Specification: AQA A-level Physics (7408) 3.3.2.3 | Paper: 1 | Data and formulae booklet: given in the exam
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📌 KEY DEFINITIONS – learn them word for word
Refractive index: the absolute refractive index of a material is n = c ÷ cs, the speed of light in a vacuum divided by the speed of light in the material. n has no unit, and n ≈ 1 for air.
Refraction: the change in direction of a wave as it crosses a boundary between two materials, because its speed changes. Light entering an optically denser material (larger n) slows down and bends towards the normal; light entering a less dense material speeds up and bends away from the normal.
Normal: the line drawn at 90° to the boundary at the point where the ray meets it. In Snell’s law, θ₁ and θ₂ are always measured from the normal, never from the surface.
The equation
refractive index of material 1 × sine of angle in material 1 = refractive index of material 2 × sine of angle in material 2 n₁ sin θ₁ = n₂ sin θ₂
| Quantity | Symbol | Unit |
|---|---|---|
| refractive index of the material the light comes from | n₁ | no unit |
| angle between the ray and the normal in material 1 | θ₁ | degrees (°) |
| refractive index of the material the light enters | n₂ | no unit |
| angle between the ray and the normal in material 2 | θ₂ | degrees (°) |
Rearranged: sin θ₂ = n₁ sin θ₁ ÷ n₂ | n₂ = n₁ sin θ₁ ÷ sin θ₂ | Find sin θ₂ first, then use sin⁻¹ with your calculator in degree mode. If an angle is given from the surface, subtract it from 90° first.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) A ray of light in air strikes a glass block at an angle of incidence of 40.0°. The refractive index of the glass is 1.50. Calculate the angle of refraction, to 3 significant figures. (n of air = 1.00)
Show solution
F: n₁ sin θ₁ = n₂ sin θ₂
I: 1.00 × sin 40.0° = 1.50 × sin θ₂, so 0.6427… = 1.50 sin θ₂
F: sin θ₂ = 0.6427… ÷ 1.50 = 0.4285…, so θ₂ = sin⁻¹ 0.4285…
A: 25.4° (3 s.f.), bent towards the normal
Q2 (Core) Light in air is incident on the surface of a liquid at 30.0° to the normal. The angle of refraction in the liquid is 22.0°. Calculate the refractive index of the liquid, to 3 significant figures.
Show solution
F: n₁ sin θ₁ = n₂ sin θ₂
I: 1.00 × sin 30.0° = n₂ × sin 22.0°, so 0.500 = n₂ × 0.3746…
F: n₂ = 0.500 ÷ 0.3746…
A: 1.33 (3 s.f.), no unit (this is water)
Q3 (Core) A ray of light travels inside a glass block of refractive index 1.52 and meets the glass–air boundary at 25.0° to the normal. Calculate the angle at which the ray leaves the glass, to 3 significant figures.
Show solution
F: n₁ sin θ₁ = n₂ sin θ₂ (here material 1 is the glass)
I: 1.52 × sin 25.0° = 1.00 × sin θ₂
F: sin θ₂ = 1.52 × 0.4226… = 0.6423…, so θ₂ = sin⁻¹ 0.6423…
A: 40.0° (3 s.f.) to the normal, bent away from the normal
Watch out: n₁ is the material the light is leaving. Putting 1.00 on the left and 1.52 on the right gives 16.1°, which would mean the light bent towards the normal on entering air.
Q4 (Core) A ray of light in air strikes a block of Perspex at an angle of 55.0° to the surface of the block. The refractive index of Perspex is 1.49. Calculate the angle of refraction, to 3 significant figures.
Show solution
Convert first: angle to the normal = 90.0° − 55.0° = 35.0°
F: n₁ sin θ₁ = n₂ sin θ₂
I: 1.00 × sin 35.0° = 1.49 × sin θ₂
F: sin θ₂ = 0.5735… ÷ 1.49 = 0.3849…, so θ₂ = sin⁻¹ 0.3849…
A: 22.6° (3 s.f.) to the normal
Watch out: AQA examiners regularly report students measuring angles from the surface instead of the normal. Using 55.0° directly gives 33.4°, which is wrong.
Q5 (Exam) A ray of light in air enters a gemstone. The angle of incidence is 50.0° and the angle of refraction is 18.4°. Calculate (a) the refractive index of the gemstone and (b) the speed of light in the gemstone, both to 3 significant figures. (c = 3.00 × 108 m s⁻¹)
Show solution
Step 1 – (a) refractive index
F: n₁ sin θ₁ = n₂ sin θ₂
I: 1.00 × sin 50.0° = n₂ × sin 18.4°, so 0.7660… = n₂ × 0.3156…
F: n₂ = 0.7660… ÷ 0.3156… = 2.426…
A: 2.43 (3 s.f.)
Step 2 – (b) speed
n = c ÷ cs, so cs = c ÷ n = 3.00 × 108 ÷ 2.426… = 1.236… × 108 m s⁻¹
A: 1.24 × 108 m s⁻¹ (3 s.f.)
Q6 (Exam) A ray of light passes from water (n = 1.33) into glass (n = 1.50). The angle of incidence in the water is 60.0°. Calculate the angle of refraction in the glass, to 3 significant figures, and state whether the ray bends towards or away from the normal.
