Learn7408Paper 1 · Paper 1WavesRefraction and optical fibres

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Refraction and Optical Fibres

Light slows down in a material; the refractive index is $n=\frac cv$ . At a boundary, $n_1\sin\theta_1=n_2\sin\theta_2$ (Snell's law). Going into a less dense medium, beyond the critical angle $\sin C=\frac{n_2}{n_1}$ all the light reflects: total internal reflection. Optical fibres trap light this way, using a core surrounded by cladding of lower refractive index.

Edexcel YPH11 WPH12AQA 7408 Paper 1≈ 40 minRevision 1 · updated 2026-10-27

Section 1 of 14

Learning objectives

Use $n=\frac cv$ and Snell's law $n_1\sin\theta_1=n_2\sin\theta_2$ .
Find critical angles and state the conditions for total internal reflection.
Explain how optical fibres work, and the effects of material and modal dispersion.

Pause and think

A straw in a glass of water looks bent at the surface. The straw isn't bent, so what is?

Section 2 of 14

Core idea

Start hereThe bent straw

A straw standing in water looks snapped at the surface. Nothing happened to the straw: light changed speed and direction as it crossed from water to air. The same bending, pushed to its limit, traps light inside the glass fibres that carry the internet.

The big idea

Light is slower in denser materials: $n=\frac cv$ . At a boundary $n_1\sin\theta_1=n_2\sin\theta_2$ . Heading into a lower- $n$ medium beyond the critical angle, no light escapes: total internal reflection.

1. Snell's law

Critical angle from Snell's law

$n_1\sin C=n_2\sin90^\circ$
At the critical angle the refracted ray runs along the boundary.
$\sin C=\frac{n_2}{n_1}$
$\sin90^\circ=1$ .
$\text{needs }n_1>n_2$
Otherwise $\sin C>1$ : no critical angle.

For glass to air,

\sin C=\frac1n

Worked exampleInto glass

Light enters glass ( $n=1.50$ ) from air at $40^\circ$ to the normal.

Goal: Angle of refraction.
Strategy: Snell's law with $n_1=1$ .

$1.00\sin40^\circ=1.50\sin\theta_2$
Snell's law.
$\sin\theta_2=0.429$
Divide.
$\theta_2=25.4^\circ$
Inverse sine.

Answer

25.4^\circ

, bent towards the normal.

✓ CheckEntering a slower medium: smaller angle ✓.

2. Optical fibres

Worked exampleCore and cladding

Core $n=1.48$ , cladding $n=1.46$ .

Goal: Critical angle at the boundary.
Strategy: $\sin C=\frac{n_2}{n_1}$ .

$\sin C=\frac{1.46}{1.48}=0.9865$
Cladding is the outer medium.
$C=80.6^\circ$
Inverse sine.

Answer

80.6^\circ

: only rays nearly parallel to the axis stay trapped.

✓ CheckClose indices give a large

C

, limiting the paths, which reduces modal dispersion ✓.

Pulse broadening in fibres

Drawing diagram…

Each kind of dispersion has its own fix.

Common misconceptionAngles to the normal If given the angle to the surface, use

90^\circ

minus it.

Key takeawayAbsorption Absorption reduces amplitude (signal strength); dispersion broadens pulses. Repeaters fix both over long distances.

Pause and think

Why does a diamond sparkle more than glass?

Summary

$n=\frac cv$ .
$n_1\sin\theta_1=n_2\sin\theta_2$ .
$\sin C=\frac{n_2}{n_1}$ .
Cladding: lower $n$ , protection.

Refractive index: Ratio $\frac cv$ .
Critical angle: Incidence angle giving 90° refraction.
Total internal reflection: Complete reflection beyond the critical angle.

HistoryLight guided by water. In 1842 Daniel Colladon showed light trapped inside a curving jet of water in a Geneva lecture hall. The same total internal reflection now carries most of the world's internet traffic.

Fun factDiamonds sparkle. Diamond has

n=2.42

, so its critical angle is only

24^\circ

. Light entering a cut diamond reflects internally many times before escaping from the top, which is why it sparkles.

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