Consider a step-index fibre of core index and cladding 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

Consider a step-index fibre of core index n1n_1 and cladding index n2n_2, with n1>n2n_1 > n_2. An axial ray and a ray that just satisfies the critical angle θc\theta_c at the core-cladding wall (where sin⁡θc=n2/n1\sin\theta_c = n_2/n_1) both travel over an axial length LL of the fibre. (a) Using the geometry of the zig-zag path, show that the critical ray's actual path length is L/sin⁡θc=Ln1/n2L/\sin\theta_c = Ln_1/n_2, and hence derive ΔtL=n1c(n1n2−1).\frac{\Delta t}{L} = \frac{n_1}{c}\left(\frac{n_1}{n_2}-1\right). (b) Write n2=n1−Δnn_2 = n_1-\Delta n, and show that the exact result can be written Δt/L=(n1/c)(Δn/n2)\Delta t/L = (n_1/c)(\Delta n/n_2), and that to first order in Δn\Delta n (using n2≈n1n_2 \approx n_1 in the denominator) this reduces to Δt/L≈Δn/c\Delta t/L \approx \Delta n/c. (c) Take n1=1.50n_1 = 1.50 and n2=1.35n_2 = 1.35 (so Δn=0.15\Delta n = 0.15, a much larger step than is used in practice). Evaluate both the exact result and the first-order approximation in nanoseconds per kilometre, state the percentage difference, and comment on whether the first-order approximation is still trustworthy. Take c=3.00×108c = 3.00\times10^8 m s−1^{-1}.

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