17-Phys-A7 Optics · December 2018
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
17-Phys-A7, Optics — National Exams, December 2018. 3 hours; closed book (approved Sharp/Casio calculator only). Total 78 marks. Questions 1–6 are mandatory; the paper then offers a choice of Question 7 or 8, and a choice of Question 9 or 10. Every question is solved in full below as a complete study resource, including both members of each either/or pair.
Reference texts. Hecht, Optics, 5th ed.; Pedrotti, Pedrotti & Pedrotti, Introduction to Optics, 3rd ed.; Griffiths, Introduction to Electrodynamics, 4th ed. (Ch. 9, EM waves in matter, for Maxwell's equations).
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.
Part (a) — cladding index from the critical ray.
Given. Core index $n_1=1.45$; the critical ray zig-zags at $\theta_t=30^\circ$ measured from the fibre axis.
Find. Cladding index $n_2$.
Approach. The interface normal at the core–cladding boundary is radial (perpendicular to the axis), so the ray's angle of incidence at that interface, measured from the normal, is $\varphi=90^\circ-\theta_t$. For the "critical ray," $\varphi$ equals the critical angle $\theta_c$, where $n_1\sin\theta_c=n_2$ (Snell's law at the glass–glass boundary with the cladding side at $90^\circ$).
Part (b) — numerical aperture.
Given. $n_1=1.45$, $n_2=1.256$ (part a); alternatively, entrance-face Snell's law with the critical ray's internal angle $\theta_t=30^\circ$ from the axis.
Find. $\mathrm{NA}$.
Approach. $\mathrm{NA}=\sqrt{n_1^2-n_2^2}$ directly from the two indices; as a check, at the flat entrance face (normal parallel to the axis) Snell's law gives $\sin\theta_{\max}=n_1\sin\theta_t$ for the ray that becomes the critical ray inside, and since the surrounding medium is air, $\mathrm{NA}=\sin\theta_{\max}$.
| Quantity | Value |
|---|---|
| Cladding index $n_2$ | $1.256$ |
| Numerical aperture | $0.725$ |
Part (c) — time difference between the axial and critical rays over $L=1$ m.
Given. $n_1=1.45$, $\theta_t=30^\circ$, $L=1$ m (axial length), $c=3\times10^8$ m/s.
Find. $\Delta t=t_{\text{critical}}-t_{\text{axial}}$.
Approach. The axial ray travels a straight path of length $L$ at speed $c/n_1$. The zig-zag ray travels at angle $\theta_t$ from the axis, so to advance the same axial distance $L$ it covers a longer geometric path $L/\cos\theta_t$, at the same speed $c/n_1$ (both rays are entirely inside the core).
| Quantity | Value |
|---|---|
| Axial-ray transit time, $L=1$ m | $4.833$ ns |
| Critical-ray transit time, $L=1$ m | $5.581$ ns |
| Time difference $\Delta t$ | $0.748$ ns per metre |
Part (d) — output pulse train after $L=1$ km. Scaling part (c) to $L=1000$ m gives a modal-dispersion spread of $\Delta t=0.748\ \text{ns/m}\times1000\ \text{m}=0.748\ \mu\text{s}$ between the fastest (axial) and slowest (critical) rays. Every launched pulse excites all ray angles between these extremes equally, so each output pulse is not a delayed copy of the $0.2\ \mu\text{s}$ input pulse but a smeared-out pulse whose width is roughly the input width plus the modal spread, $\approx0.2+0.748\approx0.95\ \mu\text{s}$ — almost the full $1\ \mu\text{s}$ pulse period. The sketch below shows the input pulse train (narrow, well separated) and the output train (each pulse broadened to nearly fill its period, its trailing edge beginning to overlap the next pulse's leading edge) — this is the classic multimode-fibre bandwidth limitation: sufficient length turns modal dispersion into inter-symbol interference.
