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When light passes from one medium to another, it changes speed. If the light enters at an angle to the boundary, this change in speed causes a change in direction — this is refraction. Under certain conditions, light can be completely reflected back into the original medium — this is total internal reflection.
Spec mapping: This lesson sits under AQA 7408 section 3.3.2 and covers refractive index (n = c/v), Snell's law (n₁ sin θ₁ = n₂ sin θ₂), the critical angle for total internal reflection (sin θ_c = n₂/n₁), and the application to step-index optical fibres including the function of the cladding and the principal causes of signal degradation (absorption, modal dispersion). The frequency-invariance and wavelength-change-on-refraction relations are also part of the spec. (Refer to the official AQA specification document for exact wording.)
Synoptic links:
- Section 3.3.1 (wave properties): the conservation of frequency at a boundary follows directly from the wavefront-matching argument introduced in the very first lesson; refraction is its quantitative consequence.
- Section 3.3.2 (Young's double slits, diffraction): wavelength inside a denser medium is reduced to λ/n, which shifts the fringe spacing of interference experiments performed in non-air media (e.g. underwater).
- Section 3.2.2 (photon energy): photon energy E = hf is also invariant on refraction (because frequency is invariant), but the speed and wavelength change — a useful synoptic check on the photon model.
When a wave crosses a boundary between two media in which it travels at different speeds, several things can happen:
At the boundary, the number of wavefronts arriving per second must equal the number leaving per second (otherwise wavefronts would pile up or disappear at the boundary). Since frequency is the number of wavefronts per second, it remains constant. Since v = fλ and f is constant, a decrease in speed means a decrease in wavelength, and vice versa.
The refractive index of a medium, n, is defined as:
n = c/v
where c is the speed of light in a vacuum (3.00 × 10⁸ m s⁻¹) and v is the speed of light in the medium.
| Medium | Approximate Refractive Index |
|---|---|
| Vacuum | 1.000 (exactly) |
| Air | 1.000 (approximately) |
| Water | 1.33 |
| Glass (crown) | 1.52 |
| Glass (flint) | 1.65 |
| Diamond | 2.42 |
| Optical fibre (core) | ~1.46–1.62 |
A higher refractive index means the light travels more slowly in that medium.
Exam Tip: The refractive index of a medium is always ≥ 1 (since v ≤ c). Air has n ≈ 1.00, so for most purposes, air and vacuum are treated as equivalent.
When light passes from medium 1 (refractive index n₁) to medium 2 (refractive index n₂), the relationship between the angles of incidence and refraction is given by Snell's law:
n₁ sin θ₁ = n₂ sin θ₂
where θ₁ is the angle of incidence (measured from the normal) and θ₂ is the angle of refraction (measured from the normal).
Worked Example 1 — Light travels from air (n = 1.00) into crown glass (n = 1.52) at an angle of incidence of 40°. Calculate the angle of refraction.
n₁ sin θ₁ = n₂ sin θ₂
1.00 × sin 40° = 1.52 × sin θ₂
sin θ₂ = sin 40°/1.52 = 0.6428/1.52 = 0.4229
θ₂ = arcsin(0.4229) = 25.0°
The light bends towards the normal as it enters the denser medium.
Worked Example 2 — Light passes from water (n = 1.33) into air (n = 1.00) at an angle of incidence of 30°. Calculate the angle of refraction.
1.33 × sin 30° = 1.00 × sin θ₂
sin θ₂ = 1.33 × 0.500 = 0.665
θ₂ = arcsin(0.665) = 41.7°
The light bends away from the normal as it enters the less dense medium.
Worked Example 3 — Light of wavelength 589 nm in vacuum enters glass of refractive index 1.52. Calculate the speed and wavelength of the light in the glass.
Speed: v = c/n = (3.00 × 10⁸)/1.52 = 1.97 × 10⁸ m s⁻¹
Wavelength in glass: λ_glass = λ_vacuum/n = 589/1.52 = 388 nm
The frequency remains the same: f = c/λ_vacuum = (3.00 × 10⁸)/(589 × 10⁻⁹) = 5.09 × 10¹⁴ Hz
When light travels from a more dense medium (higher n) to a less dense medium (lower n), the refracted ray bends away from the normal. As the angle of incidence increases, the angle of refraction increases until it reaches 90° — the refracted ray travels along the boundary.
The angle of incidence at which the refracted ray makes an angle of 90° with the normal is called the critical angle, θ_c.
Total internal reflection occurs when both conditions are met:
At the critical angle, θ₂ = 90°:
n₁ sin θ_c = n₂ sin 90° = n₂
sin θ_c = n₂/n₁
For light going from a medium of refractive index n into air (n₂ ≈ 1.00):
sin θ_c = 1/n
Worked Example 4 — Calculate the critical angle for light travelling from glass (n = 1.52) to air.
sin θ_c = 1/n = 1/1.52 = 0.6579
θ_c = arcsin(0.6579) = 41.1°
Any light hitting the glass-air boundary at an angle greater than 41.1° will be totally internally reflected.
