AQA A-Level Physics Paper 1, June 2022: Question 6
6 marks · Medium difficulty · Short Answer
Analyze refraction and total internal reflection of a light ray passing through two transparent prisms with different refractive indices.
Practise this questionQuestion
Question text
06 Two transparent prisms A and B of different refractive indices are placed in contact to
produce a rectangular block.
Figure 12 shows the path of a ray, incident normally on A, refracting as it crosses the
boundary between the prisms.
Figure 12
06.1 Explain how the path of the ray shows that the refractive index of A is greater than the
refractive index of B.
[1 mark]
06.2 Show that the angle of refraction of the ray in B is about 60°.
[2 marks]
06.3 Draw, on Figure 12, the path of the ray immediately after it reaches P.
Justify your answer with calculations.
[3 marks]
END OF SECTION A
Section B
Each of Questions 07 to 31 is followed by four responses, A, B, C and D.
*23* 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
If you want to change your answer you must cross out your original answer as shown.
If you wish to return to an answer previously crossed out, ring the answer you now wish to select
as shown.
You may do your working in the blank space around each question but this will not be marked.
Do not use additional sheets for this working.
Mark scheme
Show the mark scheme
Question Answers Additional comments/Guidelines Mark AO
As angle of refraction greater than angle of incidence with
06.1 1 AO1
reference to Snell’s law / n = sin i sin r
OR
light bends away from normal when it speeds up ✓
(Therefore nA>nB)
– A-LEVEL PHYSICS – –
06.2 Calculation of angle of incidence = 90 −43 = 47 ✓ MP1 may be seen on diagram 2 2 x AO2
Use of Snell’s law to give angle of refraction
Calculator value: 61.357 115 7
= 61(.4) cao ✓
06.3 Condone 77 but not 75 3 3 x AO3
No ecf from MP1 to MP2
Use of sin c = 1/n to get c = 48
OR Allow a range:
i = (180 – 43 − 61.4 =) 76 ✓
Other calculation and i greater than c therefore tir ✓
Ray reflecting off P to land where the top of the n of ‘not to
scale’ label meets the glass surface ✓
Total 6
How to answer it
Refraction and Total Internal Reflection in Prisms
What this question tests
This multi-part optics question tests your understanding of wave behavior at boundaries between media of different refractive indices. Key competencies assessed include applying Snell's law, interpreting ray diagrams using geometry, calculating critical angles, and determining whether total internal reflection (TIR) occurs.
Explaining Refractive Index Differences via Ray Paths
✅ Correct Answer
The ray bends away from the normal as it crosses the boundary from A into B. Since the angle of refraction is greater than the angle of incidence, Snell's law indicates that the speed of light increases, meaning the refractive index of A is greater than that of B ( nA > nB ).
💡 Key Knowledge
- Snell's Law: n₁ sin(θ₁) = n₂ sin(θ₂)
- When entering a optically less dense medium ( n₂ < n₁ ), light speeds up and bends away from the normal.
Calculating the Angle of Refraction
📐 Step-by-Step Calculation
- Find the angle of incidence (i) at the boundary:
The normal is perpendicular to the boundary (90°). From the diagram, the angle between the prism edge and the ray is 43°.
i = 90° - 43° = 47° - Apply Snell's law to find angle of refraction (r):
n₁ sin(i) = n₂ sin(r)
1.62 × sin(47°) = 1.35 × sin(r)
sin(r) = (1.62 × 0.7313) / 1.35 = 0.8776
r = sin⁻¹(0.8776) = 61.4° (or 61°)
❌ Common Errors
- Using 43° directly as the angle of incidence instead of calculating the angle relative to the normal (90° - 43°).
- Inverting refractive indices in Snell's law (e.g., putting 1.35 on the left and 1.62 on the right).
Determining Total Internal Reflection at Point P
🧠 Exam Technique & Calculations
- Calculate Critical Angle ( c ) or Incidence Angle at P:
Using sin(c) = n₂ / n₁ (where surrounding medium is air, n₂ = 1 ):
sin(c) = 1 / 1.62 → c = 38° (or using boundary with B: sin(c) = 1.35 / 1.62 → c = 56.4° depending on interface tested. Here, interface P is glass-air).
Alternatively, find angle of incidence at surface P using triangle geometry: i = 180° - 43° - 61.4° = 76° . - Compare and Conclude:
Since the angle of incidence ( 76° ) is greater than the critical angle ( 38° ), Total Internal Reflection (TIR) occurs. - Drawing requirement:
Draw a reflected ray obeying the law of reflection (angle of incidence = angle of reflection) hitting the upper boundary right where the top of the letter 'n' of the 'not to scale' label meets the glass surface.
❌ Common Errors to Avoid
- Forgetting to state the condition for TIR ( i > c ) alongside the calculations.
- Inaccurate drawing of the reflected ray—ensure the angle of reflection mirrors the angle of incidence precisely.
Topics
Physics · 3.3 Waves
Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.