AQA AS Level Physics Paper 2, June 2024: Question 4
8 marks · Hard difficulty · Short Answer
Deduce the relationship between air gap thickness and wavelength for bright fringes, determine foil thickness using fringe separation, and deduce the effect of water filling the air gap on fringe separation.
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Question text
04 Figure 13 shows an arrangement used to determine the thickness of metal foil using
interference of light.
Figure 13
Two thin glass plates are separated by the foil at one end. Monochromatic light is
incident on the glass plates from above. A series of bright fringes is observed when
viewed from above, as shown in Figure 14.
Figure 14
Figure 15 shows part of the arrangement where a bright fringe occurs.
The angle between the two plates caused by the foil in Figure 13 is very small. This
allows the following approximations to be made for Figure 15:
• the plates are parallel to each other
• the light rays travel perpendicular to both plates.
Figure 15
Some of the incident light is reflected at O. The remainder of the light is transmitted
through the air gap and is reflected at A.
The reflected light from A combines at O with the reflected light from O.
At O, there is a phase difference between the reflected light from A and the reflected
light from O.
This phase difference is caused by:
• the path difference due to the air gap
• the reflection at A.
At A the phase of the light that is reflected is changed by 180°.
At O there is no change to the phase of the light that is reflected or that is transmitted.
The thickness OA of the air gap is d.
04.1 Deduce the relationship between d and the wavelength λ of light that produces a
bright fringe.
[3 marks]
Figure 16 shows a small part of the fringe pattern as viewed from above.
Figure 16
The distance between the centre of one bright fringe and the centre of the next bright
fringe is s.
The thickness of the foil is t and the length of each glass plate up to the edge of the
foil is l as shown in Figure 17.
Figure 17
It can be shown that
t λ
21=
l 2s
04.2 Determine t using Figure 16.
λ = 590 nm
l = 6.0 cm
*20* [2 marks]
t = m
04.3 The space between the plates is now filled with water. The same light source is used,
and t and l remain the same as before.
Deduce how the distance s will change when water fills the space between the plates.
refractive index of water = 1.3
[3 marks]
END OF SECTION B
Section C
Each of Questions 05 to 34 is followed by four responses, A, B, C and D.
*21* For each question select the best response.
Only one answer per question is allowed.
For each question, completely fill in the circle alongside the appropriate answer.
CORRECT METHOD WRONG METHODS
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If you wish to return to an answer previously crossed out, ring the answer you now wish to select
as shown.
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Mark scheme
Show the mark scheme
Question Answers Additional Comments/Guidance Mark AO
04.1 Max 2 from: 1 2 3 2 × AO2
links constructive interference as cause of bright fringe a 1 × AO3
correct reference to zero phase difference or ‘in phase’ b
path difference = 2d OR (n + 0.5)λ seen c
2d = (n ± 0.5)λ 3
04.2 1 Expect s = 0.45 mm. Condone POT errors. 2 1 × AO2
determines s from 1 All substitutions must be in consistent units of
11 1 × AO3
length.
OR 1 Must be some attempt to determine s. Reject
t λ use of 5.0 mm.
substitutes their s into = with values for l and λ 1
l 2s
3.9 × 10−5 (m)
04.3 Allow either 1 or 2 if there is no reference to frequency, but If no other mark, credit use of 1.3 e.g. cwater = 2.3 × 3 1 × AO1
not both: 108 m/s.
1 × AO2
1 × AO3
wave speed decreases (and frequency is constant) 1
2 May see calculated λwater = 454 nm
(frequency is constant so) wavelength decreases 2
3 Only awarded if values of both t and l are stated,
t λ
s will decrease, with reference to = 3 or a clear reference to both of them being constant.
l 2s
3 Allow ecf from an incorrect 2
3 Reject references to e.g. double-slit equation.
Total 8
How to answer it
Interference of Light Using Thin Glass Plates
What this question tests
This question assesses your understanding of wave superposition, path difference, phase changes upon reflection, and fringe spacing in wedge interference. You will need to combine optical path conditions for constructive interference with geometric approximations to solve multi-step analytical and mathematical problems.
Part 04.1: Deriving the Fringe Condition
Deduce the relationship between d and wavelength lambda for a bright fringe. (3 marks)
✅ Correct Answer & Mark Breakdown
- Mark 1: Links constructive interference as the cause of a bright fringe.
- Mark 2: Recognises that waves are in phase or have zero phase difference considering reflections.
- Mark 3: States the final relationship: 2d = (n ± 0.5)λ (or equivalent path difference = 2d).
💡 Key Knowledge
- Light reflecting at boundary A (glass-to-air transition) undergoes a 180° (half-cycle) phase change.
- Light reflecting at boundary O (air-to-glass transition) undergoes no phase change.
- To achieve constructive interference (bright fringe), the total phase difference between the two reflected rays must be an odd multiple of 180° to compensate for the initial 180° phase shift.
🧠 Exam Technique
Carefully break down the contributions to phase difference: the air gap introduces a path difference of 2d (down and back), while the reflection at A introduces an extra half-wavelength shift ( 0.5λ ).
❌ Common Errors
Students frequently forget the 180° phase change at boundary A , leading them to write 2d = nλ (the condition for destructive interference in this specific setup instead).
Part 04.2: Calculating Foil Thickness
Determine l using Figure 16. (λ = 590 nm, l = 6.0 cm) (2 marks)
📐 Step-by-Step Calculation
- Step 1: Determine fringe spacing s .
Looking at Figure 16, 5.0 mm spans across 11 fringes (count the spaces between centres).
s = 5.0 mm / 11 = 4.545 × 10⁻⁴ m - Step 2: Use the given formula.
t / l = λ / (2s) rearranged gives t = (λ × l) / (2s) - Step 3: Substitute values with consistent units.
λ = 590 × 10⁻⁹ m , l = 6.0 × 10⁻² m
t = (590 × 10⁻⁹ × 0.060) / (2 × 4.545 × 10⁻⁴) - Step 4: Final Answer.
t = 3.89 × 10⁻⁵ m (accept 3.9 × 10⁻⁵ m )
🧠 Exam Technique & Counting Traps
When counting fringes from a diagram, always count the number of intervals (spaces), not the vertical lines/peaks. Here, measuring 5.0 mm across 11 fringe spaces is crucial.
❌ Common Errors
Dividing by the wrong number of fringes, or mixing units between millimetres, nanometres, and metres without converting them cleanly.
Part 04.3: Effect of Water Medium
Deduce how the distance s will change when water fills the space between the plates. (Refractive index of water = 1.3) (3 marks)
✅ Correct Answer & Mark Breakdown
- Mark 1: Wave speed decreases in water ( c = c₀ / n ) while frequency remains constant.
- Mark 2: Consequently, the wavelength λ of the light decreases inside the water gap ( λ_water = λ / n ).
- Mark 3: Using t / l = λ / (2s) , since t , l , and the new smaller λ dictate terms, fringe spacing s must decrease.
💡 Key Knowledge & Examiner Insights
- When light enters an optically denser medium ( n > 1 ), its speed drops proportionally to the refractive index.
- Frequency f is a property of the source and never changes when passing between media.
- Since c = fλ , a drop in speed at constant frequency forces the wavelength λ to shrink. Looking at s = (λl)/(2t) , a smaller λ directly results in a smaller fringe separation s .
❌ Common Errors
Top-level responses explicitly tied speed changes to wavelength changes via constant frequency. Weaker answers incorrectly cited standard double-slit formulas or claimed frequency changes inside water.
Topics
Physics · 3.3 Waves
Question and mark scheme from the AQA AS Level Physics examination, Paper 2, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.