Reflection, Refraction and Total Internal Reflection

When a wave meets a boundary between two different media, it can be reflected, transmitted (refracted), or both. The behaviour of light at boundaries underpins optical fibres, lenses, prisms, and a vast range of technologies. This lesson covers the laws governing these phenomena and the conditions required for total internal reflection.

Reflection

When a wave hits a surface and bounces back, it is reflected. The law of reflection states:

The angle of incidence equals the angle of reflection: θᵢ = θᵣ
The incident ray, the reflected ray, and the normal to the surface all lie in the same plane

Angles are always measured from the normal — the line perpendicular to the surface at the point of incidence — not from the surface itself. This is a common source of error in exams.

Reflection occurs for all types of wave. Light reflecting from a mirror, sound echoing from a cliff face, and water waves bouncing off a harbour wall all obey the same law.

Refraction

When a wave passes from one medium to another, it changes speed. If the wave hits the boundary at an angle (not along the normal), this change in speed causes the wave to change direction. This is refraction.

A light ray passing from air into glass slows down. Because the part of the wavefront that enters the glass first slows down while the rest is still travelling at its original speed in air, the wavefront pivots, and the ray bends towards the normal. Conversely, when light passes from glass into air it speeds up and bends away from the normal.

The key principle: when a wave enters a denser (slower) medium, it bends towards the normal. When it enters a less dense (faster) medium, it bends away from the normal.

graph TD
    A["Light ray hits\nboundary at angle"] --> B{"Is new medium\noptically denser?"}
    B -->|"Yes — slower speed"| C["Ray bends TOWARDS normal\n(angle of refraction < angle of incidence)"]
    B -->|"No — faster speed"| D["Ray bends AWAY from normal\n(angle of refraction > angle of incidence)"]
    D --> E{"Is angle of incidence\n> critical angle?"}
    E -->|"Yes"| F["TOTAL INTERNAL\nREFLECTION"]
    E -->|"No"| G["Partial refraction\n+ partial reflection"]

Refractive Index

The refractive index (n) of a medium describes how much the speed of light is reduced in that medium compared to its speed in a vacuum:

$n = \frac{c}{v}$

where c = 3.00 × 10⁸ m s⁻¹ is the speed of light in a vacuum and v is the speed of light in the medium.

A higher refractive index means a slower speed and a greater bending effect. The refractive index of a vacuum is exactly 1; for air it is very close to 1 (1.0003) and is usually taken as 1 in calculations.

Refractive Indices of Common Materials

Medium	Refractive index (n)	Speed of light / 10⁸ m s⁻¹
Vacuum	1.000	3.00
Air	1.0003	2.999
Water	1.33	2.26
Crown glass	1.52	1.97
Flint glass	1.62	1.85
Diamond	2.42	1.24
Optical fibre (core)	~1.46–1.62	~1.85–2.05
Ice	1.31	2.29
Perspex / acrylic	1.49	2.01

Snell's Law

Snell's law relates the angles and refractive indices when light passes between two media:

$n_1 \sin\theta_1 = n_2 \sin\theta_2$

where n₁ and n₂ are the refractive indices of the two media, θ₁ is the angle of incidence, and θ₂ is the angle of refraction (both measured from the normal).

Worked example 1: A ray of light passes 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 = sin⁻¹(0.4229) = 25.0°

The ray bends towards the normal as it enters the denser medium, as expected.

Worked example 2: Light travels from water (n = 1.33) into diamond (n = 2.42) at 30°. Find the angle of refraction.

1.33 × sin 30° = 2.42 × sin θ₂

sin θ₂ = (1.33 × 0.500) / 2.42 = 0.665 / 2.42 = 0.2748

θ₂ = sin⁻¹(0.2748) = 16.0°

The large refractive index of diamond causes significant bending — this is what gives diamonds their distinctive sparkle (combined with dispersion and total internal reflection at internal facets).

