Chapter 6: Losses in Optical Waveguides
Source: Hunsperger, sixth edition (2009), Problems 6.1–6.11, printed pages 125–127. Attenuation coefficients here act on power.
Problem 6.1 — Local meaning of attenuation
Working. Differentiate the assumed exponential:
Answer. Attenuation is the positive power lost per unit distance divided by the power being transmitted at that position.
Check. The dimensions are inverse length. This is a local ratio, not the cumulative power lost since the input divided by current power.
Problem 6.2 — Convert exponential loss to decibels
Working. Over a distance \(z\),
Answer. Loss per unit length is \(4.3429448\alpha\), or \(4.3\alpha\) at the precision used in the question.
Check. \(\alpha=1\,{\rm cm}^{-1}\) leaves \(e^{-1}\) of the input after 1 cm, equivalent to 4.343 dB. The factor 8.686 applies to an amplitude attenuation coefficient instead.
Problem 6.3 — Loss of the intermediate-radius bend
Data. The straight BC section is 0.5 cm; the DE, AB, and CD quarter-circle radii are 0.1, 0.2, and 0.3 cm.
Working. The straight loss is \(s=1.00/0.5=2\) dB/cm. Divide each measured bend loss by \(\pi R/2\) and subtract \(s\) to obtain radiation loss per length:
Under the fixed-mode exponential-radius model of Eq. (6.40), \(r(R)=Ae^{-BR}\). At the intermediate radius, \(r(0.2)=\sqrt{r(0.1)r(0.3)}\).
Answer. \(L_{AB}=[2+r(0.2)](0.1\pi)=1.13575\) dB.
Check. This is conditional on the book’s exponential bend-loss approximation. Two measured bend radii alone cannot determine an arbitrary radiation-loss function.
Problem 6.4 — Why tight bends radiate more
Working. Parts of a mode on the outer side of a bend must cover a longer path per unit angle than the field near its centre. Beyond a critical outer radius, the exterior field can no longer remain phase locked to the guided mode. Its evanescent barrier is effectively reduced, allowing radiation to escape.
Answer. Reducing bend radius moves the radiation region closer to the core, where more of the modal tail resides, increasing loss.
Check. A more tightly confined mode usually tolerates smaller radii than a weakly confined mode. Bending loss is distinct from absorption heating and roughness scattering.
Problem 6.5 — Minimum radius under a loss limit
Working. Subtract the straight-guide background: \(\alpha_r(0.5)=1.1\,{\rm cm}^{-1}\) and \(\alpha_r(0.3)=26.0\,{\rm cm}^{-1}\), with radii in mm. Fit \(\alpha_r=Ae^{-BR}\):
Answer. Choose \(R>0.4432\) mm to keep total \(\alpha<3\,{\rm cm}^{-1}\) under this model.
Check. The target radiation contribution is \(3-0.3=2.7\), not 3. Linear interpolation of total loss is not the assumed physical model.
Problem 6.6 — Strip loading and the material-development sequence
Working. In cross section, show a high-index guiding film on a lower-index substrate and a patterned loading layer above it; the local effective index rises beneath the strip. See the stripline diagram in Chapter 3: Theory of Optical Waveguides.
GaAs-on-GaAs can guide by carrier contrast but incurs free-carrier loss. AlGaAs provides compositional index/gap control. Adding indium permits longer-wavelength active regions, letting adjoining GaAs guides be more transparent. A quaternary system adds freedom to tune bandgap and lattice matching separately. MBE adds precise composition and thickness control.
Answer. The progression improves transparency, confinement, emitter/guide compatibility, and interface quality.
Check. “Quaternary” alone does not guarantee lattice matching; composition and the substrate still have to be chosen together.
Problem 6.7 — Cutback attenuation from a ten-percent length reduction
Working. Equal launch conditions cancel the unknown coupling factor:
Answer. Approximately \(0.912\,{\rm cm}^{-1}\).
Check. The removed distance is 0.2 cm, not 10 cm or 10 percent of the power. Shortening increases output, requiring positive attenuation.
Problem 6.8 — Separate coupling loss from propagation loss
Working. In cutback measurements, retain the same launch while measuring two or more lengths. Since \(\ln P_{\rm out}=\ln(\eta_cP_{\rm in})-\alpha L\), the slope gives propagation loss and the intercept gives coupling loss.
A second method samples calibrated side-scattered intensity along one uniform guide. If scattering collection efficiency is constant, its logarithmic slope also gives \(-\alpha\). Compare extrapolated internal power with launched power to estimate coupling.
Answer. Use length-dependent transmission or calibrated spatial decay, rather than assigning all insertion loss to propagation.
Check. Cutback fails if relaunching changes coupling; scattered-light methods fail if local scattering or collection efficiency varies strongly.
Problem 6.9 — Halve a three-centimetre guide
Working.
Answer. (a) \(0.06354\,{\rm cm}^{-1}\). (b) Approximately 0.2760 dB/cm.
Check. The 1.5 cm segment removed accounts for \(10\log_{10}(1.1)=0.41393\) dB.
Problem 6.10 — Free-carrier absorbed power
Model. Use the book’s representative GaAs estimate at 1.15 micrometres, \(\alpha_{\rm fc}\simeq10^{-18}N\) with concentration in \({\rm cm}^{-3}\). The exercise does not specify wavelength or mobility, so this assumption is necessary.
Working. The lost fraction over 5 cm is \(1-e^{-5\alpha}\).
Answer. At \(N=10^{18}\), it is \(1-e^{-5}=0.993262\), or 99.33 percent. At \(N=10^{16}\), it is \(1-e^{-0.05}=0.0487706\), or 4.877 percent.
Check. A hundredfold reduction in attenuation does not make cumulative absorbed fractions differ by exactly a hundred when the original sample is optically thick.
Problem 6.11 — Surface scattering and modal confinement
Working. Roughness perturbs the dielectric boundary. To first order, its scattering strength depends on the field sampled at that boundary and on the roughness spatial spectrum. A tightly confined field with small boundary amplitude couples less strongly to the rough interface. A field with substantial exterior penetration is often more exposed.
Answer. In the intended comparison, the weakly confined mode is more vulnerable to surface scattering.
Check. Tail length alone is not a universal loss predictor: normalize both modes to equal power and compare their boundary fields and the phase-matched roughness spectrum. Very near cutoff, changes in normalization can invalidate a simple monotonic argument.