Worked Example 5 — Calculate the critical angle for a diamond–air boundary (n_diamond = 2.42).
sin θ_c = 1/2.42 = 0.4132
θ_c = arcsin(0.4132) = 24.4°
Diamond has a very small critical angle, which means light undergoes many total internal reflections inside the stone — this is what gives diamonds their brilliance and sparkle.
Total internal reflection is the principle behind optical fibres, which are used to transmit data (light signals) over long distances with very low loss.
A step-index fibre consists of two main components:
Light enters the core and hits the core-cladding boundary at an angle greater than the critical angle, causing total internal reflection. The light bounces along the fibre, remaining within the core.
Worked Example 6 — A step-index optical fibre has a core of refractive index 1.62 and cladding of refractive index 1.52. Calculate the critical angle at the core-cladding boundary.
sin θ_c = n₂/n₁ = 1.52/1.62 = 0.9383
θ_c = arcsin(0.9383) = 69.8°
Light must hit the core-cladding boundary at an angle greater than 69.8° (measured from the normal) for total internal reflection. Since angles are measured from the normal, this means the light must travel at a fairly shallow angle within the core (close to parallel to the fibre axis).
Two main problems cause the signal to degrade over long distances:
1. Absorption Some light energy is absorbed by the glass, converting to heat. This reduces the signal intensity over distance. Solution: use very pure glass; use wavelengths where absorption is minimal (typically near-IR at 1300 nm or 1550 nm).
2. Pulse Broadening (Modal Dispersion) In a step-index multimode fibre, different rays travel at different angles, taking different path lengths through the fibre. Rays that bounce more times travel a longer path and arrive later than rays that travel almost straight. This causes short pulses to spread out (broaden), and if pulses overlap, data is lost.
Solutions:
Exam Tip: If asked to explain why optical fibres use cladding, you must give at least two reasons: (1) to maintain TIR by providing a lower refractive index boundary, and (2) to prevent cross-talk between adjacent fibres.
The refractive index of every real medium depends on wavelength — a phenomenon called dispersion. For crown glass, n_red(700 nm) ≈ 1.515 while n_violet(400 nm) ≈ 1.532, a difference of about 1%. This wavelength dependence causes white light passing through a prism to separate into a spectrum, with violet bent most and red bent least.
For red: 1.000 × sin 35° = 1.515 × sin θ_R → sin θ_R = sin 35°/1.515 = 0.5736/1.515 = 0.3786 → θ_R = 22.25°
For violet: 1.000 × sin 35° = 1.532 × sin θ_V → sin θ_V = 0.5736/1.532 = 0.3744 → θ_V = 21.99°
Angular separation = θ_R − θ_V = 22.25° − 21.99° = 0.26°
This small angular separation accumulates over the full prism geometry to produce the wide visible spectrum — and is the working principle of every prism spectrometer.
For red (n = 1.515): sin θ_c = 1/1.515 = 0.660 → θ_c = 41.3°
For violet (n = 1.532): sin θ_c = 1/1.532 = 0.653 → θ_c = 40.8°
Violet has a slightly smaller critical angle, so a wider range of incidence angles (40.8°–90° vs 41.3°–90° for red) produces TIR. This means a fibre with a wavelength-mixed signal experiences slight chromatic separation at the boundary — a contribution to chromatic dispersion in optical fibres, separate from modal dispersion.
Refraction in a continuously varying refractive index produces curved light paths rather than straight-line refraction at sharp boundaries. The classic example is the mirage: on a hot road, the air immediately above the surface is heated, becomes less dense, and develops a slightly lower refractive index (n decreases with temperature). Light from the sky descending at a low angle bends upward as it enters the cooler-air-above-warmer-air gradient, eventually undergoing total internal reflection if the angle is shallow enough. The viewer perceives this as a "reflection" — typically interpreted as a puddle of water on the road. The same physics produces inferior mirages at sea (looking down at the horizon) and superior mirages over very cold sea (where the image of distant ships appears displaced upward).
Earth's atmosphere has a refractive index of approximately 1.000293 at sea level, decreasing with altitude. A light ray from a star at low altitude is refracted by approximately 35 arcminutes (more than the Sun's apparent diameter), which is why the Sun appears to set later than it geometrically should: when the Sun's geometric position is just below the horizon, atmospheric refraction lifts its apparent position above the horizon by about half a degree. The same effect makes stars near the horizon appear higher than their true positions and causes the well-known "flattening" of the setting Sun (the lower limb is refracted more than the upper limb).
1.33 × sin 30° = 1.00 × sin θ_apparent
sin θ_apparent = 1.33 × 0.500 = 0.665
θ_apparent = 41.7°
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