Optical fibres

Compare optical-fibre types, explain total internal reflection and calculate dispersion, attenuation and decibel link budgets.

  • Advanced Physics
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An optical fibre traps light in a glass core by total internal reflection, so it can carry signals for tens of kilometres. Two things limit a fibre link: dispersion, which spreads each pulse out in time, and attenuation, which reduces its power. This lesson covers how fibres guide light and how each limit is reduced.

Structure and guiding

An optical fibre has a glass core surrounded by a cladding of slightly lower refractive index, protected by a plastic buffer coating. Light entering the end within the fibre’s acceptance cone meets the core–cladding boundary beyond the critical angle and is totally internally reflected, so it stays in the core. The half-angle of the acceptance cone in air is set by the numerical aperture, NA = sin θₘₐₓ = square root of (n₁² - n₂²), where n₁ is the core index and n₂ the cladding index.

Modes and dispersion

Scroll across the figure to read all labels.

The ray model describes multimode fibres. A single-mode fibre is better described by one guided field pattern than by a straight ray.

Step-index multimode fibre

The core is wide (typically 50–100 µm) with a uniform index, so many ray paths, or modes, are guided. A steep ray zig-zags and travels further than the axial ray, so parts of one pulse arrive at different times. This modal (intermodal) dispersion broadens pulses. If they overlap, the receiver cannot separate them, so the usable data rate falls as the fibre gets longer.

For a fibre of length L, the axial ray takes t = Ln₁/c. The steepest guided ray, which meets the boundary exactly at the critical angle, travels a path longer by the factor n₁/n₂, so the spread is

Δ t ≈ (Ln₁/c)(n₁/n₂ - 1).

Worked example 1

Modal spread in a step-index fibre

Problem

A 1.0 km step-index fibre has core index n₁ = 1.48 and cladding index n₂ = 1.46. Estimate the modal spread of a pulse and the highest pulse rate it supports.

Show full solution
  1. Find the axial transit time

    Method

    Use t = Ln₁/c.

    Reason

    The axial ray takes the shortest path.

    Working

    Answer

    Ln₁/c = (1000)(1.48)/(3.00 × 10⁸) = 4.93 μs.

  2. Find the spread

    Method

    Multiply by n₁/n₂-1.

    Reason

    The steepest guided ray travels further by the factor n₁/n₂.

    Working

    Answer

    n₁/n₂-1 = 0.0137, so Δ t ≈ (4.93 μs)(0.0137) ≈ 68 ns.

  3. State the answer

    Working

    Answer

    Δ t ≈ 68 ns. Pulses must be at least this far apart, so the fibre supports roughly 1/Δ t ≈ 15 million pulses per second over 1 km.

Graded-index multimode fibre

The core index is highest on the axis and falls gradually towards the cladding. Rays bend back towards the axis instead of reflecting sharply. The longer outer paths run through lower-index glass, where light travels faster, so the arrival times of different modes nearly match. Modal dispersion is greatly reduced, but not eliminated.

Single-mode fibre

The core is only a few micrometres across (typically about 8–10 µm for the 1310 nm and 1550 nm bands), which is small enough that only one mode is guided. That mode is not a single straight ray along the axis. It is a field pattern concentrated in the core that extends slightly into the cladding, as the figure shows. With one mode there is no modal dispersion. What remains is chromatic dispersion: any real source emits a small range of wavelengths, and these travel at slightly different speeds. Single-mode fibre therefore carries the highest data rates over the longest distances. The trade-offs are that its tiny core needs a laser source and precise alignment at joints, whereas multimode fibre accepts cheaper LED sources and is easier to connect.

Attenuation

Attenuation reduces the power without, by itself, changing the pulse shape. Its main causes are:

  • Absorption. Silica absorbs strongly in the ultraviolet, through electronic transitions, and increasingly beyond about 1.6 µm in the infrared, through molecular vibrations. Impurities add absorption peaks; residual hydroxyl (OH⁻) ions absorb strongly near 1380 nm.
  • Rayleigh scattering from tiny, frozen-in density variations in the glass. It is proportional to 1/λ⁴, so it dominates at shorter wavelengths.
  • Bending losses. A tight bend (macrobending) or small kinks from poor cabling (microbending) make rays meet the boundary below the critical angle, so light escapes into the cladding.
  • Splice and connector losses. These come from misaligned or mismatched cores and from air gaps. At an air gap, part of the light is reflected at each surface. Of the power arriving at a surface, the fractions reflected (R), transmitted (T) and absorbed (A) satisfy R + T + A = 1, counting only the light at that surface, so any reflected fraction is lost from the forward signal.

Wavelength windows

Rayleigh scattering falls as the wavelength increases, while infrared absorption rises beyond about 1.6 µm. Silica fibre therefore has a loss minimum near 1550 nm (around 0.2 dB km⁻¹), with another useful window near 1310 nm; older multimode links use 850 nm. Longer wavelengths are not always better: past the minimum, infrared absorption makes the loss rise again.

Loss in decibels is 10 log ₁₀(Pᵢₙ/Pₒᵤₜ), so losses along a link simply add. If the power falls exponentially, P = P₀e^(-μ x), the loss per unit length in dB is 4.34μ.

Worked example 2

Problem

A 40 km link uses fibre with 0.25 dB km⁻¹ loss, two connectors of 0.5 dB each and eight splices of 0.1 dB each. What fraction of the launched power arrives, and how much of a 1.0 mW signal is left?

Show full solution
  1. Add the losses

    Method

    Total the decibel losses along the link.

    Reason

    Decibel losses add because they are logarithms of power ratios.

    Working

    Answer

    40(0.25) + 2(0.5) + 8(0.1) = 10 + 1.0 + 0.8 = 11.8 dB.

  2. Convert to a power ratio

    Method

    Use Pₒᵤₜ/Pᵢₙ = 10^(-loss/10).

    Reason

    The decibel definition, 10 log ₁₀(Pᵢₙ/Pₒᵤₜ), inverts to a power of ten.

    Working

    Answer

    Pₒᵤₜ/Pᵢₙ = 10^(-1.18) = 0.066.

  3. State the answer

    Working

    Answer

    About 6.6% of the power arrives: a launched 1.0 mW leaves about 0.066 mW.

Check your understanding 1 The longest usable link

A transmitter launches 1.0 mW, and the receiver needs at least 10 µW. The fibre loses 0.35 dB km⁻¹ at 1310 nm, and connectors and splices add 3.0 dB in total. What is the longest usable link?

Show answer

The allowed loss is 10 log ₁₀(1.0 mW/10 μW) = 10 log ₁₀100 = 20 dB. Subtracting the fixed 3.0 dB leaves 17 dB for the fibre, so Lₘₐₓ = 17/0.35 ≈ 49 km. If you divided 20 dB by 0.35 dB km⁻¹, you forgot the connector and splice losses. If you used 20 log ₁₀, you treated power like a field amplitude.

Common mistakes

  • Making the cladding’s refractive index larger than the core’s. Guiding needs n_core > n_cladding.
  • Confusing dispersion, which spreads a pulse, with attenuation, which weakens it.
  • Using 20 log ₁₀ for a power ratio. Power in decibels is 10 log ₁₀(P₁/P₂).
  • Leaving out connector and splice losses from a link budget.
  • Saying single-mode fibres have no dispersion. They have no modal dispersion, but chromatic dispersion remains.

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Syllabus and review details

No official syllabus alignment is listed for this lesson.

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