Optical fibres: types, numerical aperture and losses

Why it matters

Optical fibre carries nearly all long-distance data traffic and is increasingly used in plants for immunity to electromagnetic interference, intrinsic safety in hazardous areas and distributed sensing. To design a link or a fibre sensor you must know how much light a fibre accepts (numerical aperture), how many modes it carries, how much power it loses per kilometre and how much it spreads a pulse.

Key ideas

Structure and guiding. A fibre has a core of refractive index n₁ surrounded by a cladding of slightly lower index n₂, with a protective buffer. Light striking the core–cladding boundary at more than the critical angle θ_c = sin⁻¹(n₂/n₁) is totally internally reflected and stays in the core.

Acceptance angle and numerical aperture. Rays entering the end face within a cone of half-angle θ_a are guided. From Snell's law, NA = n₀·sin θ_a = √(n₁² − n₂²) ≈ n₁√(2Δ), where Δ = (n₁ − n₂)/n₁ is the relative index difference (typically 0.3–2%). A larger NA collects more light from an LED but, in a multimode fibre, increases modal dispersion.

Modes and the V-number. Ray paths at different angles correspond to modes. The normalised frequency V = (2πa/λ)·NA (a is the core radius) sets how many modes propagate: a step-index fibre is single-mode when V < 2.405; for large V a step-index fibre carries about V²/2 modes (graded-index about V²/4).

Fibre types.

• Step-index multimode: core 50–200 µm, uniform n₁; large NA, easy coupling, strong modal dispersion; short links, sensors, plastic optical fibre.
• Graded-index multimode: core 50 or 62.5 µm, n falls parabolically from axis to cladding, so oblique rays travel faster in the lower-index outer region and modal delays nearly equalise; used for LANs and data centres up to a few km.
• Single-mode: core about 8–10 µm, Δ ≈ 0.3%; no modal dispersion, so only chromatic and polarisation-mode dispersion remain; used with lasers for long-haul links.

Attenuation (losses). Power falls exponentially; in decibels, α = (10/L)·log₁₀(P_in/P_out) dB/km.

• Absorption: intrinsic UV and infrared absorption of silica, and extrinsic absorption by impurities, mainly OH⁻ ions (peak near 1.38 µm).
• Rayleigh scattering from microscopic density fluctuations, proportional to 1/λ⁴ — the dominant loss at short wavelengths.
• Bending losses: macrobending (radiation from tight curves) and microbending (small random deformations from cabling).
• Splice, connector and coupling losses, counted separately in a link budget. These give the low-loss windows: about 2.5 dB/km at 850 nm, 0.35 dB/km at 1310 nm and 0.2 dB/km at 1550 nm.

Dispersion (pulse spreading, not a power loss).

• Intermodal: in step-index multimode fibre, the delay difference between the axial and the steepest guided ray is δT ≈ L·n₁Δ/c. Graded index reduces it by a factor of about Δ/8.
• Chromatic (intramodal): material plus waveguide dispersion, δT = |D|·L·Δλ, with D in ps/(nm·km); about zero near 1.31 µm and about 17 ps/(nm·km) at 1.55 µm in standard fibre. A broad-spectrum LED suffers far more than a narrow laser.
• A rough bit-rate limit for NRZ data is B ≈ 1/(2δT).

Link power budget. Received power (dBm) = launched power − fibre loss − splice losses − connector losses; it must exceed the receiver sensitivity by a safety margin (typically 3–6 dB).

Formulas

θ_c = sin⁻¹(n₂/n₁) θ_c critical angle; n₁ core index, n₂ cladding index.

NA = n₀·sin θ_a = √(n₁² − n₂²) ≈ n₁√(2Δ), Δ = (n₁ − n₂)/n₁ θ_a acceptance half-angle, n₀ index outside (1 for air).

V = (2πa/λ)·NA; single-mode if V < 2.405; M ≈ V²/2 (step-index modes) a core radius (m), λ free-space wavelength (m).

