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Optical Fibres

An optical fibre consists of a core (refractive index n1n_1) surrounded by a cladding (n2<n1n_2 \lt n_1). Light is guided by total internal reflection.

The numerical aperture:

NA=sinθmax=n12n22\mathrm{NA} = \sin\theta_{\mathrm{max}} = \sqrt{n_1^2 - n_2^2}

Where θmax\theta_{\mathrm{max}} is the maximum acceptance angle for light entering the fibre.

The number of modes supported depends on the V-number:

V=2πaλNAV = \frac{2\pi a}{\lambda}\mathrm{NA}

Where aa is the core radius.

  • Single-mode fibre: V<2.405V \lt 2.405. Only the fundamental HE11_{11} mode propagates.
  • Multimode fibre: V>2.405V \gt 2.405. Multiple modes propagate, causing modal dispersion.

Fibre attenuation is dominated by Rayleigh scattering (λ4\propto \lambda^{-4}) and infrared absorption peaks. The minimum attenuation for silica fibre is 0.2\sim 0.2 dB/km at λ1550\lambda \approx 1550 nm.

Dispersion broadens optical pulses as they travel, limiting the bit rate.

Modal dispersion occurs in multimode fibres because different modes travel at different speeds. This is the dominant dispersion mechanism in multimode fibres.

Chromatic dispersion arises from the wavelength dependence of the refractive index. It has two components: material dispersion (due to intrinsic glass properties) and waveguide dispersion (due to mode confinement). Standard silica fibre has zero chromatic dispersion at λ1300\lambda \approx 1300 nm.

Polarisation mode dispersion (PMD) results from birefringence in the fibre core, causing the two orthogonal polarisation components to travel at slightly different speeds.

Optical fibres are made by the Modified Chemical Vapour Deposition (MCVD) process. A thin layer of pure silica is deposited inside a rotating silica tube by passing SiCl4_4 and O2_2 through it. The tube is then collapsed into a solid preform rod at approximately 2000 ^\circC.

The preform is then placed in a drawing tower, heated to its melting point, and pulled into a thin fibre under tension. The fibre diameter is monitored precisely to maintain a 125 μ\mum outer diameter.

Doping with GeO2_2 or P2_2O5_5 increases the core refractive index, while doping with F or B2_2O3_3 decreases it for the cladding.

Step-index fibre has a uniform core refractive index with an abrupt step at the core-cladding boundary. Graded-index fibre has a core index that decreases parabolically from the centre, reducing modal dispersion significantly.

Photonic crystal fibres (PCFs) use a periodic array of air holes running along the fibre length to guide light via photonic bandgap effects. They can achieve single-mode operation over an extremely wide wavelength range and offer very high nonlinearity for supercontinuum generation.

Dispersion-shifted fibre moves the zero-dispersion wavelength to 1550 nm to coincide with the minimum attenuation window. Dispersion-flattened fibre maintains low dispersion across a broad wavelength range for WDM systems.

Telecommunications is the dominant application. Fibre links form the backbone of the internet, using wavelength-division multiplexing (WDM) to send multiple channels at different wavelengths on a single fibre, achieving Tb/s data rates.

Fibre optic sensors exploit changes in intensity, phase, polarisation, or wavelength caused by external stimuli. Fibre Bragg gratings measure strain and temperature in structural health monitoring of bridges, dams, and pipelines.

Medical applications include endoscopy (imaging via fibre bundles) and laser surgery, where high-power laser light is delivered through thin fibres.

A step-index fibre has n1=1.48n_1 = 1.48, n2=1.46n_2 = 1.46, and core radius a=25 μa = 25\ \mum. Calculate the NA, acceptance angle, and V-number at λ=850\lambda = 850 nm.

NA=1.4821.462=2.19042.1316=0.05880.242\mathrm{NA} = \sqrt{1.48^2 - 1.46^2} = \sqrt{2.1904 - 2.1316} = \sqrt{0.0588} \approx 0.242

θmax=arcsin(0.242)14.0\theta_{\mathrm{max}} = \arcsin(0.242) \approx 14.0^\circ

= \frac{2\pi \times 25}{0.85} \times 0.242 \approx 44.7$$ Since $V \gg 2.405$, this fibre is multimode. ### Worked Example 12.2 A 50 km fibre link has attenuation 0.35 dB/km at 1310 nm. Input power is 1 mW (0 dBm). Find the output power and the power lost. Total loss = $0.35 \times 50 = 17.5$ dB. Output power = $0 - 17.5 = -17.5$ dBm. Converting: $P_{\mathrm{out}} = 10^{-17.5/10} \approx 17.8\ \mu$W. The power lost is $1\ \mathrm{mW} - 17.8\ \mu\mathrm{W} \approx 0.982\ \mathrm{mW}$. ### Practice Problems 1. A fibre has NA = 0.20 and core radius 4 $\mu$m. Determine whether it supports single-mode operation at 1550 nm. 2. Calculate the pulse broadening from material dispersion for a 1 nm spectral width pulse travelling 100 km in silica fibre with $D_m = 17$ ps/(nm$\cdot$km) at 1550 nm. 3. Explain why graded-index fibres have lower modal dispersion than step-index fibres. 4. A link operates at 2.5 Gb/s over 80 km with 0.25 dB/km attenuation and 5 dB connector loss. Calculate the required input power for a receiver sensitivity of -25 dBm. 5. Derive the expression for the V-number starting from the wave equation in a step-index fibre. What condition determines the cut-off of the first higher-order mode? ### Key Takeaways - Total internal reflection at the core-cladding interface confines light in the fibre. - The numerical aperture measures the light-gathering ability of the fibre. - The V-number determines whether a fibre supports single-mode or multimode propagation. - Attenuation sets a fundamental limit on the unrepeatered transmission distance. - Dispersion broadens pulses and limits the data rate; different dispersion types require different compensation strategies. - Fibre fabrication via MCVD and drawing produces high-quality fibres with minimal loss.