Total Internal Reflection

Spec mapping: OCR H556 Module 4.4 — Waves: total internal reflection; critical angle $\sin\theta_c = n_2/n_1$ (= $1/n$ when leaving into air or vacuum); step-index optical fibres; PAG 3 anchor. (Refer to the official OCR H556 specification document for exact wording.)

When a wave passes from a denser medium into a less dense one, it bends away from the normal. As the angle of incidence increases, the angle of refraction grows even more rapidly. At a certain critical angle, the refracted ray would emerge along the boundary (angle of refraction $= 90°$ ). Beyond this, the wave simply cannot escape the denser medium: it is totally internally reflected back into the denser medium.

This phenomenon — total internal reflection (TIR) — is the operating principle of optical fibres, the technology that carries the majority of the world's internet traffic at almost the speed of light. It also underpins endoscopes, binoculars, fibre-optic illumination, prism reflectors and the sparkle of a well-cut diamond. The OCR A-Level Physics A specification requires you to derive and apply the critical-angle formula and to understand the physics of step-index optical fibres. We will see that the formula $\sin\theta_c = n_2/n_1$ is the direct limit of Snell's law with $\theta_2 = 90°$ , and that fibre optics is essentially a clever application of repeated TIR at the core-cladding boundary.

The Critical Angle

Consider a ray of light travelling in medium 1 (refractive index $n_1$ , larger) striking the boundary with medium 2 (refractive index $n_2$ , smaller). Snell's law gives:

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

As $\theta_1$ increases, so does $\theta_2$ . At some specific angle — the critical angle $\theta_c$ — the refraction angle $\theta_2$ reaches $90°$ . At this point, the refracted ray travels along the boundary itself and is said to be grazing. For angles of incidence greater than $\theta_c$ , no refracted ray can exist (Snell's law would demand $\sin\theta_2 > 1$ , which is impossible); all the light is reflected back into the denser medium.

Setting $\theta_2 = 90°$ and $\sin 90° = 1$ in Snell's law:

$n_1 \sin \theta_c = n_2 \times 1$

So:

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

This is the formula you must know. When the second medium is air (or vacuum), for which $n_2 = 1$ , the formula simplifies to:

$\sin \theta_c = \frac{1}{n_1} = \frac{1}{n}$

The general form is the one above; OCR expects you to handle both. Memorise the air-out version $\sin\theta_c = 1/n$ as the most common case, but recognise that fibre-optic problems (core-cladding) require the general $n_2/n_1$ form.

Conditions for Total Internal Reflection

TIR can occur if and only if:

The light is travelling from a medium of higher refractive index into one of lower refractive index ( $n_1 > n_2$ ), otherwise $\sin \theta_c = n_2/n_1 > 1$ , which has no solution.
The angle of incidence is greater than the critical angle ( $\theta_1 > \theta_c$ ).

Both conditions must be met. If the ray is going into a denser medium, no critical angle exists and TIR cannot happen.

Worked Example 1 — Critical Angle for Glass-Air

Q. Calculate the critical angle for light travelling in glass ( $n = 1.52$ ) meeting an interface with air ( $n = 1.00$ ).

A. Using $\sin\theta_c = n_2/n_1$ (or equivalently $1/n$ since $n_2 = 1$ ):

$\sin \theta_c = \frac{1.00}{1.52} = 0.658$ $\theta_c = \arcsin(0.658) = 41.1°$

Any ray in the glass striking the glass-air surface at more than $41.1°$ from the normal will be totally internally reflected.

Worked Example 2 — Critical Angle for Water-Air (Snell's Window)

Q. Calculate the critical angle for water ( $n = 1.33$ ) meeting air.

$\sin \theta_c = \frac{1.00}{1.33} = 0.752$ $\theta_c = \arcsin(0.752) = 48.8°$

An object underwater looking upwards sees the entire above-water world compressed into a cone of half-angle about $49°$ — Snell's window. Outside this cone, the water surface acts as a mirror, reflecting the underwater scene. Divers experience this dramatically: the sun and sky appear in a circle directly overhead, while everything outside that circle shows the bottom of the pool by TIR off the underside of the water surface.

Worked Example 3 — Critical Angle Between Two Glasses

Q. Light travels from a core of refractive index $1.50$ into a cladding of refractive index $1.47$ . Calculate the critical angle.

