Question 1 of 7: Step-Index Fiber — Acceptance Angle, Cladding Index, Modes and Dispersion

Question 1: Step-Index Fiber — Acceptance Angle, Cladding Index, Modes and Dispersion (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) $\theta_{a,\text{air}}$, (b) $\theta_{a,\text{water}}$, (c) cladding index $n_2$, (d) guided-mode count $M$, (e) intermodal dispersion in ns/km, (f) the core diameter that puts the cutoff at the single-mode limit.

Approach. Parts (a)–(c) follow directly from the definition $\mathrm{NA}=n_0\sin\theta_a=\sqrt{n_1^2-n_2^2}$ evaluated in the external medium of interest; parts (d)–(f) use the normalized frequency $V=2\pi a\,\mathrm{NA}/\lambda$ against the step-index mode-count approximation $M\approx V^2/2$ and the single-mode cutoff $V_c=2.405$.

1. Part (a) — acceptance angle in air. By definition $\mathrm{NA}=n_0\sin\theta_a$ with $n_0=1$ for air, so $$\theta_{a,\text{air}}=\arcsin(\mathrm{NA})=\arcsin(0.2)=\boxed{11.5^{\circ}}.$$
2. Part (b) — acceptance angle in water. Snell's law at the fiber's end face requires $n_0\sin\theta_a=\mathrm{NA}$ regardless of what $n_0$ is, so immersing the input face in water ($n_0=1.33$) simply rescales the angle: $$\theta_{a,\text{water}}=\arcsin\!\left(\frac{\mathrm{NA}}{n_{\text{water}}}\right) =\arcsin\!\left(\frac{0.2}{1.33}\right)=\boxed{8.65^{\circ}}.$$ The narrower cone in water is exactly what is expected: a denser entrance medium bends rays closer to the fiber axis before they even reach the core.
3. Part (c) — cladding index. Rearranging $\mathrm{NA}=\sqrt{n_1^2-n_2^2}$, $$n_2=\sqrt{n_1^2-\mathrm{NA}^2}=\sqrt{1.500^2-0.2^2}=\sqrt{2.21}=\boxed{1.487}.$$
4. Part (d) — number of guided modes. With core radius $a=50\ \mu\text{m}$, the normalized frequency is $$V=\frac{2\pi a\,\mathrm{NA}}{\lambda}=\frac{2\pi(50\times10^{-6})(0.2)}{850\times10^{-9}}=73.9.$$ Since $V\gg V_c=2.405$ the fiber is heavily multimode, and the step-index mode-count approximation gives $$M\approx\frac{V^2}{2}=\frac{73.9^2}{2}\approx\boxed{2730\ \text{modes}}.$$
5. Part (e) — intermodal dispersion. For a step-index fiber the ray-theory (meridional-ray) pulse spread per unit length is $$\frac{\Delta\tau}{L}=\frac{n_1\Delta}{c},\qquad \Delta=\frac{n_1-n_2}{n_1}=\frac{1.500-1.487}{1.500}=0.00893.$$ Substituting, $$\frac{\Delta\tau}{L}=\frac{(1.500)(0.00893)}{2.998\times10^{8}\ \text{m/s}} =4.47\times10^{-11}\ \text{s/m}=\boxed{44.7\ \text{ns/km}}.$$
6. Part (f) — single-mode diameter. Single-mode operation requires the cutoff condition $V\le V_c=2.405$ for the $\mathrm{LP}_{01}$ mode; setting $V=2.405$ and solving for the core radius, $$a_{sm}=\frac{2.405\,\lambda}{2\pi\,\mathrm{NA}}=\frac{2.405(850\times10^{-9})}{2\pi(0.2)} =1.63\ \mu\text{m}\ \Rightarrow\ 2a_{sm}=\boxed{3.25\ \mu\text{m}}.$$ This core would have to shrink to about 1/30 of the original 100 μm diameter to strip out every mode but the fundamental — the same numerical aperture forces a much smaller core if only one mode is wanted.