AQA A-Level Physics Paper 1, June 2023: Question 4

✅ Correct Answer

Use the critical angle formula: sin(c) = 1 / n

Substitute c = 41.5° to yield n = 1.51 (or higher precision confirming "about 1.5").

📐 Step-by-Step Calculation

1. Start with the formula: sin(c) = 1 / n
2. Rearrange for n: n = 1 / sin(41.5°)
3. Evaluate: n = 1 / 0.6626 = 1.509...
4. Round to at least 3 significant figures to show it equals 1.51.

❌ Common Errors & Traps

Failing to show the substitution working or rounding prematurely to 2 sig figs ( 1.5 ) without showing the intermediate 1.51 .

✅ Correct Answer

1. The angle of incidence at both the 2nd and 3rd reflecting surfaces is 45°, which is greater than the critical angle (41.5°), causing Total Internal Reflection.
2. The ray exits the block perpendicular to the boundary ( angle of incidence = 0° ), so it travels straight through without refraction, remaining parallel to the incident ray.

💡 Key Knowledge

In a 45°-45°-90° right-angled isosceles prism, normal entry on the hypotenuse results in 45° internal strikes on the shorter sides. Because 45° > 41.5° (critical angle), total internal reflection occurs perfectly.

🧠 Exam Technique

Ensure you explicitly mention both conditions for the marks: state the numerical angle of incidence at the reflecting faces, confirm it exceeds the critical angle, and explain the normal exit angle.

✅ Correct Answer

Your drawn ray must show:
- Total internal reflection taking place at the second reflecting boundary.
- The reflected ray travelling symmetrically.
- The final emergent ray leaving parallel to the initial ray path.

🧠 Exam Technique

Use a sharp pencil and a ruler! Examiners check that the reflected ray maintains geometric consistency by eye. The final emergent ray must line up visually parallel to the incoming path.

❌ Common Errors

Drawing partial refraction or letting the ray escape out of the wrong face of the prism due to incorrect angle estimation.

📐 Step-by-Step Calculation

1. Identify 2nd boundary angle: At the critical limit, the angle of incidence at the 2nd reflecting boundary is exactly the critical angle: 41.5° .
2. Find 1st boundary reflection angle: Geometry of the triangle dictates the angle of reflection at the 1st boundary is 90° - 41.5° = 48.5° .
3. Find internal refraction angle: Using geometry inside the prism vertex, the angle of refraction at entry is 90° - 45° - 41.5° = 3.5° (or equivalent angle tracking).
4. Apply Snell's Law: 1.0 × sin(θ) = 1.5 × sin(3.5°) , giving θ = 5.3° .

❌ Common Errors & Traps

Simply writing down subtractions like 90° - 41.5° = 48.5° without linking them to specific physical boundaries will lose you method marks. Always explicitly state which angle belongs to which surface.

✅ Correct Answer & Discussion

1. Equilateral Triangle Prism (Figure 9):
Does not work. The internal geometry changes such that light would no longer strike the second reflecting surface at an angle greater than the critical angle, causing light to leak/escape rather than undergo TIR.

2. Smaller Refractive Index Glass (Figure 10):
Does not work. A smaller refractive index ( n ) leads to a larger critical angle ( sin(c) = 1/n ), which reduces the allowable range for total internal reflection, thereby decreasing the value of angle θ .

💡 Key Knowledge

Remember the core relationship: lower refractive index = higher critical angle = harder to achieve total internal reflection.

🧠 Exam Technique

Structure your answer clearly into two numbered paragraphs corresponding to suggestions 1 and 2. State clearly whether each works first, followed by the optical justification to secure top-level marks.