Fibre Optics in Aircraft
Tooley ch.10 (p.162-177) · Lecture "Fiber optics" (37 slides) · Floyd ch.13 section 13-1 (p.740-745)
📚 Chapter mapping: 9 lectures ↔ textbook chapters
Tooley: Ch.10 Fibre optics (p.162-177): 10.1 advantages, 10.2 propagation (critical angle, numerical aperture, attenuation), 10.3 dispersion and bandwidth, 10.4 practical networks (ARINC 636, cable, connectors), 10.5 components; 20 MCQs (A.10, p.379-380)
Floyd: Ch.13 Data Transmission, section 13-1 Data Transmission Media (p.740-745): optical fiber cable, modes (multimode, single-mode, 50/125, 62.5/125, 8.3/125), link block diagram, connectors; Section Checkup; Problems 5 and 7 (p.793; answers A-27). Floyd has NO critical angle, numerical aperture, attenuation or dispersion: the deeper theory relies on the lecture and Tooley.
The 37-slide lecture has four calculation exercises (slides 20, 22, 27, 36); the module solves and verifies them, and only slide 36 has a printed answer. The module has 2 interactive demos (fibre calculator, Snell/TIR ray) and records the 7 mismatches found in the material.
What is an optical fibre?
An optical fibre is a waveguide along which an electromagnetic wave (light) travels with very little loss. Structure: a core (higher refractive index n₁), a reflective cladding (lower index n₂ < n₁) and a mechanical protective coating. Many fibres are bundled into a cable with aramid strength yarn and an outer jacket.
A sample optical link: electrical signal → light source (LED or laser) → fibre → optical receiver (photodiode) → electrical signal.
History: 1910, 1970, 1990
| Year | Milestone |
|---|---|
| 1910 | Hondros and Debye analyse dielectric waveguides: light can be guided in a cylindrical structure by total internal reflection |
| 1970 | Corning makes the first practical optical fibre, attenuation below 20 dB/km: opens long-distance communication |
| 1990 | Optical fibre first used in aircraft: high-speed data, EMI immunity, low weight |
| 2020 | Widely used and continually improved in modern aircraft |
Today: fibre in the cockpit and passenger cabin
Use is growing fast because of bandwidth demand.
- Cockpit: EFB, enhanced vision systems, large video-rich displays, HUD, cameras, through an optical avionics network switch.
- Passenger cabin: in-flight entertainment (on-demand movies, Wi-Fi internet, games, shopping) through a cabin network switch and a Wi-Fi access point linked by fibre.
Aircraft using standardised fibre, and where it is fitted
The lecture lists civil and military aircraft using standardised fibre: A350 XWB, A380-800, Boeing 787, A340-600, F-22, Eurofighter Typhoon, A400M, F-35, Merlin, Wildcat and Chinook helicopters, and others.
Why standardise: high bandwidth, light weight, EMI immunity, and common aerospace-standard fibres and connectors for reliability, interoperability and easier maintenance.
Typical places: passenger cabin (IFE, Wi-Fi), flight deck (EFB, video displays), network backbone (optical switches linking cockpit and cabin), tail (communications, antennas, data links), engines and wings (health monitoring, high-temperature sensors), flight control and sensors.
Tomorrow: upgrading older aircraft and fibre sensing
Optical networks are already built into the A380 and Boeing 787. Many older aircraft need wire and cable upgrades (from heavy, bandwidth-limited copper) to optical networks for bandwidth, especially entertainment.
| Next applications | Content |
|---|---|
| Data communications | High-speed data for avionics, IFE, aircraft systems |
| Secure communications | Optical links with low probability of interception |
| Fibre sensing | Strain and load (wings, fuselage), temperature (engines), impact detection and location (composites) |
Light is an electromagnetic wave
The spatial period is the wavelength λ, the time period is the period T = 1/f. Relationship: λ = v/f = vT. Visible light is about 400 to 700 nm. In vacuum v = c ≈ 3 × 10⁸ m/s; in a non-magnetic dielectric v = c/√ε_r.
Silica fibre example: n ≈ 1.44 (ε_r = n² ≈ 2.07) gives v = 3×10⁸/1.44 ≈ 2.08×10⁸ m/s.