Total Internal Reflection (TIR)

When light travels from a denser medium (higher n) to a less dense medium (lower n), it bends away from the normal. As the angle of incidence increases, the angle of refraction also increases. At a particular angle — the critical angle (θc) — the refracted ray lies exactly along the boundary (θ₂ = 90°).

For angles of incidence greater than the critical angle, no refraction occurs. All the light is reflected back into the denser medium. This is total internal reflection.

The two conditions for TIR:

Light must be travelling from a more dense to a less dense medium (from higher n to lower n)
The angle of incidence must be greater than the critical angle

The critical angle is found by setting θ₂ = 90° in Snell's law:

$\sin\theta_c = \frac{n_2}{n_1}$

(where n₁ > n₂)

Worked example 3: Calculate the critical angle for light passing from glass (n = 1.52) to air (n = 1.00).

sin θc = n₂ / n₁ = 1.00 / 1.52 = 0.6579

θc = sin⁻¹(0.6579) = 41.1°

Any ray hitting the glass-air boundary at more than 41.1° from the normal will be totally internally reflected.

Worked example 4: Calculate the critical angle for light passing from diamond (n = 2.42) to air.

sin θc = 1.00 / 2.42 = 0.4132

θc = sin⁻¹(0.4132) = 24.4°

Diamond's very small critical angle means that light entering the top of a diamond undergoes multiple total internal reflections before eventually escaping. This is why diamonds appear so brilliant — light bounces around inside the stone many times.

Optical Fibres

Optical fibres use total internal reflection to transmit light signals over long distances with very low loss. A typical optical fibre consists of:

A core made of high-refractive-index glass (n ≈ 1.62)
A cladding layer of lower-refractive-index glass (n ≈ 1.52) surrounding the core
A protective outer jacket

Light enters the fibre and hits the core-cladding boundary at an angle greater than the critical angle, so it is totally internally reflected. The light bounces along the fibre, following even gentle curves.

Why is cladding needed? Without cladding, light could escape when fibres touch or are scratched. The cladding ensures a consistent, clean boundary with a known refractive index difference, maintaining reliable TIR. It also prevents signal leakage between adjacent fibres bundled together.

Worked example 5: An 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 θc = n₂ / n₁ = 1.52 / 1.62 = 0.9383

θc = sin⁻¹(0.9383) = 69.6°

Light must hit the core-cladding boundary at more than 69.6° to the normal (i.e., at a very shallow angle to the boundary) for TIR to occur.

Signal Degradation in Optical Fibres

Two main issues reduce signal quality over long distances:

Absorption: The glass absorbs a small fraction of the light energy, converting it to heat. This reduces the signal strength (amplitude). Repeaters/amplifiers are used at intervals to regenerate the signal.

Pulse broadening (dispersion): A sharp input pulse spreads out and becomes broader as it travels through the fibre. This occurs because:

Modal dispersion: Different rays take different paths through the fibre (some travel straight, others bounce at steeper angles and travel further). Using single-mode fibres with very narrow cores (~8 μm) eliminates this.
Material dispersion: Different wavelengths travel at slightly different speeds in the glass. Using monochromatic light (e.g., from a laser) reduces this.

Pulse broadening limits the maximum data rate: if pulses overlap, the signal becomes unreadable.

Real-world application: Undersea fibre optic cables carry over 95% of intercontinental internet data. A single modern fibre can carry terabits per second over thousands of kilometres. The TAT-14 cable across the Atlantic contains multiple fibre pairs, each operating at 10 Gbit/s, with optical amplifiers every ~50 km.

Dispersion of White Light

When white light passes through a glass prism, it separates into the colours of the visible spectrum. This happens because the refractive index of glass varies slightly with wavelength — a phenomenon called dispersion.

Violet light has a shorter wavelength, a higher frequency, and a slightly higher refractive index in glass, so it is refracted more. Red light has a longer wavelength, a lower frequency, and a slightly lower refractive index, so it is refracted less.

