A longer waveguide section can transmit more power than a shorter one when reflections at its interfaces remain coherent. The extra propagation distance may increase ordinary attenuation while also changing the phase relationship between repeated transmitted contributions. Those contributions can reinforce or cancel differently at the output.
The short answer is to add the transmitted fields before calculating their power. We will derive an original single-mode model with two reflecting interfaces, calculate a passive example in which the longer section transmits more, and separate phase-dependent transmission from bulk attenuation. All lengths and coefficients are fictional teaching inputs. This is not a design prescription or a claim about a measured device.
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ROUTE 1 · CHAPTERS 1–6
Build the field-path account
Define power, propagation and coherent repeated paths.
ROUTE 2 · CHAPTERS 7–12
Work the passive examples
Compare lengths and check energy and phase interpretation.
ROUTE 3 · CHAPTERS 13–19
Average and invert
Distinguish phase means, visibility and calibrated levels.
ROUTE 4 · CHAPTERS 20–29
Test the observation model
Practise transfer and audit coherence, planes and inverse checks.
ROUTE 5 · CHAPTERS 30–31
Questions and onward routes
Resolve common questions and choose the next wave guide.
Chapter contents
Build the field-path account · Chapters 1–6
1. Define the section and the observable
2. Separate propagation loss from interface reflection
3. Track the first transmitted field
4. Write one round-trip multiplier
Work the passive examples · Chapters 7–12
7. Work a lossless longer-section example
8. Add ordinary positive propagation loss
9. Compare the coherent and single-pass ledgers
10. Check passivity from the transmission formula
11. Keep stored field and transmitted power distinct
12. Why the logarithm of transmission is not simply linear in length
Average and invert · Chapters 13–19
13. Distinguish a fixed phase from a phase average
14. Do not replace the mean denominator with a denominator of the mean
15. Calculate fringe visibility on a defined scan
16. Invert visibility without identifying too much
17. Add a calibrated absolute level
Test the observation model · Chapters 20–29
20. State the coherence and observation limits
21. Use the cavity source for the geometric-series principle
22. Try a changed-reflectivity problem
23. Check the changed-record solution
24. Plan evidence around the unresolved quantity
25. Repair, stabilise and extend the explanation
26. Recover propagation loss only after separating its factor
27. Preserve the observation plane when comparing devices
Questions and onward routes · Chapters 30–31
CHAPTER 1 OF 31 · Build the field-path account
1. Define the section and the observable
Consider one propagating mode entering a uniform section of length L and leaving through a second interface. The incoming and outgoing modes are power-normalised, so squared field-amplitude ratios represent power ratios in the defined ports. No extra output mode or aperture loss is included.
Let T be transmitted output power divided by incident input power at specified reference planes. This dimensionless transmission differs from a field-amplitude ratio, a detector voltage and the power remaining inside the section. State those reference planes before comparing different lengths.
Assume a coherent steady monochromatic input, identical lossless interfaces and a uniform passive section. We will model the interfaces through a per-interface power reflectivity R, with zero ≤ R < one. Interface phases are included in a round-trip phase defined below. These assumptions make the bookkeeping clear; a real junction can require unequal coefficients, mode conversion and a more complete scattering description.
CHAPTER 2 OF 31 · Build the field-path account
2. Separate propagation loss from interface reflection
Let α be the power attenuation coefficient of the uniform section, in inverse-length units. Define x = exp(−αL), the one-way power retention factor. For a passive section, α is nonnegative and x lies above zero and at or below one. The corresponding one-way field-amplitude retention is the square root of x.
This convention matters. If another source defines an amplitude attenuation coefficient, the power exponent contains twice that coefficient. Mixing those conventions changes the inferred loss even when the arithmetic is otherwise correct.
The interfaces contribute a different effect. Each has power reflectivity R and power transmission one − R under the stipulated lossless, power-normalised model. Their field magnitudes involve square roots, while their phases must be treated consistently. Ordinary propagation loss is represented by x; reflected feedback is represented by the repeated field paths. A measured two-port transmission generally combines both.
CHAPTER 3 OF 31 · Build the field-path account
3. Track the first transmitted field
Let β be the real propagation constant of the mode in the uniform section, expressed in radians per unit length. Apart from a fixed overall phase convention, the first transmitted contribution has amplitude ratio t0 = (one − R)√x exp(iβL).
The factor one − R comes from two interface field-transmission magnitudes, each √(one − R). The square root of x accounts for one passage through the section. Squaring the magnitude of this first contribution gives Tsp = (one − R)²x, the single-pass power term.
That term is not yet the total transmission when reflected fields return coherently. Some field can reflect at the output interface, travel back, reflect again and make another attempt to leave. The delayed contribution may overlap the same monochromatic output state. Its phase relative to the first contribution then affects the total power. Keep the first-path result, but label it as a contribution rather than the completed answer.
One extra round trip contains two reflections and two extra one-way propagation passages. Under the symmetric model, the field-magnitude multiplier is Rx: the two reflection magnitudes multiply to R, and the two propagation amplitude factors multiply to x.
