Chapter 5: Polymer and Fiber Integrated Optics
Source: Hunsperger, sixth edition (2009), Problems 5.1–5.9, printed page 103. Processing results refer to the measurements in the book.
Problem 5.1 — What controls a spun polymer-film thickness?
Working. Centrifugal flow competes with viscous resistance while solvent evaporates. Increasing polymer concentration or molecular weight normally raises viscosity and leaves a thicker film. Increasing spin speed increases outward flow and usually reduces thickness. Spin time matters until flow and evaporation largely cease. Solvent volatility, temperature, and wetting also affect the result; annealing removes residual solvent and relaxes strain.
Answer. Specify molecular weight, solution concentration and solvent, spin speed/time, and anneal temperature/time. These form a process recipe, not one universal thickness law.
Check. The concentration unit in Table 5.2 is solvent volume per gram of polymer: increasing that number means a more dilute solution and normally a thinner film.
Problem 5.2 — Arrayed-waveguide path difference and a consistency check
Working. Direct substitution in the book’s printed Eq. (5.1), using micrometres throughout, gives
However Eq. (5.3) predicts the number of nonoverlapping output positions:
Thus the supplied geometry does not accommodate 20 channels in one free spectral range. A standard dispersive phase calculation independently gives
where \(n_{g,a}\) is the array-waveguide group index, not the number of channels.
Answer. The printed equation evaluates to 14.16 micrometres, but the complete stated design is inconsistent. For example, assuming \(n_{g,a}=1.67\) gives a physical path difference of 169.53 micrometres and a free spectral range about 8.49 nm, insufficient for twenty channels spaced 1 nm apart.
Check. The group index and phase-order condition must also be specified for a realizable AWG. The two meanings of \(N\) cannot be interchanged to repair the printed expression silently.
Problem 5.3 — Increase AWG channel count
Working. The spatial range condition is \(N_{\rm ch}\le\lambda_0f/(n_s dD)\). With fixed centre wavelength and channel spacing, doubling capacity from an already consistent baseline requires doubling \(f/(n_s dD)\). Possible geometric changes are doubling focal radius, halving the array pitch, halving output pitch, or combining smaller changes. An index change can contribute but is constrained by the material platform.
Answer. Enlarge spatial free spectral range and ensure the spectral free spectral range \(\lambda_0^2/(n_{g,a}\Delta L)\) covers the enlarged channel band. Adjust path increment/order and increase the illuminated array aperture as needed to retain resolution and low crosstalk.
Check. Merely adding more output guides does not remove order overlap. The original 5.2 geometry first needs correction even to reach 20 channels; its geometric capacity is only about 8.49.
Problem 5.4 — Polymer advantages over III–V materials
Working. Solution deposition and low-temperature patterning can avoid the demanding epitaxial growth required for crystalline III–V stacks. Polymers can cover large areas, coat different substrates, and form flexible or mechanically compliant interconnects. Composition and dopants can tailor index, fluorescence, or electro-optic response. Lower index contrast can also help match a fiber’s mode size.
Answer. Processing simplicity, material/geometry flexibility, and potentially inexpensive large-area fabrication are key advantages.
Check. These are not universal performance victories. Optical absorption, thermal stability, aging, and attainable active-device performance can favor semiconductors.
Problem 5.5 — Different polymer emission colours
Working. Emission energy depends on the electronic structure of the emitting molecule or conjugated segment, not solely on the bulk host. Different fluorescent dopants or locally modified conjugation lengths can therefore create regions with different transition energies. Electrons and holes recombine in those emissive regions.
Answer. Pattern different emissive species or compositions in one polymer platform. Each emits near \(\lambda=hc/\Delta E\).
Check. Distinct colours do not establish lasing. Energy transfer between dopants and reabsorption can alter the observed spectrum, so the emitters need suitable isolation and excitation.
Problem 5.6 — Recognize polymer lasing
Working. Measure output and spectrum while increasing pump strength. Look for a threshold-like change in output slope, strong spectral narrowing into cavity-selected lines, and directional coherent emission. Test whether those features track an available optical feedback path.
Answer. The combination of gain overcoming cavity loss and feedback-selected coherent oscillation establishes lasing.
Check. Amplified spontaneous emission can also narrow a spectrum and produce superlinear output. Line narrowing alone is insufficient; resolve cavity modes or measure coherence and feedback dependence.
Problem 5.7 — Fiber Bragg gratings and three fabrication routes
Working. An FBG is a longitudinal periodic effective-index modulation. Forward and backward guided waves couple when \(2\beta=2\pi m/\Lambda\), giving \(\lambda_B=2N\Lambda/m\).
Three fabrication routes are two-beam interferometric UV exposure of photosensitive fiber, phase-mask UV exposure, and focused point-by-point laser inscription. The first two expose a periodic interference pattern; the last writes individual perturbations directly.
Answer. Each method creates the required longitudinal index periodicity rather than merely an absorbing stripe pattern.
Check. \(\lambda_B\) uses the guided effective index and the index-modulation period, not automatically the writing laser wavelength.
Problem 5.8 — Read a process-specific film thickness
Working. Match every supplied recipe parameter to Table 5.2: 280,000 molecular weight, 10 ml solvent per gram, 3000 rpm, 40 s, and 110 degrees Celsius for 45 min. These identify sample 2. Its measured average thickness is 10,710 angstroms.
Answer. Approximately 1.071 micrometres.
Check. This is an empirical recipe lookup, not an independent prediction from spin speed alone. Samples with different concentration cannot be substituted simply because their spin speeds match.
Problem 5.9 — Stern–Volmer constant from fluorescence data
Working. Use the straightened plot in Fig. 5.14(b):
The plotted line rises from roughly 1 at zero chloride to about 5.5 at 100 mM. Its slope is approximately \((5.5-1)/100=0.045\,{\rm mM}^{-1}\).
Answer. \(K_{\rm SV}\simeq0.045\,{\rm mM}^{-1} =45\,{\rm M}^{-1}\), a graph-reading estimate.
Check. At 50 mM this predicts \(I/I_0\simeq1/3.25\), consistent with the strongly quenched fluorescence. Using raw fluorescence versus concentration as the linear plot would be incorrect.