IB · C · extended response
A fringe investigation that tests two hypotheses
Wave behaviour · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 1
- Answer
- Complete
Fringe spacing versus screen distance
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Mean fringe spacing / mm
Horizontal axis (x): D / m
Read the exact data as a table
| D / m | Mean fringe spacing / mm |
|---|---|
| 0.5 | 1 |
| 1 | 2 |
| 1.5 | 3 |
| 2 | 4 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Two coherent slits of separation 0.30 mm illuminate a screen. For each screen distance D, the separation between the centres of the first and eleventh bright fringes is measured and divided by ten. These mean fringe-spacing data are supplied. The small-angle condition applies. Supplied model: Use fringe spacing w = λD/d under the small-angle approximation. Supplied data (D / m; Mean fringe spacing / mm): (0.5, 1), (1, 2), (1.5, 3), (2, 4).
- (a)
Infer the light wavelength and predict fringe spacing if slit separation is doubled at D = 1.5 m.
3 marks - (b)
Explain how the method reduces relative reading uncertainty, and design a check that the pattern is interference rather than a screen ruler artefact.
3 marks
Ready to self-mark?Reveal the detailed answer guide
(a)
Infer the light wavelength and predict fringe spacing if slit separation is doubled at D = 1.5 m.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
For two-slit interference, w = λD/d. The slope w/D is 2.0 mm m⁻¹ = 0.0020.
- 2
λ = slope×d = 0.0020(0.00030) = 6.0×10⁻⁷ m.
- 3
Doubling d halves w at fixed λ and D. At 1.5 m, the original 3.0 mm spacing becomes 1.5 mm.
(b)
Explain how the method reduces relative reading uncertainty, and design a check that the pattern is interference rather than a screen ruler artefact.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
First-to-eleventh fringe centres contain ten intervals, not eleven. A 30 mm span is less sensitive to a fixed endpoint error than one 3 mm interval.
- 2
Keep wavelength and slit separation fixed while changing D; proportional spacing supports the wave model. Then change d while keeping D fixed and test the inverse relationship.
- 3
Record repeat endpoint readings, keep the screen perpendicular to the central ray and shield stray light. A ruler artefact would not track both controlled physical changes.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- For two-slit interference, w = λD/d. The slope w/D is 2.0 mm m⁻¹ = 0.0020.
- λ = slope×d = 0.0020(0.00030) = 6.0×10⁻⁷ m.
- Doubling d halves w at fixed λ and D. At 1.5 m, the original 3.0 mm spacing becomes 1.5 mm.
- First-to-eleventh fringe centres contain ten intervals, not eleven. A 30 mm span is less sensitive to a fixed endpoint error than one 3 mm interval.
- Keep wavelength and slit separation fixed while changing D; proportional spacing supports the wave model. Then change d while keeping D fixed and test the inverse relationship.
- Record repeat endpoint readings, keep the screen perpendicular to the central ray and shield stray light. A ruler artefact would not track both controlled physical changes.
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