Question 5 of 6: Optical-fibre attenuation, Rayleigh scattering, modal dispersion, and numerical aperture
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
98-Phys-A7, Optics — National Exams, December 2016. 3 hours; closed book (formula sheet supplied on pages 6–8). Notes 1–7 on the cover page require Q1 and Q2 (mandatory) plus any two of Q3–Q6 for a complete paper; every question is solved below as a full study resource, including all four of Q3–Q6.
Part (c) — longest/shortest ray paths in the step-index fibre.
Given. Core index $n_1=1.46$, cladding index $n_2=1.45$, core diameter $50\ \mu$m (radius $a=25\ \mu$m), fibre perfectly straight; no fibre length $L$ is stated in the source, so lengths/times are reported per unit fibre length $L$.
Find. Path length and transit time of the shortest possible ray (axial) and the longest possible GUIDED ray (the meridional ray at the maximum angle still totally internally reflected).
Approach. The shortest path is the axial ray, travelling straight down the core: length $=L$. The longest guided path is the ray that zig-zags at exactly the critical angle at the core–cladding interface (any steeper angle escapes by refraction instead of reflecting); by geometry, that ray's path length over an axial run $L$ is $L/\cos\theta_z$, where $\theta_z$ is its angle from the axis, fixed by $\cos\theta_z=n_2/n_1$ (the interface normal is radial, so the ray's angle of incidence there is $90^\circ-\theta_z$, and the critical condition is $\sin(90^\circ-\theta_z)=n_2/n_1$).
Fig. Q5(c) — the axial ray (green) is the shortest guided path; the meridional ray reflecting at exactly the critical angle at the core–cladding wall (red) is the longest guided path.
Maximum ray angle from the axis. $\cos\theta_z=\dfrac{n_2}{n_1}=\dfrac{1.45}{1.46}=0.99315\Rightarrow\theta_z=6.75^\circ$.
Shortest path (axial ray). Length $=\boxed{L}$; travelling entirely in the core at speed $c/n_1$, transit time $t_{\min}=\boxed{\dfrac{n_1 L}{c}}$.
Longest guided path (critical-angle ray). Length $=\dfrac{L}{\cos\theta_z}=L\left(\dfrac{n_1}{n_2}\right)=\boxed{1.00690\,L}$; transit time (same core speed $c/n_1$ along this longer path) $t_{\max}=\dfrac{n_1}{c}\cdot L\dfrac{n_1}{n_2}=\boxed{\dfrac{n_1^2 L}{c\,n_2}}$.
Illustrative delay per unit length. $\dfrac{t_{\max}-t_{\min}}{L}=\dfrac{n_1}{c}\!\left(\dfrac{n_1}{n_2}-1\right)=\dfrac{1.46}{2.998\times10^8}(0.006897)=\boxed{33.6\ \text{ns/km}}$ — a typical multimode step-index intermodal-dispersion figure, quoted here per km purely as a concrete illustration since no fibre length was given.
Ray
Path length
Transit time
Shortest (axial)
$L$
$n_1L/c$
Longest (critical-angle)
$1.00690\,L$
$n_1^2L/(c\,n_2)$
Delay difference (per km, illustrative)
$33.6$ ns/km
Part (d) — numerical aperture and acceptance angle.
Given. $n_1=1.46$, $n_2=1.45$ (from part c); fibre end face in air ($n_0=1$).
Find. NA and the maximum acceptance half-angle $\theta_{\max}$.
Approach. $\mathrm{NA}=\sqrt{n_1^2-n_2^2}$ (from requiring the refracted ray inside the core to strike the core–cladding wall at or beyond the critical angle); the acceptance angle in air follows from $\sin\theta_{\max}=\mathrm{NA}/n_0$.
Fig. Q5(d) — the marginal accepted ray enters at θmax in air and refracts to exactly the critical angle at the core wall; NA = sinθmax sets the fibre's light-gathering cone.