$\sin \theta_c = \frac{n_2}{n_1} = \frac{1.47}{1.50} = 0.980$ $\theta_c = \arcsin(0.980) = 78.5°$

This is much larger than for a glass-air interface, because the two refractive indices are close. This is the characteristic geometry of a step-index optical fibre — only rays striking the core-cladding boundary at angles very close to grazing ( $> 78.5°$ from the normal, or equivalently very small angles to the fibre axis) will be totally internally reflected and trapped in the core. Rays entering at steeper angles to the axis hit the boundary below the critical angle, refract out into the cladding, and are lost.

Optical Fibres

An optical fibre is a long, thin strand of glass (or sometimes plastic) along which light is transmitted by repeated total internal reflection. Modern optical fibres are at the heart of telecommunications and medicine.

flowchart LR
    S[Light enters fibre] --> C1[Travels in core: n1 = 1.50]
    C1 --> B1[Strikes core-cladding boundary]
    B1 -- angle > theta_c --> TIR1[Total internal reflection]
    TIR1 --> C2[Back into core]
    C2 --> B2[Strikes other side]
    B2 -- angle > theta_c --> TIR2[TIR again]
    TIR2 --> E[Exits at far end]

Step-Index Fibre

A step-index fibre consists of:

A core of high-refractive-index glass (typically n ≈ 1.50), where the light propagates.
A cladding of slightly lower-refractive-index glass (typically n ≈ 1.47) surrounding the core.
An outer protective coating (plastic jacket) for mechanical protection.

The refractive index "steps" abruptly at the core-cladding boundary (hence "step-index"). Light entering the core at a sufficiently shallow angle to the fibre axis will always strike the core-cladding boundary at an angle greater than the critical angle, and so will be totally internally reflected. The light is trapped in the core and travels along the fibre with almost no loss.

Why Cladding?

You might wonder why you cannot just use an uncoated glass core with air outside. There are two essential reasons for the cladding.

Protection of the total-reflection surface. If the glass surface is touched, scratched or contaminated by dust, grease or moisture, TIR breaks down at those points and light leaks out. The cladding protects the core from surface contamination by keeping the TIR surface inside the fibre.
Prevention of cross-talk between fibres. If two uncoated fibres touched, light could cross from one to the other (the fibres would act as a single enlarged core at the contact point). Cladding keeps each fibre's light safely contained.
Defined geometry. The exact refractive index of the cladding controls the critical angle and therefore the range of angles ("acceptance cone") over which light can enter and still be guided — this is fundamental to designing the optical system feeding the fibre.

Step-Index vs Graded-Index

Beyond step-index fibres, there are also graded-index fibres in which the refractive index decreases smoothly from a maximum at the core axis to a minimum at the cladding. Graded-index fibres reduce modal dispersion (see below) and are important for higher-performance applications. For OCR A-Level, you mainly need to understand step-index fibres, but you should be aware that the graded-index design exists.

Modal Dispersion

In a step-index fibre, different rays enter at different angles and therefore take different path lengths through the fibre. A ray travelling straight down the axis has the shortest path, equal to the length of the fibre. A ray that enters at the steepest acceptance angle bounces frequently off the core-cladding boundary and therefore travels a longer total distance. Both rays travel at the same speed (v = c/n) in the core, but the longer-path ray takes more time.

The result is that a short light pulse entering the fibre spreads out in time as it travels — the first arrival is from the axial ray, the last arrival is from the most-bounced ray. This is called modal dispersion (or sometimes "multipath dispersion").

Why It Matters

Modal dispersion limits the rate at which information can be sent down a fibre. If two successive pulses are too close in time, they will overlap at the far end and become indistinguishable. For long-distance high-bit-rate communications this is a serious problem.

How It Is Mitigated

Several strategies reduce modal dispersion:

Single-mode fibres — the core is made so thin (~8 μm) that only the axial ray can propagate; all other ray paths are cut off. This eliminates modal dispersion almost completely but makes the fibre harder to couple to. All long-distance telecoms fibres are single-mode.
Graded-index fibres — the refractive index decreases gradually from the axis outward, so rays travelling at steeper angles (through the outer parts of the core) travel in regions of lower refractive index and therefore higher speed. This compensates for their longer paths and brings all rays back to approximately the same travel time.
Using shorter fibre lengths or lower bit rates where high bandwidth is not required.

Exam Tip: If an OCR exam asks about "why optical fibres have cladding", give at least two of the following reasons: (i) to protect the TIR surface from scratches and contamination; (ii) to prevent light leaking into adjacent fibres; (iii) to define a well-known critical angle and acceptance angle. If asked about "limitations" or "dispersion", explain modal dispersion — different paths take different times, so pulses broaden.