Show solution
F: n₁ sin θ₁ = n₂ sin θ₂
I: 1.33 × sin 60.0° = 1.50 × sin θ₂, so 1.1518… = 1.50 sin θ₂
F: sin θ₂ = 1.1518… ÷ 1.50 = 0.7678…, so θ₂ = sin⁻¹ 0.7678…
A: 50.2° (3 s.f.), bent towards the normal because the glass has the larger refractive index
Watch out: the light starts in water, not air. Using n₁ = 1.00 gives 35.3°, which is wrong.
Q7 (Exam) A ray of light in air is incident on a glass surface (n = 1.55) at 65.0° to the normal. Calculate the angle through which the ray is deviated (turned) as it enters the glass, to 3 significant figures.
Show solution
Step 1 – angle of refraction
F: n₁ sin θ₁ = n₂ sin θ₂
I: 1.00 × sin 65.0° = 1.55 × sin θ₂
F: sin θ₂ = 0.9063… ÷ 1.55 = 0.5847…, so θ₂ = 35.78…°
Step 2 – deviation
deviation = θ₁ − θ₂ = 65.0° − 35.78…°
A: 29.2° (3 s.f.)
Q8 (Exam) Light travels at 2.20 × 108 m s⁻¹ in material A and at 1.95 × 108 m s⁻¹ in material B. A ray in A meets the A–B boundary at 45.0° to the normal. Calculate the angle of refraction in B, to 3 significant figures. (c = 3.00 × 108 m s⁻¹)
Show solution
Step 1 – refractive indices
n = c ÷ cs: nA = 3.00 × 108 ÷ 2.20 × 108 = 1.363…; nB = 3.00 × 108 ÷ 1.95 × 108 = 1.538…
Step 2 – Snell’s law
F: n₁ sin θ₁ = n₂ sin θ₂
I: 1.363… × sin 45.0° = 1.538… × sin θ₂
F: sin θ₂ = 1.363… × 0.7071… ÷ 1.538… = 0.6267…, so θ₂ = sin⁻¹ 0.6267…
A: 38.8° (3 s.f.), towards the normal because light slows down in B
Shortcut: sin θ₁ ÷ sin θ₂ = c₁ ÷ c₂ = 2.20 ÷ 1.95, so c = 3.00 × 108 cancels out
Q9 (Stretch) A ray of light in air strikes a parallel-sided glass block of thickness 6.0 cm at an angle of incidence of 50.0°. The refractive index of the glass is 1.50. (a) Calculate the angle of refraction inside the block. (b) State the angle at which the ray leaves the opposite face. (c) Calculate the lateral (sideways) shift between the incident ray and the emerging ray, to 2 significant figures.
Show solution
Step 1 – (a) refraction at the first face
F: n₁ sin θ₁ = n₂ sin θ₂
I: 1.00 × sin 50.0° = 1.50 × sin θ₂
F: sin θ₂ = 0.7660… ÷ 1.50 = 0.5106…, so θ₂ = 30.71…°
A: 30.7° (3 s.f.)
Step 2 – (b) second face
the faces are parallel, so the ray meets the second face at 30.71…° to the normal. 1.50 × sin 30.71…° = 1.00 × sin θ, so sin θ = 0.7660… and θ = 50.0°: the ray leaves parallel to the incident ray
Step 3 – (c) lateral shift
path length in the glass = thickness ÷ cos θ₂ = 6.0 ÷ cos 30.71…° = 6.0 ÷ 0.8597… = 6.978… cm
lateral shift = path length × sin(θ₁ − θ₂) = 6.978… × sin 19.28…° = 6.978… × 0.3303…
A: 2.305… cm = 2.3 cm (2 s.f.)
Watch out: the shift is measured at 90° to the rays, not along the face of the block. You need the path length inside the glass, not just the thickness.
Q10 (Stretch) A ray of light in air is incident at 45.0° to the normal on one face of an equilateral glass prism (apex angle 60.0°, n = 1.50). Calculate (a) the angle of refraction at the first face, (b) the angle of incidence at the second face, (c) the angle at which the ray emerges from the second face and (d) the total deviation of the ray. Give angles to 3 significant figures.
Show solution
Step 1 – (a) first face
F: n₁ sin θ₁ = n₂ sin θ₂
I: 1.00 × sin 45.0° = 1.50 × sin θ₂
F: sin θ₂ = 0.7071… ÷ 1.50 = 0.4714…, so θ₂ = 28.12…°
A: 28.1° (3 s.f.)
Step 2 – (b) geometry inside the prism
for a prism, the two angles inside the glass add up to the apex angle, so angle at the second face = 60.0° − 28.12…° = 31.87…° = 31.9° (3 s.f.)
Step 3 – (c) second face
I: 1.50 × sin 31.87…° = 1.00 × sin θ
F: sin θ = 1.50 × 0.5280… = 0.7920…, so θ = sin⁻¹ 0.7920…
A: 52.4° (3 s.f.) to the normal
Step 4 – (d) total deviation
deviation = 45.0° + 52.38…° − 60.0° = 37.38…° = 37.4° (3 s.f.)
Watch out: check the ray can leave. The critical angle for this glass is 41.8°, and 31.9° is smaller, so the ray refracts out rather than totally internally reflecting.
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