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
Absolute refractive index: the refractive index of a substance, n, is the speed of light in a vacuum divided by the speed of light in the substance, n = c ÷ cₛ. It is a ratio of two speeds, so it has no unit, and it is always greater than or equal to 1.
Refractive index of air: light travels only very slightly slower in air than in a vacuum (n ≈ 1.0003), so for calculations the refractive index of air is taken as 1.
Relative refractive index: the refractive index of medium 2 relative to medium 1 is ₁n₂ = n₂ ÷ n₁ = c₁ ÷ c₂, the ratio of the speeds of light in the two media. When light enters a medium, its speed and wavelength change but its frequency does not.
The equation
refractive index = speed of light in a vacuum ÷ speed of light in the substance n = c ÷ cₛ
| Quantity | Symbol | Unit |
|---|---|---|
| absolute refractive index of the substance | n | no unit (a ratio) |
| speed of light in a vacuum | c | metres per second (m s⁻¹) |
| speed of light in the substance | cₛ | metres per second (m s⁻¹) |
Rearranged: cₛ = c ÷ n | c = ncₛ | A bigger refractive index means light travels more slowly in that material. If your n comes out less than 1, you have divided the wrong way round.
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) Light travels through a glass block with a refractive index of 1.50. Calculate the speed of light in the glass. (c = 3.00 × 108 m s⁻¹)
Show solution
F: n = c ÷ cₛ
I: 1.50 = 3.00 × 108 ÷ cₛ
F: cₛ = c ÷ n = 3.00 × 108 ÷ 1.50
A: 2.00 × 108 m s⁻¹
Q2 (Core) Water has a refractive index of 1.33. Calculate the speed of light in water, to 3 significant figures. (c = 3.00 × 108 m s⁻¹)
Show solution
F: n = c ÷ cₛ
I: 1.33 = 3.00 × 108 ÷ cₛ
F: cₛ = c ÷ n = 3.00 × 108 ÷ 1.33
A: 2.255… × 108 m s⁻¹ = 2.26 × 108 m s⁻¹ (3 s.f.)
Q3 (Core) The speed of light in diamond is 1.24 × 108 m s⁻¹. Calculate the refractive index of diamond, to 3 significant figures. (c = 3.00 × 108 m s⁻¹)
Show solution
F: n = c ÷ cₛ
I: n = 3.00 × 108 ÷ 1.24 × 108
F: n is already the subject
A: 2.419… = 2.42 (3 s.f., no unit)
Q4 (Core) Light travels through a plastic at 197 000 km s⁻¹. Calculate the refractive index of the plastic, to 3 significant figures. (c = 3.00 × 108 m s⁻¹)
Show solution
Convert first: 197 000 km s⁻¹ = 197 000 × 103 = 1.97 × 108 m s⁻¹
F: n = c ÷ cₛ
I: n = 3.00 × 108 ÷ 1.97 × 108
F: n is already the subject
A: 1.522… = 1.52 (3 s.f.)
Watch out: giving refractive index a unit such as m s⁻¹. It is a ratio of two speeds, so the units cancel.
Q5 (Exam) The refractive index of air at room temperature is 1.0003. (a) Calculate the speed of light in air, to 4 significant figures. (b) Calculate how much slower light travels in air than in a vacuum, to 2 significant figures. (c) Explain why n for air is usually taken as 1. (c = 3.00 × 108 m s⁻¹)
Show solution
Step 1 – (a) speed in air
F: n = c ÷ cₛ
I: 1.0003 = 3.00 × 108 ÷ cₛ
F: cₛ = 3.00 × 108 ÷ 1.0003 = 2.9991… × 108 m s⁻¹
A: 2.999 × 108 m s⁻¹ (4 s.f.)
Step 2 – (b) difference
3.00 × 108 − 2.9991… × 108 = 89 973 m s⁻¹ = 9.0 × 104 m s⁻¹ (2 s.f.)
Step 3 – (c)
this is only 0.03% of c, smaller than the uncertainty in most measurements, so nair ≈ 1 and light in air is treated as travelling at c
Q6 (Exam) Light passes from water (n = 1.33) into glass (n = 1.52). (a) Calculate the refractive index of the glass relative to the water, to 3 significant figures. (b) State, with a reason, whether the light speeds up or slows down.
Show solution
Step 1 – (a) relative refractive index
F: ₁n₂ = n₂ ÷ n₁
I: ₁n₂ = 1.52 ÷ 1.33
F: ₁n₂ is already the subject
A: 1.142… = 1.14 (3 s.f.)
