Why does light bend when it changes medium?

Its speed changes. One edge of a wavefront reaches the boundary first and changes speed before the other, swinging the wavefront — and so the ray — round.

Why can there be no critical angle going into a denser medium?

Snell’s law would need C = n_2/n_1 > 1, which is impossible. Light entering a denser medium always refracts.

What is the cladding for in an optical fibre?

It keeps total internal reflection reliable, protects the core from damage, prevents light crossing between fibres, and — with an index close to the core’s — reduces modal dispersion.

Does refraction change the frequency of light?

No. Frequency is set by the source. The speed and wavelength both decrease in the same ratio when light enters a denser medium.

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5.1WavesCore

Refraction and Total Internal Reflection

Light changes direction when it crosses from one material into another because its speed changes. The $refractive index$ measures that slowing, and from it follow Snell’s law, the critical angle, and the total internal reflection that makes optical fibres work.

35 min Topic 5 · Waves and the Particle Nature of Light

What you'll be able to do

Define refractive index as a ratio of speeds and use $n = c / v$
Apply Snell’s law $n_{1} sin θ_{1} = n_{2} sin θ_{2}$
Derive and use the critical angle $sin C = n_{2} / n_{1}$
Explain total internal reflection and its use in optical fibres

Refractive index

The absolute refractive index of a material is the ratio of the speed of light in a vacuum to its speed in the material: $n = c / v$ . Because light is always slower in a material, $n \geq 1$ — air is about 1.0003 (usually taken as 1), water about 1.33, glass about 1.5.

A higher refractive index means light travels more slowly, and the material is described as $optically denser$ — which has nothing to do with its mass density.

Frequency is unchanged at a boundary, so as the speed drops the wavelength shortens in the same proportion.

n = \frac{c}{v}

Refractive index as a speed ratio.

v = \frac{c}{n} = \frac{3.00 \times 1 0 ^{8}}{1.52}

Answer

1.97 \times 1 0^{8} m s^{- 1}

Snell’s law

At a boundary, $n_{1} sin θ_{1} = n_{2} sin θ_{2}$ , with angles measured from the $normal$ — the line perpendicular to the surface.

Entering an optically denser material ( $n_{2} > n_{1}$ ), light bends $towards$ the normal. Leaving one, it bends away. Travelling along the normal, it does not bend at all.

The physical reason is the change in speed: one side of a wavefront reaches the boundary and slows before the other, which swings the wavefront round.

n_{1} sin θ_{1} = n_{2} sin θ_{2}

Snell’s law, angles from the normal.

1.00 \times sin 4 0^{\circ} = 1.33 sin θ_{2}

sin θ_{2} = \frac{0.6428}{1.33} = 0.4833

Answer

θ_{2} = 28. 9^{\circ}

— towards the normal, as expected.

Tip — Angles must be measured from the normal, not the surface. If a diagram gives the angle to the surface, subtract it from $9 0^{\circ}$ first.

The critical angle

Light leaving a denser medium bends away from the normal. As the angle of incidence increases, the refracted ray swings towards the surface until, at the $critical angle$ $C$ , it travels exactly along the boundary at $9 0^{\circ}$ .

Setting $θ_{2} = 9 0^{\circ}$ in Snell’s law gives $sin C = n_{2} / n_{1}$ . For glass into air, $sin C = 1/1.5$ and $C \approx 4 2^{\circ}$ .

A critical angle only exists when light travels from a higher to a lower refractive index. In the other direction the ratio would exceed 1, which no sine can equal.

sin C = \frac{n _{2}}{n _{1}}

Critical angle, with

n_{1} > n_{2}

sin C = \frac{1.00}{2.42} = 0.413

Answer

C = 24. 4^{\circ}

Total internal reflection and optical fibres

If the angle of incidence exceeds the critical angle, no refracted ray is possible and $all$ the light is reflected back into the denser medium — total internal reflection. Two conditions must both hold: light is going from higher to lower $n$ , and the angle of incidence is greater than $C$ .

An optical fibre has a glass $core$ surrounded by $cladding$ of slightly lower refractive index. Light entering the core at a shallow enough angle meets the core–cladding boundary above the critical angle and is reflected along the fibre repeatedly.

The cladding protects the core from scratches (which would let light escape) and prevents signal crossing between touching fibres. Its refractive index being only slightly lower gives a large critical angle, which limits the range of ray paths and so reduces $modal dispersion$ — pulses spreading out because different rays travel different distances.

sin C = \frac{1.46}{1.50} = 0.9733

Answer

C = 76. 7^{\circ}

Tip — When explaining total internal reflection, state both conditions. Omitting "from a higher to a lower refractive index" is the usual lost mark.

Equation recap

n = c / v

Refractive index.

n_{1} sin θ_{1} = n_{2} sin θ_{2}

Snell’s law.

sin C = n_{2} / n_{1}

Critical angle.

θ > C \Rightarrow TIR

Plus higher → lower

n

Common mistakes to avoid

Measuring angles from the surface.

Snell’s law uses angles from the normal.

Expecting total internal reflection when light enters a denser medium.

It only occurs going from higher to lower refractive index.

Writing

sin C = n_{1} / n_{2}

It is the smaller index over the larger:

sin C = n_{2} / n_{1}

with

n_{1} > n_{2}

Saying optically denser means heavier.

It means higher refractive index — slower light — unrelated to mass density.

Key takeaways

Refractive index $n = c / v$ is a dimensionless speed ratio, always at least 1.
Snell’s law $n_{1} sin θ_{1} = n_{2} sin θ_{2}$ uses angles from the normal.
Critical angle: $sin C = n_{2} / n_{1}$ , only from higher to lower index.
Beyond $C$ , light undergoes total internal reflection.
Optical fibres use a core and lower-index cladding to guide light with little loss.

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