AQA AS Level Physics Paper 2, November 2021: Question 18
1 mark · Medium difficulty · Multiple Choice
Determine the angle of incidence of a light ray at the right-hand boundary of a rectangular glass block given the angle of refraction and refractive index.
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Question text
18 A ray of light is incident on the internal boundary of a rectangular glass block in air.
Part of the light refracts out of the block at an angle of 30°.
Some of the remaining light reflects within the block to become incident on the right-hand
boundary.
refractive index of glass = 1.48
What is the angle of incidence of the ray at the right-hand boundary?
[1 mark]
A 20°
B 42°
C 48°
D 70°
Mark scheme
Show the mark scheme
18 D 70°
How to answer it
Optics: Refraction & Internal Reflection in a Glass Block
This question assesses your ability to apply Snell's Law of refraction, understand the relationship between angles of incidence and refraction at a boundary, and use basic geometry (angles on a straight line and alternate interior angles) to trace light paths through a rectangular block.
Question 18: Multiple Choice Analysis
Determining the angle of incidence at the right-hand boundary
✅ Correct Answer: D (70°)
The correct option is D. Following a step-by-step calculation using Snell's Law and block geometry yields an angle of incidence of 70° at the right-hand boundary.
💡 Key Knowledge
- Snell's Law: n₁ sin(θ₁) = n₂ sin(θ₂)
- For light entering air from glass, n_glass = 1.48 and n_air = 1.00.
- Angles of incidence and refraction are always measured relative to the normal (perpendicular to the boundary).
- The geometry of a rectangle means internal reflected angles correspond predictably with angles inside the block.
🧠 Exam Technique
Multi-step optics questions require working backwards from the given output. Break the problem into chronological milestones:
- Find the angle inside the glass at the bottom boundary.
- Use triangle or straight-line geometry to find the internal angle facing the right-hand boundary.
❌ Common Errors
- Measuring from the surface: Using the 30° angle directly as the angle in air instead of measuring from the normal.
- Inverting refractive indices: Forgetting which medium is 1 and which is 2 in Snell's Law.
- Geometric slips: Subtracting angles from 90° incorrectly when transitioning from the bottom boundary to the side boundary.
📐 Step-by-Step Calculation
- Step 1: Calculate the angle of incidence at the bottom internal boundary.
Using Snell's Law: n_glass sin(θ_glass) = n_air sin(θ_air)
1.48 × sin(θ_glass) = 1.00 × sin(30°)
sin(θ_glass) = sin(30°) / 1.48 = 0.5 / 1.48 ≈ 0.3378
θ_glass = sin⁻¹(0.3378) ≈ 19.76° - Step 2: Relate the transmitted angle to the internally reflected angle.
By the law of reflection, the angle of reflection equals the angle of incidence (19.76°). - Step 3: Determine the angle at the right-hand boundary.
Looking at the triangle formed inside the glass block, the normal to the bottom surface is perpendicular (90°) to the surface, and the right-hand boundary is perpendicular to the bottom surface.
The ray travels up toward the right-hand boundary. The complementary angle to the normal at the bottom is: 90° − 19.76° = 70.24°.
Alternatively, considering the geometry of the normal at the right-hand boundary (which is horizontal):
Angle of incidence at the right-hand boundary = 90° − 19.76° ≈ 70° (matching option D).
Topics
Physics · 3.3 Waves
Question and mark scheme from the AQA AS Level Physics examination, Paper 2, November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.