Step 2 – (b) speed
the glass has the larger refractive index (₁n₂ > 1), so the light slows down; its speed in water is 1.14 times its speed in glass
Watch out: the relative index is the second medium divided by the first. Writing 1.33 ÷ 1.52 = 0.875 would suggest the light speeds up.
Q7 (Exam) A signal is sent along an optical fibre 1.5 km long. The core has a refractive index of 1.48. (a) Calculate the time taken for light to travel along the fibre (assume it travels parallel to the axis), to 2 significant figures. (b) How much longer is this than the time for light to travel 1.5 km in a vacuum? (c = 3.00 × 108 m s⁻¹)
Show solution
Convert first: 1.5 km = 1500 m
Step 1 – speed in the core
F: n = c ÷ cₛ, so cₛ = c ÷ n
I: cₛ = 3.00 × 108 ÷ 1.48 = 2.027… × 108 m s⁻¹
Step 2 – (a) time
t = distance ÷ speed = 1500 ÷ 2.027… × 108
A: 7.4 × 10−6 s (7.4 μs, 2 s.f.)
Step 3 – (b) compare with a vacuum
t = 1500 ÷ 3.00 × 108 = 5.0 × 10−6 s, so the extra time is 7.4 − 5.0 = 2.4 × 10−6 s (2.4 μs, 2 s.f.)
Q8 (Exam) A very short pulse of light takes 0.760 ns to travel straight through a glass block 15.0 cm thick. Calculate the refractive index of the glass, to 3 significant figures. (c = 3.00 × 108 m s⁻¹)
Show solution
Convert first: 15.0 cm = 0.150 m; 0.760 ns = 7.60 × 10−10 s
Step 1 – speed in the glass
cₛ = distance ÷ time = 0.150 ÷ 7.60 × 10−10 = 1.973… × 108 m s⁻¹
Step 2 – refractive index
F: n = c ÷ cₛ
I: n = 3.00 × 108 ÷ 1.973… × 108
F: n is already the subject
A: 1.52 (3 s.f.)
Watch out: ns means 10−9 s. Keep the unrounded speed in your calculator for the second step.
Q9 (Stretch) Red light from a helium–neon laser has a wavelength of 633 nm in air. It enters water (n = 1.33). Take n for air as 1. Calculate (a) the frequency of the light, (b) its speed in water and (c) its wavelength in water, each to 3 significant figures. (c = 3.00 × 108 m s⁻¹)
Show solution
Convert first: 633 nm = 6.33 × 10−7 m
Step 1 – (a) frequency (the same in both media)
f = c ÷ λ = 3.00 × 108 ÷ 6.33 × 10−7 = 4.739… × 1014 Hz
A: 4.74 × 1014 Hz (3 s.f.)
Step 2 – (b) speed in water
cₛ = c ÷ n = 3.00 × 108 ÷ 1.33 = 2.255… × 108 m s⁻¹
A: 2.26 × 108 m s⁻¹ (3 s.f.)
Step 3 – (c) wavelength in water
λ = cₛ ÷ f = 2.255… × 108 ÷ 4.739… × 1014 = 4.759… × 10−7 m (or simply 633 ÷ 1.33)
A: 476 nm (3 s.f.)
Watch out: the frequency is set by the source and does not change at a boundary. Only the speed and wavelength fall, both by the factor n.
Q10 (Stretch) Monochromatic light has a wavelength of 520 nm in water (n = 1.33). It passes into glass (n = 1.52). (a) Calculate the wavelength of the light in the glass, to 3 significant figures. (b) Calculate the wavelength of the light in a vacuum and its frequency, each to 3 significant figures. (c) Explain why the light looks the same colour in both materials. (c = 3.00 × 108 m s⁻¹)
Show solution
Step 1 – (a) wavelength in glass
f is constant, so λ ∝ cₛ = c ÷ n; hence λ ∝ 1 ÷ n and n₁λ₁ = n₂λ₂
I: 1.33 × 520 = 1.52 × λglass
F: λglass = 691.6 ÷ 1.52
A: 455 nm (3 s.f.)
Step 2 – (b) vacuum wavelength and frequency
λvacuum = nλwater = 1.33 × 520 = 691.6 nm = 692 nm (3 s.f.)
f = c ÷ λvacuum = 3.00 × 108 ÷ 6.916 × 10−7 = 4.337… × 1014 Hz = 4.34 × 1014 Hz (3 s.f.)
Step 3 – (c) colour
colour depends on the frequency, which does not change when light crosses a boundary, so the light looks the same colour (red) even though its wavelength changes
Watch out: multiplying 520 by 1.52 ÷ 1.33. The wavelength gets shorter in the denser medium, so the answer must be below 520 nm.
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