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Refraction

The entire internet runs, in part, on a single optical trick: light bouncing along inside a glass fibre, guided by nothing more than a well-chosen angle. Understanding exactly why light bends at a boundary — and when it stops crossing that boundary altogether — explains both rainbows and fibre-optic cables with the same handful of equations.

40 min Video by Kit Betts-Masters Module 4.4 Waves

Snell’s Law (Refraction) — A Level and GCSE Physics RevisionWatch the full walkthrough before the notes below.

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What you'll be able to do

Define refractive index in terms of the speed of light
Use Snell’s law
Explain total internal reflection and the critical angle
Describe the structure and purpose of a step-index optical fibre

Refractive index

The $refractive index$ , $n$ , of a material is the ratio of the speed of light in a vacuum to its speed in that material. It is always $\geq 1$ , since light never travels faster in a medium than in a vacuum; a larger $n$ means light is slowed more.

n = \frac{c}{c _{s}}

Refractive index = speed of light in vacuum ÷ speed in the medium.

1Use

n = \frac{c}{c _{s}}

n = \frac{3.00 \times 1 0 ^{8}}{1.97 \times 1 0 ^{8}}

Answer

n \approx 1.52

Snell’s law

At a boundary between two media, the angles of incidence and refraction (both measured from the normal) are related by Snell’s law. Light bends towards the normal entering a denser medium (larger $n$ ), and away from the normal entering a less dense medium.

n_{1} sin θ_{1} = n_{2} sin θ_{2}

Snell’s law.

1Snell’s law:

1.00 sin 50° = 1.33 sin θ_{2}

sin θ_{2} = \frac{sin 50°}{1.33} = \frac{0.766}{1.33}

θ_{2} = sin^{- 1} (0.576)

Answer

θ_{2} \approx 35.2°

Total internal reflection

Travelling from a denser to a less dense medium, as the angle of incidence increases, the refracted ray bends further from the normal. At the $critical angle$ , $θ_{c}$ , the refracted ray runs exactly along the boundary. Beyond this angle, no refracted ray exists at all — the light undergoes $total internal reflection$ instead.

sin θ_{c} = \frac{n _{2}}{n _{1}} (n_{1} > n_{2})

Critical angle.

1For a boundary with air (

n_{2} = 1

sin θ_{c} = \frac{1}{n _{1}} = \frac{1}{1.55}

sin θ_{c} = 0.645

θ_{c} = sin^{- 1} (0.645)

Answer

θ_{c} \approx 40.2°

Tip — Total internal reflection needs BOTH conditions at once: travelling into a less dense medium, AND an angle of incidence greater than the critical angle.

Optical fibres

A step-index optical fibre has a high-refractive-index glass core surrounded by lower-refractive-index cladding. Light entering at a shallow angle repeatedly undergoes total internal reflection at the core–cladding boundary, guiding it along the fibre even around gentle bends. The cladding provides the necessary refractive-index step for TIR, protects the core, and prevents light leaking (crossing over) between adjacent fibres.

Equation recap

n = \frac{c}{c _{s}}

Refractive index.

n_{1} sin θ_{1} = n_{2} sin θ_{2}

Snell’s law.

sin θ_{c} = \frac{n _{2}}{n _{1}}

Critical angle for TIR.

Common mistakes to avoid

Measuring angles from the surface rather than the normal.

All angles in refraction (incidence, refraction, critical) are measured from the NORMAL, the line perpendicular to the boundary.

Believing TIR can occur going into a denser medium.

TIR only occurs travelling from a denser to a LESS dense medium, above the critical angle.

Using sinθc = n for a glass-air boundary.

It is sinθc = 1/n (equivalently n_air/n_glass) — forgetting the reciprocal gives an impossible value greater than 1.

Key takeaways

Refractive index n = c/cs; Snell’s law: n₁sinθ₁ = n₂sinθ₂.
Total internal reflection occurs beyond the critical angle, travelling into a less dense medium: sinθc = n₂/n₁.
Step-index optical fibres guide light by repeated total internal reflection; the cladding enables TIR, protects the core, and prevents crossover.

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Refraction

40 min Video by Kit Betts-Masters Module 4.4 Waves