17-Phys-A7 Optics · Undated paper

Question 2 of 10: Step-index multimode fiber — ray angles and modal dispersion

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 2: Step-index multimode fiber — ray angles and modal dispersion (9 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. Core $n_1=1.450$; cladding $n_2=1.400$; surrounding air $n_{air}=1.000$; core radius $r_1=1$ mm; outer (cladding) radius $r_2=3$ mm; fiber length $L=100$ m; ray 2 (core critical ray) is given at $\theta_2=15.1^\circ$ from the axis; $c=3\times10^8$ m/s.

Part (a) — angle of incidence of the cladding ray at the cladding–air critical angle.

Find. $\varphi_{c,3}$, the (normal-referenced) angle of incidence at which ray 3 strikes the cladding–air interface at that interface’s critical angle.

Approach. At the critical angle, the refracted ray grazes the interface (refraction angle $90^\circ$), so Snell’s law $n_2\sin\varphi_c=n_{air}\sin90^\circ$ gives the incidence angle directly.

1. Snell’s law at the cladding–air interface, refraction angle 90°. $\sin\varphi_{c,3}=\dfrac{n_{air}}{n_2}=\dfrac{1.000}{1.400}=0.7143\ \Rightarrow\ \varphi_{c,3}=\boxed{45.58^\circ}$ (measured from the interface normal — this is the critical angle of the cladding–air boundary).

Part (b) — cladding ray’s propagation angle θ₃ (relative to the fiber axis).

Find. $\theta_3$, the angle ray 3 makes with the fiber axis as it zig-zags through the cladding.

Approach. The cladding–air interface runs parallel to the fiber axis, so the angle a ray makes with that interface’s normal and the angle it makes with the axis are complementary — exactly the relationship already given for ray 2, where the printed axis angle $\theta_2=15.1^\circ$ is the complement of that ray’s own core–cladding critical angle from the normal ($\sin^{-1}(n_2/n_1)=\sin^{-1}(1.400/1.450)=74.9^\circ$, and $90^\circ-74.9^\circ=15.1^\circ$, confirming the convention). Apply the same complementary relation to ray 3’s normal-incidence critical angle from part (a).

1. Convert from normal-referenced to axis-referenced angle. $\theta_3=90^\circ-\varphi_{c,3}=90^\circ-45.58^\circ=\boxed{44.42^\circ}.$

Part (c) — transit time of each ray over L = 100 m.

Find. $t_1,t_2,t_3$ and the pairwise time differences.

Approach. A ray that zig-zags at angle $\theta$ from the axis covers geometric path length $L/\cos\theta$ to advance an axial distance $L$, travelling at speed $c/n$ in its medium; the axial ray has $\theta=0$ so its path is simply $L$.

1. Axial ray (1), straight path in the core. $t_1=\dfrac{n_1L}{c}=\dfrac{1.450(100)}{3\times10^8}=4.833\times10^{-7}\ \text{s}=\boxed{483.3\ \text{ns}}.$
2. Core critical ray (2), zig-zag in the core at $\theta_2=15.1^\circ$. $t_2=\dfrac{n_1L}{c\cos\theta_2}=\dfrac{1.450(100)}{3\times10^8\cos15.1^\circ}=\boxed{500.6\ \text{ns}}.$
3. Cladding critical ray (3), zig-zag in the cladding at $\theta_3=44.42^\circ$. $t_3=\dfrac{n_2L}{c\cos\theta_3}=\dfrac{1.400(100)}{3\times10^8\cos44.42^\circ}=\boxed{653.3\ \text{ns}}.$
4. Time differences (modal dispersion). $\Delta t_{21}=t_2-t_1=17.3\ \text{ns}$; $\Delta t_{31}=t_3-t_1=170.0\ \text{ns}$; $\Delta t_{32}=t_3-t_2=\boxed{152.7\ \text{ns}}.$ The axial ray arrives first, the core critical ray a little later, and the cladding ray arrives last and by far the largest margin — even though the cladding has the lower index, its much shallower axis angle ($44.4^\circ$ vs $15.1^\circ$) more than offsets that, giving it the longest path and the largest spread. This total spread ($\approx$170 ns over the full set of guided/cladding rays) is the fiber’s intermodal (modal) dispersion.

Part (d) — expected output pulse train.

Each input pulse (0.2 μs wide) launches power into all ray angles between the axial ray and the cladding critical ray, so by the time it reaches the far end it has smeared into a burst spanning roughly $483$–$653$ ns of extra spread ($\approx0.17\ \mu\text{s}$) on top of its original 0.2 μs width — each output pulse is therefore about $0.2+0.17\approx0.37\ \mu\text{s}$ wide, lower in peak amplitude (the same energy is spread over a longer time), and rounded rather than rectangular (rays are launched with a continuum of angles between the two extremes, not just the three shown, so the pulse edges blur smoothly rather than stepping). Because the 1 μs pulse period is far larger than the $\approx0.17$ μs of spreading, the broadened output pulses remain fully resolved — there is no inter-symbol overlap at this length, though a much longer fiber (or a higher pulse rate) would eventually cause adjacent pulses to run together.