Question 9.18: RAYLEIGH SCATTERING LIMIT What is the attenuation due to Ray...
RAYLEIGH SCATTERING LIMIT What is the attenuation due to Rayleigh scattering at around the λ = 1.55 μm window given that pure silica (SiO_{2}) has the following properties: T_{f} = 1730 °C (softening temperature), \beta _{T} = 7 \times 10^{−11} m^{2} N^{−1} (at high temperatures), n = 1.4446 at 1.5 μm?
We simply calculate the Rayleigh scattering attenuation using
\alpha_{R} \approx \frac{8 \pi^{3}}{3 \lambda^{4}}\left(n^{2}-1\right)^{2} \beta_{T} k T_{f}
so
\begin{aligned} \alpha_{R} & \approx \frac{8 \pi^{3}}{3\left(1.55 \times 10^{-6}\right)^{4}}\left(1.4446^{2}-1\right)^{2}\left(7 \times 10^{-11}\right)\left(1.38 \times 10^{-23}\right)(1730+273) \\ &=3.27 \times 10^{-5} m ^{-1} \quad \text { or } \quad 3.27 \times 10^{-2} km ^{-1} \end{aligned}
Attenuation in dB per km is then
\alpha_{ dB }=4.34 \alpha_{R}=(4.34)\left(3.27 \times 10^{-2} km ^{-1}\right)=0.142 dB km ^{-1}
This represents the lowest possible attenuation for a silica glass fiber at 1.55 μm.