α(dB/km) = (10/L)·log₁₀(P_in/P_out) L length (km), P in W or mW (same units).

δT_modal ≈ L·n₁Δ/c (step index), δT_chrom = |D|·L·Δλ c = 3 × 10⁸ m/s; D dispersion parameter (ps/(nm·km)); Δλ source spectral width (nm).

B_max ≈ 1/(2δT)

P_rx(dBm) = P_tx(dBm) − αL − Σ splice and connector losses (dB)

Worked examples

Example 1 (standard). A step-index fibre has n₁ = 1.48, n₂ = 1.46. Find NA, the acceptance angle in air, Δ and the critical angle.

1. NA = √(1.48² − 1.46²) = √(2.1904 − 2.1316) = √0.0588 = 0.2425.
2. θ_a = sin⁻¹(0.2425) = 14.0°.
3. Δ = (1.48 − 1.46)/1.48 = 0.0135.
4. θ_c = sin⁻¹(1.46/1.48) = 80.6°. Answer: NA ≈ 0.243, θ_a ≈ 14.0°, Δ ≈ 1.35%, θ_c ≈ 80.6°.

Example 2 (GATE level). The same fibre has a core diameter of 50 µm and is used at 850 nm. Find V, the approximate number of modes, the modal pulse spread per km and the bit-rate–length limit. What core radius would make a fibre with NA = 0.12 single-mode at 1310 nm?

1. V = 2π × 25 × 10⁻⁶ × 0.2425/(0.85 × 10⁻⁶) = 44.8.
2. Modes ≈ V²/2 = 44.8²/2 ≈ 1004.
3. δT/L ≈ n₁Δ/c = 1.48 × 0.0135/(3 × 10⁸) = 6.67 × 10⁻¹¹ s/m = 66.7 ns/km.
4. B·L ≈ 1/(2 × 66.7 ns) = 7.5 Mbit/s·km.
5. Single-mode condition: a < 2.405λ/(2π·NA) = 2.405 × 1.31/(2π × 0.12) µm = 4.18 µm. Answer: V ≈ 44.8, about 1000 modes, ≈ 67 ns/km, ≈ 7.5 Mbit/s·km; a < 4.18 µm (core diameter < 8.4 µm).

Example 3 (power budget). A 1310 nm link launches −3 dBm into 40 km of fibre (0.35 dB/km) with 8 splices of 0.1 dB and 2 connectors of 0.5 dB. The receiver needs −30 dBm. Find the received power and margin.

1. Fibre loss = 0.35 × 40 = 14 dB; splices = 0.8 dB; connectors = 1.0 dB; total 15.8 dB.
2. P_rx = −3 − 15.8 = −18.8 dBm.
3. Margin = −18.8 − (−30) = 11.2 dB. Answer: −18.8 dBm received, 11.2 dB margin.

Common mistakes

• Using NA = n₁ − n₂, or forgetting the square root.
• Using the core diameter instead of the radius in V.
• Calling dispersion a loss: it spreads pulses and limits bit rate but does not remove power.
• Dividing powers in the wrong order in the attenuation formula (a negative dB/km means you inverted it).
• Assuming graded-index fibre is single-mode; it is multimode with reduced modal dispersion.
• Adding losses in linear units; add them in dB.

For GATE IN

• NA, acceptance angle, critical angle and Δ numericals.
• V-number, single-mode cut-off wavelength or radius, number of modes.
• Attenuation in dB/km from input and output power, and link power budgets.
• Modal and chromatic pulse spreading, and the resulting bit-rate limit.
• Conceptual comparison of step-index, graded-index and single-mode fibres.

Quick check

1. A fibre loses half its power over 10 km. What is α?
2. NA = 0.3. What is the acceptance half-angle in air?
3. V = 2.0 at 1550 nm. Single-mode or multimode?
4. Which loss mechanism varies as 1/λ⁴? Answers: 1. 0.301 dB/km. 2. 17.5°. 3. Single-mode. 4. Rayleigh scattering.