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Optical Fibres

01Optical Fibres

Fibre structure

This section covers step-index optical fibre structure, total internal reflection and the function of the cladding.

Guiding light through a fibre

Optical fibres guide light using total internal reflection. Light refracts as it enters the fibre, repeatedly reflects at the core–cladding boundary, and refracts again when it leaves the fibre.

For total internal reflection at the core–cladding boundary, the core must have a higher refractive index than the cladding and the angle of incidence must be greater than the critical angle.

A step-index optical fibre consists of:

  • a glass or plastic core with the higher refractive index;
  • lower-refractive-index cladding around the core;
  • an outer sheath.

In the step-index arrangement considered here, the refractive index is higher in the core than in the material surrounding it.

cladding,ncladcladding,ncladouter sheathncore>ncladcore,ncorei>µclight

Function of the cladding

  • Its lower refractive index provides the boundary needed for total internal reflection.
  • It shields the core surface from physical damage that could interfere with the confinement of light.
  • It helps keep the light inside the core, reducing deterioration of the transmitted signal.
  • It separates neighbouring fibre cores, reducing unwanted transfer of information between them.

Optical fibres are used for communication signals such as telephone and internet transmission, and for carrying light in medical imaging systems such as endoscopes.

Worked example: The speed of light is 2.027×108 m s−12.027\times10^8\text{ m s}^{-1} in the core and 2.055×108 m s−12.055\times10^8\text{ m s}^{-1} in the cladding. A ray enters the end of the fibre and then reaches the core–cladding boundary at the critical angle. Calculate its angle of incidence as it enters from air.

1. Find the two refractive indices.

ncore=3.00×1082.027×108=1.48n_{\mathrm{core}}=\frac{3.00\times10^8}{2.027\times10^8}=1.48 nclad=3.00×1082.055×108=1.46n_{\mathrm{clad}}=\frac{3.00\times10^8}{2.055\times10^8}=1.46

2. Calculate the critical angle.

sin⁡θc=ncladncore=2.027×1082.055×108\sin\theta_c=\frac{n_{\mathrm{clad}}}{n_{\mathrm{core}}} =\frac{2.027\times10^8}{2.055\times10^8} θc≈80.5∘\theta_c\approx80.5^\circ

The ray inside the core therefore makes an angle of

90∘−80.5∘=9.5∘90^\circ-80.5^\circ=9.5^\circ

with the normal to the end face.

3. Apply Snell's law at the air–core boundary.

1×sin⁡θ=1.48sin⁡9.5∘1\times\sin\theta =1.48\sin9.5^\circ

sin⁡θ=0.2443\sin\theta=0.2443 θ≈14.1∘\theta\approx14.1^\circ

So the entry angle is about 14.1∘14.1^\circ.

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