Refractive index and the speed of light in a material
The refractive index n = √ε_r tells how much light is slowed relative to vacuum: v = c/n.
| Material | n | v (m/s) |
|---|---|---|
| Vacuum | 1.00 | 3.00×10⁸ |
| Air | 1.0003 | ≈ 3.00×10⁸ |
| Water | ≈ 1.33 | ≈ 2.25×10⁸ |
| Typical glass | ≈ 1.50 | ≈ 2.00×10⁸ |
| Acrylic (PMMA) | ≈ 1.60 | ≈ 1.87×10⁸ |
Snell's law
When a ray meets the boundary between two dielectrics of different refractive index it changes direction (refraction): n₁ sin θi = n₂ sin θr, with θi and θr measured from the normal.
- Into a higher index (n₂ > n₁): θr < θi, the ray bends toward the normal.
- Into a lower index (n₂ < n₁): θr > θi, the ray bends away from the normal.
Critical angle and total internal reflection
For n₁ > n₂ the refraction angle is always larger than the incidence angle. When θr = 90° the ray runs along the boundary; the corresponding incidence angle is the critical angle: sin θc = n₂/n₁.
| Condition | Result |
|---|---|
| θi < θc | Part refracts into medium 2, part reflects |
| θi = θc | The refracted ray runs along the boundary (θr = 90°) |
| θi > θc | No refracted ray: total internal reflection (TIR) |
Fibre structure and the index profile
| Layer | Role | Typical size |
|---|---|---|
| Core (n₁) | Transparent dielectric, high index, carries the light | 5 to 100 µm (e.g. 9, 50 µm) |
| Cladding (n₂) | Lower index, keeps light in the core by TIR | about 125 µm |
| Coating | Silicone rubber, protects the fibre | 250 to 300 µm |
| Outer jacket | Plastic, mechanical protection | about 0.9 to 3 mm |
Index profile: the core is higher than the cladding (n₁ > n₂).
Guiding light in a fibre by total internal reflection
Light entering the core meets the core-cladding boundary at an angle greater than or equal to the critical angle and is totally reflected, then meets the opposite boundary... Through this chain of reflections it runs along the core, even when the fibre is bent. A ray with an incidence angle below θc passes into the cladding and is lost by scattering or absorption.
Condition: n₁ > n₂ and θi ≥ θc with sin θc = n₂/n₁.
Acceptance angle and numerical aperture
Only rays entering the fibre inside the acceptance cone (half-angle θa) are guided. Applying Snell at the fibre end (air to core): n₀ sin θ₁ = n₁ sin θ₂. The limit occurs when the ray inside the core meets the boundary exactly at the critical angle, giving:
NA = n₀ sin θa = √(n₁² − n₂²), in air NA = sin θa
NA describes the light-gathering ability of the fibre.
Exercise 1 (slide 20): θc, θa and NA
Task: a fibre has n₁ = 1.53 (SiO₂) and n₂ = 1.50 (SiO₂ with GeO₂ doping). Compute the critical angle θc, the acceptance angle θa and the numerical aperture NA.
The module's solution (the slide prints no answer):
- θc = sin⁻¹(1.50/1.53) = sin⁻¹(0.9804) = 78.64°
- NA = √(1.53² − 1.50²) = √0.0909 = 0.3015
- θa = sin⁻¹(0.3015) = 17.55° (in air)
Propagation modes, normalised frequency V and mode count
Light in the core travels as many separate rays, each on a different path called a mode; every mode satisfies the TIR condition. Low-order modes have small angles (near the axis), high-order modes have large angles (near the critical angle) and travel longer paths, giving more modal dispersion.
V = (π d/λ) √(n₁² − n₂²) = (π d/λ) NA; for V ≫ 1 (multimode step-index): N ≈ V²/2.
Exercise 2 (slide 22): V, mode count and θa
Task: d_core = 62.5 µm, NA = 0.275, λ = 1300 nm. Compute V, the mode count N and θa.
The module's solution:
- V = π d NA / λ = π × 62.5×10⁻⁶ × 0.275 / 1300×10⁻⁹ = 41.54
- N ≈ V²/2 = 1725.2/2 = 863 modes
- θa = sin⁻¹(0.275) = 15.96°
Three propagation limits: attenuation, modal and chromatic dispersion
- Attenuation: power falls with distance due to absorption, scattering and bending loss. Typically 0.2 to 0.5 dB/km at 1550 nm (slide).
- Modal (temporal) dispersion: modes have different path lengths and arrive at different times, so the output pulse broadens.
- Chromatic dispersion: different wavelengths have different group velocities, so the output pulse broadens.
