3.3.2.3WavesCore

Refraction at a Plane Surface

Light changes speed when it crosses from one material to another, and if it hits the boundary at an angle it changes direction — it $refracts$ . The $refractive index$ measures how much a material slows light down, and beyond a certain angle light stops crossing the boundary altogether: $total internal reflection$ , the effect that guides signals along optical fibres.

40 min Video by Science Shorts 3.3.2 Refraction, diffraction and interference

Refraction & TIR — A-Level & GCSE Physics (full version)Watch 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 $n_{1} sin θ_{1} = n_{2} sin θ_{2}$
Explain total internal reflection and calculate the critical angle
Describe the structure and purpose of a step-index optical fibre
Explain the role of the cladding and applications of optical fibres

Refractive index

The $refractive index$ $n$ of a material is the ratio of the speed of light in a vacuum, $c$ , to its speed in the material, $c_{s}$ . It is always $\geq 1$ (light never travels faster than in a vacuum), and a bigger $n$ means the material slows light down more — it is “optically denser”.

For air, $n \approx 1.00$ ; for water, $n \approx 1.33$ ; for glass, $n \approx 1.5$ . As light slows on entering a denser medium it bends $towards$ the normal; speeding up into a less dense medium it bends $away$ from the normal.

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

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

1Use

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

n = \frac{3.0 \times 1 0 ^{8}}{2.0 \times 1 0 ^{8}}

n = 1.5

Answer

n = 1.5

Tip — Optically denser means a higher refractive index — not necessarily a higher mass density.

Snell’s law

At a boundary, the angles of incidence and refraction (both measured from the $normal$ ) are linked by $Snell’s law$ . It relates the refractive indices and the angles on each side of the surface.

Going from a less dense to a denser medium ( $n_{2} > n_{1}$ ) bends the ray towards the normal, so $θ_{2} < θ_{1}$ . The reverse — denser to less dense — bends it away from the normal.

n_{1} sin θ_{1} = n_{2} sin θ_{2}

Snell’s law — angles measured from the normal.

1Snell’s law:

1.00 sin 40° = 1.33 sin θ_{2}

sin θ_{2} = \frac{sin 40°}{1.33} = \frac{0.643}{1.33} = 0.483

θ_{2} = sin^{- 1} (0.483) = 28.9°

Answer

θ_{2} \approx 29°

(bent towards the normal)

Total internal reflection

When light travels from a denser to a less dense medium and 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 ( $θ_{2} = 90°$ ). Beyond $θ_{c}$ there is no refracted ray at all — the light is entirely reflected back, which is $total internal reflection (TIR)$ .

Two conditions are needed for TIR: the light must be travelling into a less optically dense medium, and the angle of incidence must exceed the critical angle. Setting $θ_{2} = 90°$ in Snell’s law gives the critical-angle formula.

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

Critical angle; for a boundary with air (n₂ = 1) this is sin θc = 1/n₁.

1For a boundary with air,

sin θ_{c} = \frac{1}{n} = \frac{1}{1.50}

sin θ_{c} = 0.667

θ_{c} = sin^{- 1} (0.667) = 41.8°

Answer

θ_{c} \approx 41.8°

Tip — TIR needs two things: going into a less dense medium AND an angle bigger than the critical angle. Miss either and the light refracts out.

Optical fibres

A $step-index optical fibre$ has a highly transparent glass $core$ of high refractive index, surrounded by $cladding$ of lower refractive index. Light enters at a shallow angle and repeatedly undergoes total internal reflection at the core–cladding boundary, so it is guided along the fibre even as it bends.

The cladding does several jobs: it lowers the refractive index of the boundary so TIR occurs, protects the core from scratches, and prevents light leaking (or “crossing over”) between adjacent fibres, which would blur the signal. Optical fibres carry telephone and internet data and are used in medical endoscopes.

Tip — A common exam answer: the cladding gives a refractive-index step for TIR, protects the core, and stops signal crossover between fibres.

Equation recap

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

Refractive index from the speed of light.

n_{1} sin θ_{1} = n_{2} sin θ_{2}

Snell’s law at a boundary.

sin θ_{c} = \frac{n _{2}}{n _{1}}

Critical angle for TIR (n₁ > n₂).

Common mistakes to avoid

Measuring angles from the surface instead of the normal.

All angles in refraction (incidence, refraction, critical) are measured from the normal — the line perpendicular to the surface.

Thinking total internal reflection can happen going into a denser medium.

TIR only occurs going from a denser to a less dense medium, at angles above the critical angle.

Using sin θc = n for a glass–air boundary.

It is sin θc = 1/n (= n_air/n_glass). Forgetting the reciprocal gives a value greater than 1, which is impossible.

Key takeaways

Refractive index n = c/cs measures how much a medium slows light (n ≥ 1).
Snell’s law n₁sinθ₁ = n₂sinθ₂ gives the refracted angle, all measured from the normal.
Above the critical angle (sin θc = n₂/n₁) light undergoes total internal reflection.
Step-index fibres guide light by TIR; the cladding enables TIR, protects the core and prevents crossover.

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Refraction at a Plane Surface

Light changes speed when it crosses from one material to another, and if it hits the boundary at an angle it changes direction — it

refracts

. The

refractive index

measures how much a material slows light down, and beyond a certain angle light stops crossing the boundary altogether:

total internal reflection

, the effect that guides signals along optical fibres.

40 min Video by Science Shorts 3.3.2 Refraction, diffraction and interference