Colour	Wavelength / nm	Refractive index in crown glass
Red	~700	1.514
Orange	~600	1.517
Yellow	~580	1.519
Green	~550	1.522
Blue	~470	1.528
Violet	~400	1.536

Common exam mistake: Students sometimes say "blue light bends more because it has more energy." While blue light does have more energy per photon (E = hf), the reason it bends more is that the refractive index of glass is higher for shorter wavelengths. Energy is not the direct cause of greater refraction.

Summary of Key Equations

Equation	Meaning
θᵢ = θᵣ	Law of reflection
n = c/v	Definition of refractive index
n₁ sin θ₁ = n₂ sin θ₂	Snell's law
sin θc = n₂/n₁	Critical angle (n₁ > n₂)

A-Level Deep Dive: Reflection, Refraction and Total Internal Reflection

Spec mapping

Edexcel 9PH0 specification Topic 5 — Waves and the Particle Nature of Light, in the sub-strands covering reflection and refraction at plane boundaries, requires students to use the law of reflection ( $\theta_i = \theta_r$ ); define and apply absolute refractive index $n = c/v$ ; apply Snell's law in the form $n_1\sin\theta_1 = n_2\sin\theta_2$ ; and derive and apply the critical-angle condition $\sin\theta_c = n_2/n_1$ for total internal reflection together with its application to optical fibres (refer to the official Pearson Edexcel specification document for exact wording). The same fluencies are tested again in the Topic 5 sub-strand on the photon and EM-spectrum context (whenever wavelength changes across a boundary while frequency does not), and they support the geometry questions in Topic 6 (further mechanics) that involve directional energy delivery, and Topic 12 (turning points) for the de-Broglie analogy of refraction-as-momentum-change. The Edexcel data-and-formulae booklet lists $n = c/v$ , $n_1\sin\theta_1 = n_2\sin\theta_2$ and $\sin\theta_c = 1/n$ (for the dense-to-air case), but does not list the explicit two-medium critical-angle form $\sin\theta_c = n_2/n_1$ — that one must be remembered, along with the strict condition $n_1 > n_2$ .

Worked example with full mark scheme

Question (8 marks):

A semicircular glass block has refractive index $n_g = 1.50$ . A narrow ray of monochromatic light is directed at the curved surface so that it passes undeviated to the centre of the flat face, where it strikes the glass–air boundary at an angle of incidence of $35.0°$ measured from the normal.

(a) Calculate the angle of refraction in the air. (3)

(b) Determine the critical angle for the glass–air boundary. (2)

(c) The ray is now redirected so that it strikes the flat face at an angle of incidence of $45.0°$ . State and explain what happens to the light, and calculate the path of the reflected ray relative to the normal. (3)

Solution with mark scheme:

(a) Step 1 — apply Snell's law at the glass–air boundary.

$n_g \sin\theta_g = n_{\text{air}}\sin\theta_{\text{air}}$ $1.50 \times \sin 35.0° = 1.00 \times \sin\theta_{\text{air}}$

M1 — correct application of Snell's law with refractive indices on the correct sides (denser medium paired with smaller angle expectation; here the smaller angle is in glass).

Step 2 — evaluate.

$\sin\theta_{\text{air}} = 1.50 \times 0.5736 = 0.8604$ $\theta_{\text{air}} = \sin^{-1}(0.8604) = 59.4°$

M1 — correct rearrangement and substitution; A1 — final value $59.4°$ (accept $59°$ to $59.5°$ ). A common slip is to write $\sin\theta_{air} = \sin 35° / 1.50 = 0.382$ , giving $\theta_{air} = 22.5°$ — that is the wrong direction (light leaving glass for air should bend away from the normal, so $\theta_{air} > \theta_g$ ).

(b) Step 1 — set up the critical-angle condition.

At the critical angle the refracted ray grazes the boundary, so $\theta_{air} = 90°$ and $\sin\theta_{air} = 1$ .

