Step-index fibre — acceptance, cladding index, mode count and intermodal dispersion: 22-Elec-B10 Electro-Optical Engineering, December 2017, Q1

Question 1: Step-index fibre — acceptance, cladding index, mode count and intermodal 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. A silica step-index fibre characterised by its numerical aperture, core index and core size, operating in the first telecom window.

Given data — Question 1
Quantity	Symbol	Value
Operating wavelength	$\lambda$	850 nm
Numerical aperture	$\mathrm{NA}$	0.200
Core refractive index	$n_{1}$	1.500
Core diameter (radius)	$2a$ ($a$)	100 µm (50 µm)
External medium, part (b)	$n_{0}$	1.33 (water)

Find. The acceptance half-angle in air and in water, the cladding index, the number of guided modes, the intermodal delay spread per kilometre, and the core diameter that would make the fibre single-moded at 850 nm.

Meridional-ray picture of the step-index fibre. Only rays inside the acceptance cone refract into the core steeply enough to strike the core–cladding boundary beyond the critical angle and be guided by total internal reflection.

Approach. The numerical aperture is the single quantity that ties the launch geometry ($\mathrm{NA} = n_{0}\sin\theta_{a}$) to the guide construction ($\mathrm{NA}^{2} = n_{1}^{2} - n_{2}^{2}$) and to the normalised frequency $V = 2\pi a\,\mathrm{NA}/\lambda$, from which the mode count, the intermodal spread and the single-mode cutoff all follow.

Acceptance half-angle in air. The definition of numerical aperture for a ray launched from a medium of index $n_{0}$ is $$\mathrm{NA} = n_{0}\sin\theta_{a}\qquad\Rightarrow\qquad \sin\theta_{a} = \frac{\mathrm{NA}}{n_{0}} = \frac{0.200}{1.000} = 0.200$$ so that $$\boxed{\theta_{a} = \arcsin(0.200) = 11.54^\circ}$$ The full acceptance cone, which is what a connector or a launch lens actually has to fill, is $2\theta_{a} = 23.07^\circ$.
Acceptance half-angle in water. The numerical aperture is a property of the fibre; immersing the end face changes only the medium the ray arrives from, so $$\sin\theta_{a}' = \frac{\mathrm{NA}}{n_{0}} = \frac{0.200}{1.33} = 0.1504 \qquad\Rightarrow\qquad \theta_{a}' = 8.65^\circ$$ Immersion therefore narrows the cone (to $17.30^\circ$ full angle) because the denser medium refracts the incoming ray less steeply at the end face. Note that the guided power for a fully filled launch is unchanged; only the external cone shrinks.
Cladding index. Inverting the construction relation, $$\mathrm{NA} = \sqrt{n_{1}^{2} - n_{2}^{2}}\qquad\Rightarrow\qquad n_{2} = \sqrt{n_{1}^{2} - \mathrm{NA}^{2}} = \sqrt{1.500^{2} - 0.200^{2}} = \sqrt{2.2100}$$ $$\boxed{n_{2} = 1.4866}$$ The corresponding relative index difference is $\Delta = (n_{1}-n_{2})/n_{1} = 0.01339/1.500 = 8.93\times10^{-3}$, i.e. about 0.89 % — a typical weakly-guiding multimode value.
Normalised frequency and mode count. With the core radius $a = 50\ \mu\text{m}$, $$V = \frac{2\pi a}{\lambda}\,\mathrm{NA} = \frac{2\pi (50\times10^{-6})(0.200)}{850\times10^{-9}} = 73.9$$ Because $V \gg 2.405$ the fibre is strongly multimoded, and the standard large-$V$ estimate for a step-index guide (counting both polarisations) gives $$M \approx \frac{V^{2}}{2} = \frac{73.9^{2}}{2} \qquad\Rightarrow\qquad \boxed{M \approx 2.73\times10^{3}\ \text{modes}}$$
Intermodal dispersion. The worst case is the difference between the axial ray and the ray launched at the critical angle. Their transit times over a length $L$ differ by $$\frac{\delta\tau}{L} = \frac{n_{1}\Delta}{c} = \frac{n_{1}-n_{2}}{c} = \frac{1.500 - 1.4866}{2.998\times10^{8}} = 4.466\times10^{-11}\ \text{s/m}$$ $$\boxed{\frac{\delta\tau}{L} = 44.7\ \text{ns/km}}$$ Using instead the small-$\Delta$ form with $\Delta = \mathrm{NA}^{2}/(2n_{1}^{2})$ gives 44.5 ns/km; the two agree to better than half a percent, which is well inside the precision of the quoted numerical aperture. A spread of this size limits the fibre to roughly $0.35/(44.7\ \text{ns/km}) \approx 8\ \text{Mb/s}$ over one kilometre.
Core diameter for single-mode operation. Only the $\mathrm{LP}_{01}$ mode propagates when $V \le 2.405$, so $$a \le \frac{2.405\,\lambda}{2\pi\,\mathrm{NA}} = \frac{2.405 (850\times10^{-9})}{2\pi (0.200)} = 1.63\ \mu\text{m}$$ $$\boxed{2a \le 3.25\ \mu\text{m}}$$ The core would have to shrink by a factor of about 31 in diameter. In practice a single-mode fibre at 850 nm is built with a much smaller numerical aperture (typically 0.1) so that a manufacturable 5–6 µm core still satisfies the cutoff.