Attenuation: the exponential law and dB/km
P(z) = P_in e^(−αz): power falls exponentially with distance (α in Np/m). It is usually expressed as A = −(1/L)·10 log₁₀[P(L)/P(0)] (dB/km). Loss includes absorption, scattering, bending loss and connector loss.
Conversion: A (dB/km) = 4.343 × α (Np/km).
The 1310 nm and 1550 nm transmission windows
The silica fibre attenuation curve versus wavelength has three contributions: Rayleigh scattering (∝ 1/λ⁴, dominant below about 1 µm), OH⁻ absorption peaks (water, around 1.24 and 1.39 µm) and lattice absorption (dominant above 1.6 µm).
Minima near 1310 nm (≈ 0.35 dB/km) and 1550 nm (≈ 0.18 dB/km); hence most optical communication systems operate at these two wavelengths.
Exercise 3 (slide 27): maximum fibre length
Task: input power 1 mW, wavelength 1310 nm, minimum measurable output power 50 µW, SiO₂ fibre, using the attenuation read from the graph (slide 26: ≈ 0.35 dB/km at 1310 nm). Compute the maximum length l_max.
The module's solution: budget = 10 log₁₀(1 mW / 0.05 mW) = 10 log₁₀(20) = 13.01 dB. L_max = 13.01 / 0.35 = 37.2 km.
Modal dispersion (slide 28)
In a multimode fibre, a ray at angle θ to the axis travels the path L(θ) = L/cos θ, longer than the axial ray, so it arrives later. With group velocity v_g ≈ c/n₁, the time difference between the longest and shortest rays is Δt = ΔL/v_g.
Result: a narrow input pulse becomes a wide output pulse, and two nearby pulses may overlap, limiting the bit rate.
Chromatic dispersion (slide 29)
The refractive index of the material depends on wavelength: n = n(λ) (material dispersion), so each wavelength travels at a different speed v(λ) = c/n(λ). A pulse made of many wavelengths (spectral width Δλ) broadens by Δt = L·D(λ₀)·Δλ, with dispersion parameter D(λ) = (1/L)(dt_g/dλ) in ps/(nm·km). At the zero-dispersion wavelength λ₀: D(λ₀) = 0.
The three types of optical fibre
| Multimode step-index (MM-SI) | Multimode graded-index (MM-GI) | Single-mode (SM-SI) | |
|---|---|---|---|
| Core diameter | 50 to 200 µm | 50 to 100 µm | 8 to 10 µm |
| Index profile | Step (uniform) | Parabolic | Step (uniform) |
| Number of modes | Many (hundreds) | Many (tens to hundreds) | One |
| Modal dispersion | High | Lower | Negligible |
| Bandwidth | Low | Higher | Highest |
| Typical use | Short, low-cost links (LAN, data centres) | Medium distance (campus networks) | Long distance, high speed (telecom) |
The fibre-optic cable and colour coding
A typical cable (Tooley Figure 10.9, lecture slide 31) comprises: five optical fibres and two filler strands (polyester, non-optical, keep the cable shape), polyester separator tape (keeps fibres organised, makes the cable flexible), aramid (Kevlar) yarn for tensile strength and a thermoplastic outer jacket against moisture and for insulation.
Each fibre has a protective colour-coded buffer: blue, red, green, yellow, white (fibres 1 to 5).
The system: transmitter, connector, fibre, receiver
| Block | Function |
|---|---|
| Optical transmitter | Converts electrical signals to light with an LED or laser diode (typically 850, 1310, 1550 nm) |
| Connector | Couples light from the transmitter into the fibre and from the fibre to the receiver; minimises coupling loss and alignment error |
| Optical fibre | High-index core n₁, low-index cladding n₂, guided by TIR, low-loss long-distance transmission |
| Optical receiver | Converts light back to electrical signals with a photodiode (PIN or APD), then amplifies and processes |
Optical connectors: requirements and construction
Connectors must be reliable, robust, precise and repeatable even after many matings, installable without specialist tooling, low loss and low cost. Connector loss is often quoted as an equivalent fibre length: six connectors × 0.5 dB = 3 dB, equal to several kilometres of low-loss fibre.
Construction (Tooley Figure 10.10): alignment keys and grooves, guide pins and cavities, coloured bands, three start threads; ceramic contacts.
