17-Phys-A7 Optics · December 2013

Question 5 of 6: Fibre attenuation, modal dispersion and a symbolic dispersion derivation

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

Notes on this paper

Paper format. National Examinations, December 2013, 98-Phys-A7 Optics. Three hours, closed book, one approved Casio or Sharp calculator. Question 1 (20 short parts) is mandatory; the rubric asks for any four of Questions 2–6 — all six are worked here, because the set is a study resource.

Reference texts. E. Hecht, Optics, 5th ed.; F. L. Pedrotti, L. M. Pedrotti and L. S. Pedrotti, Introduction to Optics, 3rd ed.

Question 5: Fibre attenuation, modal dispersion and a symbolic dispersion derivation (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.

Part (a).

1. Rayleigh scaling. Loss coefficient $\alpha\propto1/\lambda^4$, so $$\alpha(1500)=\alpha(900)\left(\frac{900}{1500}\right)^4=(1.5\text{ dB/km})(0.6)^4=\boxed{0.194\text{ dB/km}}$$ — the classic argument for why telecom fibre moved from 850/900 nm to the 1300/1550 nm windows.

Part (b). Given. $n_1=1.46$ (core), $n_2=1.45$ (cladding), $L=1$ km.

1. Shortest path. The axial ray travels straight down the core: $$t_{short}=\frac{L\,n_1}{c}=\frac{(1000\text{ m})(1.46)}{2.998\times10^8}=\boxed{4.870\ \mu\text{s}}$$
2. Longest path. The ray guided at the critical angle grazes the core–cladding boundary; its angle from the fibre AXIS satisfies $\cos\theta_{max}=n_2/n_1=1.45/1.46$, giving $\theta_{max}=6.71^\circ$ and a zig-zag path length $L/\cos\theta_{max}=L\,n_1/n_2$: $$t_{long}=\frac{L\,n_1}{n_2}\cdot\frac{n_1}{c}=\frac{L\,n_1^2}{c\,n_2}=\frac{(1000)(1.46)^2}{(2.998\times10^8)(1.45)}=\boxed{4.904\ \mu\text{s}}$$ (path length 1006.9 m).
3. Modal spread. $$\Delta t=t_{long}-t_{short}=\boxed{33.6\text{ ns}}$$ over 1 km — the intermodal dispersion that limits step-index-fibre bandwidth.

Part (c). No numeric $S_o$, $\lambda_o$ are given — the part is answered symbolically.

1. (i) Units of $S_o$. $n$ is dimensionless and $\lambda^2-\lambda_o^4/\lambda^2$ has units of length$^2$, so $\boxed{[S_o]=\text{length}^{-2}}$ (e.g. nm$^{-2}$).
2. (ii) Zero dispersion and $D(\lambda)$. Material dispersion arises because $d^2n/d\lambda^2\ne0$ makes different wavelengths travel at different group velocities, spreading a pulse; the zero-dispersion wavelength $\lambda_o$ is where the dispersion parameter $D(\lambda)=-(\lambda/c)\,d^2n/d\lambda^2$ vanishes. Differentiating twice, $$\frac{dn}{d\lambda}=\frac{S_o}{4}\left(\lambda+\frac{\lambda_o^4}{\lambda^3}\right),\qquad \frac{d^2n}{d\lambda^2}=\frac{S_o}{4}\left(1-\frac{3\lambda_o^4}{\lambda^4}\right)$$ $$\boxed{D(\lambda)=-\frac{\lambda}{c}\frac{d^2n}{d\lambda^2}=\frac{S_o}{4c}\left(\frac{3\lambda_o^4}{\lambda^3}-\lambda\right)}$$
3. (iii) Group index. $n_g=n-\lambda\,dn/d\lambda$: substituting the two expressions above and simplifying, $$\boxed{n_g(\lambda)=-\frac{S_o}{8}\left(\lambda^2+\frac{3\lambda_o^4}{\lambda^2}\right)}$$
4. (iv) Transit-time difference. Group velocity $v_g=c/n_g$, so over a length $L$ the transit time is $t=Ln_g/c$; for two wavelengths $\lambda_1,\lambda_2$, $$\boxed{\Delta t=\frac{L}{c}\Big[n_g(\lambda_1)-n_g(\lambda_2)\Big]}=\frac{L S_o}{8c}\left[\left(\lambda_2^2-\lambda_1^2\right)+3\lambda_o^4\left(\frac{1}{\lambda_1^2}-\frac{1}{\lambda_2^2}\right)\right]$$

Final results

Quantity asked	Result
(a) attenuation at 1500 nm	0.194 dB/km
(b) shortest transit time	4.870 μs (axial ray)
(b) longest transit time	4.904 μs (critical-angle ray, 1006.9 m path)
(b) modal dispersion	33.6 ns over 1 km
(c)(i) units of S₀	length−²
(c)(ii) D(λ)	(S₀/4c)[3λ₀⁴/λ³ − λ]
(c)(iii) n_g(λ)	−(S₀/8)[λ² + 3λ₀⁴/λ²]
(c)(iv) Δt	(L/c)[n_g(λ₁) − n_g(λ₂)]

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