AQA AS Level Physics Paper 2, June 2023: Question 4
10 marks · Medium difficulty · Short Answer
Analyze refraction and total internal reflection of light in a glass block refractometer to determine sugar solution concentration.
Practise this questionQuestion
Question text
04 Figure 12 shows a type of refractometer.
A semi-circular glass block is arranged so that its semi-circular faces are vertical.
A drop of liquid is placed at the centre of the flat horizontal surface of the block.
Figure 12
Light enters the block through the curved surface and is incident on the midpoint of
the horizontal surface at angle of incidence θ.
Light that reflects at the glass–liquid boundary is detected on a screen that lies
parallel to the horizontal surface.
04.1 Explain why the light ray in Figure 12 does not change direction as it enters the block.
[1 mark]
04.2 The refractometer is calibrated using a drop of liquid.
When θ = 15°, light is partially refracted at the glass–liquid boundary.
Calculate the angle of refraction at this boundary.
refractive index of glass block = 1.84
refractive index of liquid = 1.33
[2 marks]
16 = °
angle of refraction
The refractometer is used to determine the critical angle θc at the glass–liquid
boundary.
*15* Figure 13 shows dimensions of the arrangement.
Figure 13
The intensity of the light ray on the screen is observed as θ is increased from 15°.
When θ = θc the intensity of the light ray is seen to increase sharply at a point T on the
screen.
The distance between the left-hand edge of the screen and T is x.
04.3 Explain why the intensity of the light ray on the screen increases at T.
[2 marks]
04.4 The liquid is replaced with a drop of sugar solution.
The refractive index of the sugar solution is greater than 1.33
Deduce how this change affects the position at which the sharp increase in intensity is
observed on the screen.
[2 marks]
*16* 18
04.5 The refractometer in Figure 13 is used to determine the concentration of a sugar
solution.
Figure 14 shows the variation of refractive index with concentration of sugar solution.
Figure 14
For a drop of a particular sugar solution, x = 69 mm.
Determine the percentage concentration of the sugar solution.
refractive index of glass block = 1.84
[3 marks]
percentage concentration =
END OF SECTION B
Section C
Each of Questions 05 to 34 is followed by four responses, A, B, C and D.
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/Guidance Mark AO
04.1 ray is incident along/at the normal owtte Allow perpendicular/90° to the surface/block. 1 AO1
Allow ‘angle of incidence is 0°’.
Reject ‘towards the normal’.
04.2 correct use of Snell’s law Expect to see 1.84 sin15 = 1.33 sinθ 2 AO2
21° Calculator value for θ is 20.9814°
04.3 (intensity increases because) total internal reflection occurs Reject ‘more total internal reflection occurs’. 2 AO2
owtte
‘There is more reflection’ is insufficient.
MP2 must be in terms of reflected, not refracted,
idea that the lower intensity to left of T is due to partial light.
reflection
Allow answers in terms of x.
04.4 T will move to right OR x will increase MP1 requires some relevant justification 2 AO2
𝑛𝑛2
16 (because) critical angle will increase e.g. ratio will be closer to 1.
𝑛𝑛1
04.5 uses tan θ = 69 to calculate correct θc to 2 sf min θ 3 AO3
MP1: Calculator value for c is 48.99091°
𝑛𝑛
uses sin θc = 1.84 to determine refractive index n 1.388.
MP2: Allow ecf from MP1. Expect =
uses their refractive index to at least 4 sf to obtain the MP3: Allow ecf from MP2. Expect 33.5%.
concentration
Total 10
How to answer it
Refraction and Total Internal Reflection Study Guide
What this question tests
This question assesses your understanding of ray optics, Snell's Law of refraction, the conditions required for total internal reflection (TIR), critical angle calculations, and interpreting graphical data to determine solution concentrations. You will need strong geometric reasoning combined with core wave equations.
Part 04.1: Normal Incidence
Explain why the light ray in Figure 12 does not change direction as it enters the block. [1 mark]
✅ Correct Answer
The ray is incident along (or at) the normal to the surface.
💡 Key Knowledge
- When angle of incidence θ = 0° relative to the normal, sin(0) = 0 .
- Snell's law ( n₁ sinθ₁ = n₂ sinθ₂ ) yields sinθ₂ = 0 , meaning no deviation occurs.
❌ Common Errors
- Stating the ray travels "towards the normal" (it is already *on* the normal line).
- Vague statements like "it hits it straight" without referencing the normal.
Part 04.2: Applying Snell's Law
Calculate the angle of refraction when θ = 15° . ( n_glass = 1.84 , n_liquid = 1.33 ) [2 marks]
📐 Step-by-Step Calculation
- State Snell's Law: n₁ sinθ₁ = n₂ sinθ₂
- Substitute values: 1.84 × sin(15°) = 1.33 × sinθ₂
- Rearrange: sinθ₂ = (1.84 × sin(15°)) / 1.33 = 0.3582
- Inverse sine: θ₂ = sin⁻¹(0.3582) = 20.98°
Final Answer: 21° (to 2 sig figs)
🧠 Exam Technique
Always show your substitution explicitly before rearranging. Ensure your calculator is set to Degrees mode, not Radians!
Part 04.3: Total Internal Reflection & Intensity
Explain why the intensity of the light ray on the screen increases at T when θ = θ_c . [2 marks]
✅ Correct Answer
- Total internal reflection (TIR) occurs at the boundary.
- All light is reflected rather than split, so lower intensity to the left of T is due to previous partial reflection.
❌ Common Errors
- Writing "more total internal reflection occurs" (TIR is a binary state: it either occurs or it doesn't).
- Discussing refracted light instead of focusing on the sudden transition to total reflection.
Part 04.4: Effect of Higher Refractive Index Liquid
Deduce how replacing the liquid with a sugar solution ( n > 1.33 ) affects the position of the sharp increase in intensity. [2 marks]
✅ Correct Answer
T will move to the right (or x will increase) because the critical angle will increase.
💡 Key Knowledge
The critical angle formula is sinθ_c = n_liquid / n_glass . If n_liquid increases, the ratio n_2 / n_1 gets closer to 1, causing θ_c to increase. A larger critical angle requires a larger incident angle, pushing the reflection point further along the screen.
Part 04.5: Determining Concentration from Geometry
Given x = 69 mm , height dimensions from Figure 13 ( 60 mm vertical, 50 mm offset), and n_glass = 1.84 , determine the percentage concentration. [3 marks]
📐 Step-by-Step Calculation
- Find critical angle using geometry:
From Fig 13, total distance from center is 50 mm + x = 50 + 69 = 119 mm . Wait, check dimensions: vertical height is 60 mm (from center to screen base offset). Using trigonometry, tan(θ_c) = opposite / adjacent = 69 / 60 = 1.15 ?
*Note from mark scheme:* Uses tanθ_c = 69 / 60 gives θ_c = 48.99° . - Calculate refractive index of solution:
sinθ_c = n_liquid / 1.84
n_liquid = 1.84 × sin(48.99°) = 1.388 - Read concentration from graph (Figure 14):
Cross-referencing n = 1.388 on the calibration curve gives approximately 33.5% concentration.
Final Answer: 33.5% (accept 33% to 34%)
🧠 Top-Level Exam Strategy
This is a synoptic AO3 question combining trigonometry, optics, and data interpretation. Carry forward intermediate values to at least 4 significant figures to avoid rounding errors before reading off the final graph axis.
Topics
Physics · 3.3 Waves
Question and mark scheme from the AQA AS Level Physics examination, Paper 2, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.