Question 1 of 7: Silica Step-Index Fiber — Acceptance Angle, Modes, Dispersion, Bend Loss and PMD

Question 1: Silica Step-Index Fiber — Acceptance Angle, Modes, Dispersion, Bend Loss and PMD (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) acceptance angle $\theta_a$, (b) mode count $M$, (c) modal dispersion, (d) chromatic dispersion, (e) RZ bandwidth–length product, (f) critical bend radius $R_c$, (g) polarization-mode dispersion (PMD).

Approach. Parts (a)–(b) use $\mathrm{NA}=\sqrt{n_1^2-n_2^2}$ and the normalized frequency $V=2\pi a\,\mathrm{NA}/\lambda$; (c)–(e) combine ray-theory modal delay with the given chromatic-dispersion coefficient to get a bandwidth–length product; (f) is a ray-geometry argument for a fiber bent to radius $R$; (g) is conceptual.

1. Part (a) — numerical aperture and acceptance angle. With $n_1=1.460$, $n_2=1.459$, $$\mathrm{NA}=\sqrt{n_1^2-n_2^2}=\sqrt{1.460^2-1.459^2}=0.0540,$$ $$\theta_a=\arcsin(\mathrm{NA})=\boxed{3.10^\circ}.$$ The index step is tiny ($\Delta n=0.001$), so this is a very weakly guiding, low-NA fiber — the acceptance cone is barely 3° wide.
2. Part (b) — number of guided modes. Core radius $a=31.25\ \mu\text{m}$, so the normalized frequency is $$V=\frac{2\pi a\,\mathrm{NA}}{\lambda}=\frac{2\pi(31.25\times10^{-6})(0.0540)}{1550\times10^{-9}}=6.84,$$ and for a step-index fiber $M\approx V^2/2$: $$M\approx\frac{6.84^2}{2}=\boxed{23\ \text{modes}}.$$
3. Part (c) — modal (intermodal) dispersion. The ray-theory delay difference between the fastest (axial) and slowest (critical-angle) meridional rays per unit length is $$\frac{\Delta\tau_{\text{modal}}}{L}=\frac{n_1\Delta}{c}\approx\frac{n_1-n_2}{c}, \qquad \Delta=\frac{n_1-n_2}{n_1},$$ $$\frac{\Delta\tau_{\text{modal}}}{L}=\frac{0.001}{2.998\times10^{8}\ \text{m/s}} =\boxed{3.34\ \text{ns/km}}.$$
4. Part (d) — chromatic dispersion. The source's own linewidth sets the pulse spreading through the given coefficient $D$: $$\frac{\Delta\tau_{\text{chrom}}}{L}=D\,\Delta\lambda =15\ \frac{\text{ps}}{\text{nm}\cdot\text{km}}\times120\ \text{nm} =\boxed{1.80\ \text{ns/km}}.$$
5. Part (e) — RZ bandwidth–length product. The two mechanisms are statistically independent, so the total pulse spread combines in quadrature: $$\frac{\Delta\tau}{L}=\sqrt{\left(\frac{\Delta\tau_{\text{modal}}}{L}\right)^2 +\left(\frac{\Delta\tau_{\text{chrom}}}{L}\right)^2} =\sqrt{3.34^2+1.80^2}=3.79\ \text{ns/km}.$$ An RZ pulse occupies roughly half the bit slot, so it tolerates about twice the spreading an NRZ pulse would for the same power penalty; using the design rule $B\,\Delta\tau\le0.4$ (double the usual NRZ $0.2$ criterion), $$B\cdot L\approx\frac{0.4}{\Delta\tau/L}=\frac{0.4}{3.79\times10^{-9}\ \text{s/km}} =\boxed{105.5\ \text{Mb/s}\cdot\text{km}}.$$
6. Part (f) — critical bend radius.
   Ray traveling along the fiber axis meets the sidewall at the critical angle when R = a n2/(n1-n2).
   Consider a ray that travels exactly along the axis before the bend begins. Relative to the center of curvature $O$, that ray is tangent to the axis circle of radius $R$, so its perpendicular distance from $O$ stays fixed at $R$ while its distance from $O$ grows with arc length $s$ travelled: $d(s)=\sqrt{R^2+s^2}$. It meets the outer core–cladding wall (radius $R+a$) when $d(s)=R+a$, and the angle of incidence there (measured from the local radial normal) satisfies $\cos\theta_i=s/(R+a)$. Setting $\theta_i=\theta_c$ (guiding just barely holds, $\sin\theta_c=n_2/n_1$) and solving the resulting quadratic in $R$ collapses to the clean closed form $$R_c=\frac{a\,n_2}{n_1-n_2}=\frac{(31.25\times10^{-6}\ \text{m})(1.459)}{0.001} =\boxed{45.6\ \text{mm}}.$$ Because this fiber's index step is so small, the guiding margin is thin and the fiber must stay straighter than a ${\sim}4.6\ \text{cm}$ radius almost everywhere along its length — any tighter bend leaks the axial ray into the cladding.
7. Part (g) — polarization-mode dispersion. A "single-mode" fiber actually guides two orthogonally polarized versions of the fundamental mode. A perfectly circular, stress-free core would keep them degenerate, but real fiber has residual core ellipticity and frozen-in mechanical stress, which makes the two polarization axes see slightly different effective indices (birefringence $\Delta n\sim10^{-7}$–$10^{-6}$). The two polarizations then travel at different group velocities and arrive with a differential group delay that grows, for a fiber with random, slowly-varying birefringence along its length, as $\sqrt{L}$ rather than linearly — the signature that distinguishes PMD from ordinary chromatic or modal dispersion. Typical modern single-mode fiber has a PMD coefficient of order $0.1$–$1\ \text{ps}/\sqrt{\text{km}}$, small next to the modal/chromatic terms above but still a limiting impairment at multi-Gb/s rates over long spans.