$\sin\theta_c = \frac{n_{\text{air}}}{n_g} = \frac{1.00}{1.50} = 0.6667$

M1 — correct critical-angle expression with the lower refractive index in the numerator.

Step 2 — evaluate.

$\theta_c = \sin^{-1}(0.6667) = 41.8°$

A1 — final value $41.8°$ (accept $41.8°$ to $41.9°$ ).

The angle of incidence is $45.0°$ and the critical angle is $41.8°$ . Since $45.0° > 41.8°$ and the light is travelling from a denser to a less dense medium, total internal reflection occurs.

B1 — explicit comparison with the critical angle and statement that TIR occurs (both required — stating only "TIR" without comparison loses the mark).

Step 2 — apply the law of reflection.

By the law of reflection $\theta_r = \theta_i = 45.0°$ from the normal on the glass side of the boundary.

M1 — correct application of $\theta_r = \theta_i$ . A1 — reflected ray at $45.0°$ to the normal, on the opposite side of the normal to the incident ray, in the same plane.

Total: 8 marks (M3 A3 B1 + M1). Note the breakdown: the AO1-heavy substitution work in (a) and (b) is worth 5 marks; the AO2/AO3 reasoning in (c) — comparing the calculated critical angle with a given angle of incidence and selecting the correct law — is worth 3.

Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Question (6 marks): A step-index optical fibre has a core of refractive index $1.50$ and a cladding of refractive index $1.45$ .

(a) Calculate the critical angle at the core–cladding boundary. (2)

(b) Light enters the flat end of the fibre from air. Calculate the largest angle, measured from the fibre's central axis, at which a ray can enter the core and still undergo total internal reflection at the core–cladding boundary. (3)

Mark scheme decomposition by AO:

(a)

M1 (AO1) — substitution into $\sin\theta_c = n_{cladding}/n_{core} = 1.45/1.50$ .
A1 (AO1) — $\theta_c = 75.2°$ (accept $75.1°$ to $75.3°$ ).

(b)

M1 (AO2) — geometry: the angle the ray makes with the normal to the core–cladding boundary is the complement of its angle to the central axis. The ray must hit the core–cladding boundary at $\theta_c = 75.2°$ or greater, so its angle to the central axis inside the core must be at most $90° - 75.2° = 14.8°$ .
M1 (AO2) — Snell's law applied at the entry face (air to core): $1.00 \times \sin\theta_{air} = 1.50 \times \sin 14.8°$ .
A1 (AO2) — $\sin\theta_{air} = 1.50 \times 0.2554 = 0.3831$ , so $\theta_{air} = 22.5°$ . (This angle is the acceptance angle of the fibre.)

(c)

B1 (AO1) — single-mode fibres have a very narrow core that supports only one ray path, eliminating modal dispersion (pulse broadening from rays of different path length), so pulses remain sharp at higher data rates over longer distances.

Total: 6 marks split AO1 = 3, AO2 = 3. This question is typical of the Edexcel pattern of stacking a one-step calculation (a) with a two-step geometric application (b) and a short qualitative explanation (c). The AO2 marks in (b) reward the candidate who recognises that the angle of incidence at the core–cladding boundary is not the angle the ray makes with the fibre's axis but the complement of it.

Synoptic links

Connects to:

Topic 5, sub-strand on the wave equation: when light enters a denser medium its speed drops and, because frequency is set by the source and is conserved across the boundary, the wavelength must also drop in proportion. The combination $v = f\lambda$ together with $n = c/v$ gives $\lambda_{medium} = \lambda_{vacuum}/n$ — a relationship implicit in every diffraction-grating and thin-film calculation. Without this link, the constancy of frequency at a boundary is just a memorised rule rather than a derivable consequence.
Topic 5, sub-strand on standing waves and resonance: the partial reflection that accompanies refraction at any boundary is the mechanism by which standing waves form on a string fixed at one end; the same boundary physics that produces TIR in a fibre produces the resonant cavity of a laser. Recognising "fixed boundary → reflection → standing wave" connects geometric optics to acoustics.