Final results — Question 1
Quantity	Symbol	Result
(a) Acceptance half-angle in air (full cone)	$\theta_{a}$	11.54° (23.07°)
(b) Acceptance half-angle in water	$\theta_{a}'$	8.65°
(c) Cladding refractive index	$n_{2}$	1.4866
(d) Normalised frequency / guided modes	$V$ / $M$	73.9 / ≈ 2732
(e) Intermodal delay spread	$\delta\tau/L$	44.7 ns/km
(f) Single-mode core diameter at 850 nm	$2a$	≤ 3.25 µm

Topic. Ray-optics description of a step-index fibre: acceptance, guidance condition and modal content.

Key relations. $\mathrm{NA} = n_{0}\sin\theta_{a} = \sqrt{n_{1}^{2}-n_{2}^{2}} \approx n_{1}\sqrt{2\Delta}$; $V = 2\pi a\,\mathrm{NA}/\lambda$; $M \approx V^{2}/2$ (step index); $\delta\tau/L = n_{1}\Delta/c$; single-mode cutoff $V \le 2.405$ (the first zero of $J_{0}$).

Why this works. A meridional ray is guided only if it strikes the core–cladding interface beyond the critical angle $\theta_{c}$ with $\sin\theta_{c} = n_{2}/n_{1}$. Applying Snell’s law at the end face and then the total-internal-reflection condition at the boundary eliminates the internal angle entirely and leaves $n_{0}\sin\theta_{a} = \sqrt{n_{1}^{2}-n_{2}^{2}}$, so a single measured number — the numerical aperture — captures both the launch geometry and the index step. The same combination scaled by the core circumference in wavelengths is the normalised frequency $V$, and because each guided mode occupies a fixed cell of the transverse phase space, the mode count grows as $V^{2}$.

Common pitfall. Two recur. First, using $n_{1}\sin\theta_{a}$ instead of $n_{0}\sin\theta_{a}$ in part (b), which wrongly makes the acceptance angle grow on immersion; the numerical aperture belongs to the fibre and does not change when the fibre is dunked. Second, putting the core diameter into $V$ — the formula takes the radius, and the resulting factor-of-two error doubles $V$ and quadruples the mode count.

Source. Senior, Optical Fiber Communications, §2.3 (acceptance and numerical aperture) and §2.4–2.5 (modes, $V$-number, cutoff); Keiser, Optical Fiber Communications, Ch. 2.

Question 1 of 7: Step-index fibre — acceptance, cladding index, mode count and intermodal dispersion

Question 1: Step-index fibre — acceptance, cladding index, mode count and intermodal dispersion (20 marks)