🧪 Demo 1: fibre calculator (runs in your browser)
Defaults are the fibre of the slide 20 exercise (n₁ = 1.53; n₂ = 1.50; SiO₂) with a 62.5 µm core at 1300 nm. Change the numbers to see the critical angle, numerical aperture, normalised frequency V, number of modes and the maximum length from the attenuation budget (slide 27 exercise: 1 mW in, 50 µW minimum, 0.35 dB/km).
🧪 Demo 2: a light ray at the boundary of two media (Snell's law and total internal reflection)
The incident medium n₁ is below the boundary. Drag the slider past the critical angle to see the refracted ray vanish and only the reflected ray remain.
📜 Theory and historical background
According to the lecture, in 1910 Hondros and Debye analysed dielectric waveguides theoretically and showed that light can be guided in a cylindrical structure by total internal reflection. In 1970 Corning made the first practical optical fibre with attenuation below 20 dB/km, opening the way to long-distance communication. Optical fibre entered aircraft around 1990 for high-speed data, EMI immunity and low weight. Per Tooley, the Boeing 777 was the first commercial aircraft to enter production with an optical-fibre LAN, developed in the 1980s, comprising an AVLAN in the flight deck and electrical equipment bay and a CABLAN in the cabin roof, conforming to ARINC 636, adapted from FDDI to reach 100 Mbps.
🔎 Case study: which fibre for an aircraft link, and how long can it be?
Suppose a transmitter of 1 mW, a receiver that can measure at least 50 µW, working at 1310 nm on SiO₂ fibre with 0.35 dB/km attenuation (the slide 27 exercise). The power budget is 10 log₁₀(1 mW / 50 µW) = 13.01 dB, so the maximum length is 13.01 / 0.35 ≈ 37.2 km. An aircraft only needs a few tens of metres, so this budget is generous; connectors matter more: Tooley counts six connectors at 0.5 dB each, 3 dB in total, equal to several kilometres of fibre. For a multimode fibre with a 62.5 µm core, NA = 0.275 at 1300 nm, V = 41.5 and about 863 modes (the slide 22 exercise), so modal dispersion is the main limit on speed. That is why today's aircraft networks still use a lot of multimode fibre (easier to cut, splice and connect), while single-mode has higher bandwidth but is harder to align (Tooley 10.2.2).
⚠️ Common traps (and mismatches found in the material)
General trap: mixing up the critical angle θc (computed at the core-cladding boundary) with the acceptance angle θa (computed at the fibre input, from air). With n₁ = 1.53 and n₂ = 1.50, θc = 78.6° but θa is only 17.5°. That is why slide 19, which writes "θ₁ = θc = θa", is wrong: the two angles are measured in different places and are not equal; the correct limiting condition is that the angle φ at the boundary (measured from the normal) equals θc, equivalently that the ray angle to the fibre axis inside equals 90° − θc. The symbol θ₁ is also used differently on slide 17 (angle in air) and slide 19 (angle in the core). Slide 9 lists water with ε_r ≈ 80 but n ≈ 1.33; √80 ≈ 8.9, not 1.33: ε_r = 80 is the static permittivity, while at light frequencies ε_r ≈ 1.77 (= 1.33²). Slide 28 writes Δt ≈ (n₁ − n₂)L/c, missing a factor n₁/n₂ (the correct expression for the critical ray is n₁(n₁ − n₂)L/(c n₂)), and gives a typical 10 to 50 ns/km whereas n₁ − n₂ = 0.02 to 0.04 gives about 67 to 137 ns/km. Slide 30 labels the graded-index fibre "MM-SI" while the comparison table below says MM-GI. Slide 24 gives 0.2 to 0.5 dB/km at 1550 nm while slide 26 gives 0.18 dB/km. Finally, Tooley section 10.2 states the refraction rule backwards: the book says that going from higher to lower refractive index the ray is "bent towards the normal", but by Snell's law it bends away from the normal (lecture slide 12 states it correctly); the book's wording contradicts Snell's law and the lecture, so do not use it for study.
📓 Companion Python notebook
Companion notebook: recomputes the critical angle, NA, V and mode count for all four lecture exercises (slides 20, 22, 27, 36) and compares them with the printed answer on slide 36; attenuation budget and the dB/km to Np/m conversion; Snell and TIR simulation; checks of the mismatches (θc versus θa, water ε_r, inter-mode Δt, the 20 Tooley chapter 10 questions).