Other Uses of Total Internal Reflection

Binoculars and SLR cameras — prisms use TIR to fold the light path and correct image orientation. More efficient than mirrors.
Endoscopes — flexible bundles of fibres carry images out of the body for medical diagnosis, and separate bundles carry illumination in.
Bicycle reflectors and cat's-eye road studs — corner-cube reflectors use three TIR surfaces to send light back towards its source.
Fibre lamps and decorative lighting — plastic fibres carry light from a single source to multiple output points.
Water fountains with illuminated arcs — TIR along the jet traps the light within the water.
Natural phenomena — rainbows involve refraction plus TIR inside water droplets.

Evanescent Waves (Enrichment)

Strictly speaking, when TIR happens, an evanescent wave extends a tiny distance (about one wavelength) into the second medium. This is how some optical devices (like couplers and fibre sensors) work: they exploit this evanescent wave to couple light in or out of a fibre without ever breaking it. This is beyond A-Level but is the reason a second fibre placed within a wavelength of the first can pick up some of the signal.

Synoptic Links

Refraction (previous lesson, Module 4.4) — the critical-angle formula $\sin\theta_c = n_2/n_1$ is the direct limit of Snell's law with $\theta_2 = 90°$ . TIR is what happens when Snell's law would demand $\sin\theta_2 > 1$ — i.e. when the wave cannot escape into the less dense medium.
Wave-particle duality and electron diffraction (Module 4.5) — the wave-mode condition for confinement in an optical fibre (TIR at the core boundary) has a quantum analogue in electron confinement in a quantum well: the electron de Broglie wave is "totally internally reflected" at the well walls when its energy is below the barrier.
Polarisation by reflection (earlier in this module, Module 4.4) — at the Brewster angle the reflected light is completely polarised; at the critical angle the transmitted (refracted) ray vanishes entirely. Both are angle-dependent boundary phenomena governed by Snell's law and the Fresnel reflection coefficients.

Specimen question modelled on the OCR H556 paper format

Question (9 marks): A step-index optical fibre has a core of refractive index $1.50$ surrounded by a cladding of refractive index $1.47$ .

(a) State two physical reasons (beyond mechanical protection) why optical fibres use a cladding rather than leaving the core bare. [2]

(b) Calculate the critical angle at the core-cladding boundary, and state what happens to a ray that strikes this boundary at exactly that angle. [3]

(c) Light pulses sent through a long step-index fibre arrive at the far end broadened in time. Name this effect and explain its origin in terms of ray paths within the core. [2]

(d) Suggest one design modification that reduces this broadening, and briefly explain how it works. [2]

AO breakdown

Mark	AO	Awarded for
1	AO1	Protects TIR surface from contamination/scratches
2	AO1	Prevents cross-talk between adjacent fibres
3	AO2	$\sin\theta_c = 1.47/1.50 = 0.980$
4	AO2	$\theta_c = 78.5°$
5	AO1	At $\theta_c$ the refracted ray grazes the boundary
6	AO1	Names effect as modal dispersion (or multipath dispersion)
7	AO2	Different ray paths have different lengths $\Rightarrow$ different travel times
8	AO1	Names a mitigation (single-mode fibre or graded-index fibre)
9	AO2	Brief mechanism: single-mode kills all but one ray path / graded-index compensates faster speed for outer rays

AO split: AO1 = 5, AO2 = 4, AO3 = 0.

Mid-band response (4/9)

(a) The cladding protects the surface from being scratched, and stops light leaking between fibres.

(b) $\sin\theta_c = 1.47/1.50 = 0.98$ , so $\theta_c = 78.5°$ . The ray will travel along the boundary.

(d) A single-mode fibre.

Total Internal Reflection

Total Internal Reflection

The Critical Angle

Conditions for Total Internal Reflection

Worked Example 1 — Critical Angle for Glass-Air

Worked Example 2 — Critical Angle for Water-Air (Snell's Window)

Worked Example 3 — Critical Angle Between Two Glasses

Optical Fibres

Step-Index Fibre

Why Cladding?

Step-Index vs Graded-Index

Modal Dispersion

Why It Matters

How It Is Mitigated

Other Uses of Total Internal Reflection

Evanescent Waves (Enrichment)

Synoptic Links

Specimen question modelled on the OCR H556 paper format

AO breakdown

Mid-band response (4/9)

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