A straight step-index fibre has a uniform core of index and cladding of index

PHYSICS · Subjective · Fermi University Physics · Optics & Photonics · Beams, resonators and guided-wave photonics · Beams, resonators and guided-wave photonics

Optics and Photonics, Ch Optical fibres in communication and endoscopy

Problem

A straight step-index fibre has a uniform core of index n1n_1 and cladding of index n2<n1n_2 < n_1. Light in the core travels at c/n1c/n_1. Consider meridional rays only and use the ray picture of the zig-zag: a guided ray bounces at the wall at an angle of incidence at least the critical angle θc\theta_c, where sin⁡θc=n2/n1\sin\theta_c = n_2/n_1. (a) Show that, over an axial length LL, the extreme guided ray (the one incident at exactly θc\theta_c) travels a geometrical path L n1/n2L\,n_1/n_2, and hence that the spread in arrival time between it and the axial ray is ΔtL=n1c(n1n2−1).\frac{\Delta t}{L} = \frac{n_1}{c}\left(\frac{n_1}{n_2} - 1\right). (b) Writing the fractional index step as Δ=(n1−n2)/n1\Delta = (n_1 - n_2)/n_1, show that the exact result is (n1/c) Δ/(1−Δ)(n_1/c)\,\Delta/(1-\Delta) and that to first order in Δ\Delta it becomes Δt/L≈n1Δ/c=(n1−n2)/c\Delta t/L \approx n_1\Delta/c = (n_1 - n_2)/c. (c) For n1=1.46n_1 = 1.46 and n2=1.44n_2 = 1.44, with c=3.00×108c = 3.00\times10^{8} m s−1^{-1}, evaluate the exact spread and the first-order spread, both in ns km−1^{-1}.

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