9.1 Physics of Light Transmission and Optical Fibres

9.1 Physics of Light Transmission and Optical Fibres

Commission Implementing Regulation (EU) 2023/989 sets Appendix I topic 5.10, Fibre optics, at level 2 for B2/B2L and level 1 for B1. The current text supplies no detailed content list; this guide develops the former detailed Appendix I scope as historical study material. Categories A and B3 are not examined on 5.10. Module 5 is a multiple-choice paper only: B2 sits 72 questions in 90 minutes, B1 sits 40 questions in 50 minutes, and A/B3 sit 20 questions in 25 minutes, with a 75% pass mark, no negative marking, and no essay. The live paper uses three options; the practice items in this chapter use four.

An optical fibre is a dielectric waveguide. Information travels as a modulated light wave, not as a metallic current. The waveguide is a core of glass or polymer, a cladding of slightly lower refractive index, then a primary coating, buffer and outer jacket. Light that enters the core and meets the core–cladding wall at a sufficiently grazing angle undergoes total internal reflection (TIR) and is guided along the run. The core physical fact behind that guidance is simple and exam-useful: the core refractive index is higher than the cladding.

Refractive index, Snell's law and the critical angle

Refractive index n is the ratio of the speed of light in vacuum c to the speed in the material v:

n = c / v

Fused-silica cores used in avionics jumpers typically sit near n = 1.46 to 1.48. The cladding is doped, or is a different polymer, so that its index is a few parts in a thousand lower. At a dielectric boundary, Snell's law relates the angles measured from the surface normal:

n1 sin θ1 = n2 sin θ2

When a ray is in the denser core (n1 > n2) and θ1 is increased, the refracted ray in the cladding bends away from the normal. At the critical angle θc the refracted ray skims the interface (θ2 = 90°):

θc = arcsin(n2 / n1)

For any incidence greater than θc, still measured from the normal, no real refracted ray exists in the cladding. The wave is totally internally reflected and remains in the core. A frequent exam trap is to reverse that inequality because aircraft “shallow angle” language is taken from the fibre axis. The two pictures agree: a ray that stays close to the axis has a large angle from the normal, so it satisfies TIR.

Worked example — critical angle

A silica core has n1 = 1.480 and cladding n2 = 1.460.

θc = arcsin(1.460 / 1.480) = arcsin(0.9865) ≈ 80.6°

A ray that meets the wall at 82° from the normal is guided. A ray at 75° from the normal is too steep relative to the axis, refracts into the cladding, and is lost as a radiation or cladding mode that the jacket then strips. If a question swaps the indices (cladding higher than core), TIR in the core is physically impossible: the ray would bend toward the normal and leak on the first bounce.

Acceptance cone and numerical aperture

Not every ray that hits the end face meets the core–cladding wall above θc. The acceptance angle θa in the outside medium (usually air, n0 ≈ 1.00) is the largest launch angle from the fibre axis that still produces TIR. Numerical aperture (NA) is the usual teaching figure of merit, even though Appendix I does not name NA as its own bullet:

NA = n0 sin θa = √(n_core² − n_cladding²)

A larger NA means a fatter acceptance cone, easier coupling from an LED, and more guided modes. A smaller NA is typical of single-mode fibre and needs more precise connector geometry.

Worked example — NA

Using the same indices:

NA = √(1.480² − 1.460²) = √(2.1904 − 2.1316) = √0.0588 ≈ 0.242

In air, θa = arcsin(0.242) ≈ 14°. An LED whose emission cone is much wider than 14° wastes power as unguided light in the first millimetre of fibre. That is coupling loss, not cable attenuation along the run.

A contaminated or poorly seated connector throws light outside the acceptance cone even when the glass along the run is intact. That is why a reported “fibre fault” is often an end-face problem, not a change in core index.

Wavelength, delay and why EMI does not couple

Avionics links commonly use 850 nm LEDs or VCSELs on multimode cable for short runs, and 1310 nm or 1550 nm lasers on single-mode cable where distance or bandwidth is larger. The transmitter is an electro-optic converter; the receiver is a photodiode (PIN or avalanche) that restores a voltage or current. Between those transducers there is no metallic signal current, which is why the guided wave is immune to the electromagnetic environment that topic 5.14 treats for copper.

Group velocity in the glass is v = c / n. For n = 1.48, v ≈ 2.03 × 10⁸ m/s, so a 20 m jumper delays light by about 99 ns. That delay is stable compared with copper geometry effects, but it still matters when a high-speed bus time-stamps frames. End-face Fresnel reflection (about 4% at an uncoated glass–air interface) is a separate loss and a source of back-reflection into a laser; physical-contact polishes exist to cut that reflection (section 9.3).

Advantages of fibre versus electrical wire

Compared with a copper twisted pair or coaxial run of the same information capacity:

1. EMI immunity. Photons in a dielectric do not form a conducting loop. Generator magnetic fields, HIRF and lightning-induced transients do not induce a signal voltage on the optical path the way they do on a metallic pair.
2. Weight and volume. Glass plus a thin jacket undercuts an equivalent high-speed copper harness, which matters on long in-flight entertainment and backbone runs.
3. No short-circuit sparks. A crushed or open fibre does not dump bus current into structure or fuel vapour. Optical isolation also removes ground-loop currents between distant racks.
4. Bandwidth. One fibre supports data rates that would need many copper pairs or a bulky coaxial bundle.
5. Tap resistance and galvanic isolation. Extracting a usable optical signal without an intrusive coupler is difficult, and the boxes remain electrically isolated from each other on the data path.

Disadvantages versus electrical wire

The former detailed scope calls for the complementary list, not a claim that fibre always wins:

1. Minimum bend radius. Macrobends leak guided modes into the cladding. Routing around a former or through a tight clamp below the manufacturer’s radius — often about ten times the cable diameter, or a stated 25–38 mm for a jumper — produces a sudden insertion-loss step that looks like a bad connector.
2. Contamination. A micrometre-scale dust particle or a fingerprint on a 9 µm single-mode core can scatter most of the light. Copper pins with similar dirt still conduct.
3. Repair skill. Cleave, polish, fusion splice and microscope inspection are specialised. A crimp-and-solder copper repair is not an analogue. Unqualified termination is a common source of high loss and intermittent faults.
4. No aircraft power. Fibre does not deliver 28 V DC or 115 V AC; boxes still need copper power and often a separate discrete path.
5. Fragility and test cost. Glass fatigue, fibre pistoning in the ferrule, and the need for an optical source and power meter raise the maintenance burden relative to a milliohm check.

[!NOTE] Exam trap: “Fibre is immune to lightning” is too strong. The optical path does not carry induced current, but the LRU, connector shells, composite structure and any metallic strength member still need the bonding and shielding practices of topic 5.14. Fibre removes one coupling mechanism; it does not delete the aircraft electromagnetic environment.

B1 candidates at level 1 must recognise TIR, the index relationship, and the advantage/disadvantage list. B2 candidates at level 2 must also handle θc and NA, explain why a ray is lost, and connect EMI immunity to the absence of a conducted signal current.