 | L doubled from Exp. A | |
| E | 0.25 | 2.0 | 5.0 | Unknown wavelength; calculate λ |
1.1 Calculate the missing Δx values for Experiments A, B, C and D, and calculate the unknown λ for Experiment E. Show full working including unit conversions. 5 marks (1 per experiment)
1.2 Compare Experiments A and B. Explain why blue light produces narrower fringe spacing than green light in the same apparatus. 2 marks
1.3 Identify the colour of the light in Experiment E (visible spectrum: violet 380–450 nm, blue 450–495 nm, green 495–570 nm, yellow 570–590 nm, orange 590–620 nm, red 620–750 nm) and state which experiment it is most similar to. 1 mark
1.4 A student claims: “Experiment D gives the same fringe spacing as Experiment C because doubling L has the same numerical effect as halving d.” Evaluate this claim using your calculated values. 2 marks
2. Interpret graph — fringe spacing versus slit separation
A student used a green laser (λ = 532 nm) with a fixed screen distance of L = 1.80 m and varied the slit separation d from 0.10 mm to 0.60 mm. The graph below shows the measured fringe spacing. 7 marks
Figure 2.1. Fringe spacing (Δx) vs slit separation (d) for λ = 532 nm (green laser), L = 1.80 m. Illustrative data.
2.1 Describe the shape of the graph and identify the mathematical relationship between Δx and d. 2 marks
2.2 Use the data point at d = 0.20 mm to verify the fringe-spacing equation Δx = λL/d. Show your working. 2 marks
2.3 Estimate from the graph what fringe spacing would be observed if d were increased to 0.80 mm, and explain why such a small fringe spacing might become difficult to measure in a school laboratory. 3 marks
3. Compare and contrast — particle model vs wave model predictions
Complete the table comparing what Newton’s particle model and Young’s wave model predict for the double slit experiment. 8 marks (1 per cell)
| Feature | Particle model (Newton) | Wave model (Huygens / Young) |
|---|---|---|
| Pattern on screen | ||
| Number of bright regions | ||
| Effect of closing one slit | ||
| Does wavelength affect the pattern? | ||
| Is the observed result explained? |
4. Predict and justify — double slit in water
A student performs a double slit experiment in air with green light (λair = 532 nm), d = 0.20 mm, L = 1.50 m, obtaining a fringe spacing of 3.99 mm. The apparatus is then moved into a tank filled with water (refractive index n = 1.33). 5 marks
Useful relationship: the wavelength of light in a medium of refractive index n is λmedium = λair/n.
4.1 Calculate the wavelength of green light in water. 1 mark
4.2 Calculate the new fringe spacing in water. 1 mark
4.3 Predict whether the fringes become wider or narrower in water, and justify your prediction by referencing the fringe-spacing equation. 2 marks
4.4 Explain whether the number of fringes visible on the screen changes when the experiment is moved into water. 1 mark
Q1.1 — Calculations
Exp A: Δx = (550×10−9 × 1.5) / (0.20×10−3) = 4.125×10−3 m = 4.1 mm
Exp B: Δx = (450×10−9 × 1.5) / (0.20×10−3) = 3.375×10−3 m = 3.4 mm
Exp C: Δx = (550×10−9 × 1.5) / (0.40×10−3) = 2.0625×10−3 m = 2.1 mm (half of Exp A)
Exp D: Δx = (550×10−9 × 3.0) / (0.20×10−3) = 8.25×10−3 m = 8.3 mm (double Exp A)
Exp E: λ = Δx × d / L = (5.0×10−3 × 0.25×10−3) / 2.0 = 6.25×10−7 m = 625 nm (red, 620–750 nm)
Q1.2 — Blue vs green fringe spacing
Since Δx = λL/d and L and d are identical for both, the fringe spacing is directly proportional to λ. Blue light (λ = 450 nm) has a shorter wavelength than green (λ = 550 nm), so Δxblue < Δxgreen: the fringes are more closely spaced. [1 for identifying λ in numerator; 1 for correct direction of change]
Q1.3 — Colour identification
λ = 625 nm lies in the red range (620–750 nm). It is most similar to Experiment A (green, 550 nm) in apparatus but produces wider fringes than A due to the longer wavelength.
Q1.4 — Evaluating the student’s claim
The claim is correct in numerical outcome but imprecise in reasoning. Both Exp C (d doubled: Δx halved = 2.1 mm) and Exp D (L doubled: Δx doubled = 8.3 mm) produce the same magnitude factor of 2 applied to Δx, but in opposite directions — C halves the spacing while D doubles it. The claim that they give the “same” fringe spacing is therefore false. [1 for comparing numerical values; 1 for identifying the error — opposite direction effects]
Q2.1 — Shape of graph
The graph shows a hyperbolic (inverse) curve: as d increases, Δx decreases rapidly at first then more slowly. This reflects the inverse proportionality Δx ∝ 1/d from the equation Δx = λL/d. [1 mark: hyperbolic or inverse shape; 1 mark: states inverse proportionality or 1/d relationship]
Q2.2 — Verification at d = 0.20 mm
Δx = λL/d = (532×10−9 × 1.80) / (0.20×10−3) = 4.788×10−3 m ≈ 4.8 mm. This matches the graph value at d = 0.20 mm, confirming the equation. [1 mark for correct substitution; 1 mark for calculated value matching graph]
Q2.3 — Extrapolation to d = 0.80 mm and measurement difficulty
From Δx = λL/d: Δx = (532×10−9 × 1.80) / (0.80×10−3) = 1.197×10−3 m ≈ 1.2 mm. The graph trend (extrapolating the curve) gives approximately 1.2 mm, consistent with the calculation. [1 mark for estimated or calculated value ∼1.2 mm] Such small fringe spacing would be difficult to measure in a school lab because: (1) the fringes are closer together than typical ruler resolution (1 mm) so individual fringes are hard to resolve; (2) diffraction effects and imperfect coherence broaden and blur the fringes. [2 marks: 1 per valid reason]
Q3 — Particle vs wave model comparison
Pattern on screen: Particle: two bright patches (one behind each slit). Wave: multiple alternating bright and dark fringes (interference pattern). Number of bright regions: Particle: two. Wave: many (central maximum + symmetric fringes either side). Effect of closing one slit: Particle: one bright patch disappears. Wave: the interference pattern disappears; a single-slit diffraction pattern appears. Effect of wavelength: Particle: wavelength has no effect on pattern positions. Wave: longer wavelength gives wider fringe spacing (Δx = λL/d). Does it explain observations?: Particle: No — cannot explain multiple fringes or dependence on λ. Wave: Yes — predicts and quantitatively matches all observed features.
Q4.1 — Wavelength in water
λwater = λair / n = 532 / 1.33 = 400 nm.
Q4.2 — Fringe spacing in water
Δxwater = λwater × L / d = (400×10−9 × 1.50) / (0.20×10−3) = 3.00×10−3 m = 3.0 mm.
Q4.3 — Prediction: narrower fringes in water
The fringes become narrower. Water reduces the wavelength of light from 532 nm to 400 nm. Since Δx = λL/d and λ is in the numerator, reducing λ reduces Δx: 3.0 mm < 3.99 mm. [1 mark: correct prediction (narrower); 1 mark: explicit link to λ in numerator of formula]
Q4.4 — Number of fringes
The number of visible fringes increases slightly because the reduced fringe spacing means more fringes fit within the same screen width. However, the overall diffraction envelope also narrows slightly, and the brightness of outer fringes decreases — so in practice the total number of clearly visible fringes does not change dramatically. Accept: more fringes fit on the screen because Δx is smaller. [1 mark]