✅ Self-check quiz
Book 1: Mike Tooley, Aircraft Digital Electronic and Computer Systems (end-of-chapter MCQs, independently recomputed to verify each answer)
Book 2: Thomas Floyd, Digital Fundamentals (True/False Quiz + Self-Test, cross-checked against the book’s own printed answer key)
Additional practice questions (AI-authored in the same style, answers self-verified by computation)
✍️ Free-response questions (source: Floyd, Digital Fundamentals)
Floyd Digital Fundamentals 11th ed., Chapter 13 section 13-1: Problems 5 and 7 (p.793, PDF 794; answers A-27). Lecture Fiber optics: exercises on slides 20, 22, 27 (no printed answer, the module's solutions) and slide 36 (with a printed answer). Tooley ch.10 Test Your Understanding 10.2 (no printed answer).
Floyd has only two odd fibre problems (5 and 7); problems 4, 6, 8 are even so have no printed answers. Lecture items state the answer source: an image cut from slide 36, while slides 20, 22, 27 and Tooley 10.2 are the module's own solutions, verified in the notebook.
5. Describe the single mode in an optical fiber.
The printed answer matches section 13-1 (p.743): single-mode has a much smaller core than multimode, light travels nearly straight as one ray, more bandwidth and distance but higher cost.
Floyd Ch.13 · Problem 5
👁️ Show the book's original answer
7. Draw a basic block diagram of a fiber optics communications link.
The printed answer is Figure P-100: electrical signal source, electrical-to-light conversion, optical connectors, fibre-optic cable, light-to-electrical conversion and receiver; the same as Figure 13-8 and lecture slide 23.
Floyd Ch.13 · Problem 7
👁️ Show the book's original answer
S20. For a fibre with n₁ = 1.53 (SiO₂) and n₂ = 1.50 (SiO₂ with GeO₂ doping), compute the critical angle θc, the acceptance angle θa and the numerical aperture NA.
The slide prints no answer; this is the module's solution, recomputed in code in the notebook (not a printed answer image).
Lecture Fiber optics · slide 20
👁️ Show the book's original answer
θc = sin⁻¹(n₂/n₁) = sin⁻¹(1.50/1.53) = 78.64°.
NA = √(n₁² − n₂²) = √(2.3409 − 2.25) = √0.0909 = 0.3015.
θa = sin⁻¹(NA) = 17.55° (n₀ = 1).
S22. d_core = 62.5 µm, NA = 0.275, λ = 1300 nm. Calculate the normalised frequency V, the number of modes N of the fibre and the angle θa.
The slide prints no answer; the module's solution, verified in the notebook.
Lecture Fiber optics · slide 22
👁️ Show the book's original answer
V = π d NA / λ = π × 62.5×10⁻⁶ × 0.275 / 1300×10⁻⁹ = 41.54.
N ≈ V²/2 = 863 modes.
θa = sin⁻¹(0.275) = 15.96°.
S27. Calculate the maximal length l_max of the optical fibre (SiO₂): input power 1 mW, wavelength 1310 nm, minimum measurable output power 50 µW; use the attenuation coefficient given by the previous graph.
The slide prints no answer; the result depends on the α read from the graph.
Lecture Fiber optics · slide 27
👁️ Show the book's original answer
From the slide 26 graph: α ≈ 0.35 dB/km at 1310 nm. Budget 10 log₁₀(1 mW / 50 µW) = 13.01 dB. l_max = 13.01 / 0.35 = 37.2 km. (With 0.18 dB/km at 1550 nm: 72.3 km.)
S36. A step-index multimode fiber with a NA of 0.20 supports approximately 1000 modes at an 850 nm wavelength. What is the diameter of its core? How many modes does the fiber support at 1320 nm and 1550 nm?
Printed answer on the slide: d = 60.49 µm; at 1320 nm V = 28.78, M = 414; at 1550 nm V = 24.51, M = 301. Recomputed: V = √(2M) = 44.72; a = Vλ/(2π NA) = 30.25 µm so d = 60.50 µm; V(1320) = 28.80, M = 414.7; V(1550) = 24.52, M = 300.7. They agree within rounding.
Lecture Fiber optics · slide 36
👁️ Show the book's original answer
T10.2. Determine the critical angle for a glass-air interface (values of refractive index for glass and air are respectively 1.5 and 1.0).
The book prints no answer; the module's solution. The image above is the critical-angle slide.
Tooley ch.10 · Test Your Understanding 10.2 (and lecture slide 32)
👁️ Show the book's original answer
sin θc = n₂/n₁ = 1.0/1.5 = 0.6667, so θc = 41.81°.