Let δ be the full round-trip phase, including propagation and the relevant reflection phases. Write the complex multiplier as q = Rx exp(iδ). In our numerical example, choose the fixed combined reflection phase as zero, so δ = 2βL. This is a stipulated phase convention for the example, not a statement that every real interface has zero reflection phase.
The successive transmitted fields are t0, t0q, t0q² and so on. Because Rx < one in the stated domain, the geometric series converges. Describing q as a field multiplier is essential: its magnitude is Rx, whereas its power multiplier is R²x². Confusing these two factors produces an incorrect denominator.
CHAPTER 5 OF 31 · Build the field-path account
5. Sum fields before taking the squared magnitude
Adding all transmitted contributions gives t = t0/(one − q). Therefore T = (one − R)²x/[one + R²x² − 2Rx cosδ]. This is the coherent power transmission of the stipulated two-interface section.
The cosine term is the trace of interference between repeated field contributions. It would be absent if one simply added their separate powers. The denominator can be small when the contributions reinforce, but the passive model still constrains total transmission. A large internal circulating field does not imply creation of net output energy.
Check the no-reflection limit: at R = zero, T = x, so the result reduces to ordinary one-way propagation. Check also that x is a power factor and that δ is dimensionless. These tests are simpler than a full numerical example and catch many transcription errors before a physical story is attached to the curve.
CHAPTER 6 OF 31 · Build the field-path account
6. Identify resonance and antiresonance
For fixed R and x, the denominator is smallest when cosδ = one. Then Tmax = (one − R)²x/(one − Rx)². It is largest when cosδ = minus one, giving Tmin = (one − R)²x/(one + Rx)².
These values describe constructive and destructive alignment of the repeated transmitted contributions. In a lossless section, x = one and Tmax = one. The minimum is [(one − R)/(one + R)]². Transmission can therefore vary substantially with phase even though neither interface nor the bulk section absorbs power.
The phase dependence is not a new attenuation coefficient. It is a change in how fields combine at the output. When interpreting a measurement, ask whether a minimum reflects absorption, reflection back toward the input, coupling into an omitted mode or a mixture of effects. Our simplified model includes reflection and propagation loss but excludes additional modes and detector acceptance changes.
CHAPTER 7 OF 31 · Work the passive examples
7. Work a lossless longer-section example
Set R = one-half, α = zero and β = π radians per metre. Choose the combined reflection phase as zero. For L = half a metre, δ = π and the section is at the stated antiresonance. Its transmission is (0.5/1.5)² = one-ninth, about 0.111111.
For L = one metre, δ = 2π and the section is at resonance. Its transmission is one. The longer section transmits nine times as much power relative to the same incident power at its reference plane, even though its interfaces have the same reflectivity.
There is no negative attenuation in this example; α is zero throughout. The difference is phase. A claim that a longer passive guide must always give less measured transmission would be true for a simple propagation-only model with fixed coupling, but it misses the coherent interface feedback included here. Identify the complete measured path before applying an exponential length rule.
CHAPTER 8 OF 31 · Work the passive examples
8. Add ordinary positive propagation loss
Retain R = one-half and β = π radians per metre, but set α = 0.1 per metre. The half-metre section has x = exp(−0.05), about 0.951229. Its phase remains π, and the coherent formula gives T about 0.109214.
The one-metre section has x = exp(−0.1), about 0.904837. Its phase is 2π and T is about 0.754420. The longer path loses a larger fraction during each single passage, yet its total transmitted power is still much greater because its repeated contributions reinforce.
Both transmissions are below one, as required for this passive model. Positive propagation attenuation and a longer-section transmission increase can coexist. The comparison does not determine the bulk loss coefficient from its direction alone. A learner should be able to point separately to the x factor and the phase-dependent denominator and explain which changed in the two-length comparison.
CHAPTER 9 OF 31 · Work the passive examples
9. Compare the coherent and single-pass ledgers
The table uses the same fictional R = 0.5, α = 0.1 per metre and β = π per metre. Transmission values are power ratios, with matched incident normalisation at the defined input reference plane.
| Section length | One-way power factor x | Round-trip phase | Single-pass term | Coherent total T |
|---|---|---|---|---|
| 0.5 m | 0.951229 | π | 0.237807 | 0.109214 |
| 1.0 m | 0.904837 | 2π | 0.226209 | 0.754420 |
The single-pass column decreases with length, while the coherent total increases. At half a metre, later contributions partly cancel the first transmitted field. At one metre, they reinforce it. Neither case is represented by simply multiplying the single-pass term by a common positive gain.
Read the two columns as different quantities rather than competing arithmetic answers. The first counts one path. The second includes all repeated coherent paths permitted by the model. Which one corresponds to an actual measurement depends on its temporal, spectral and modal observation conditions.
CHAPTER 10 OF 31 · Work the passive examples
10. Check passivity from the transmission formula
The largest transmission at a fixed R and x occurs at resonance. Comparing its denominator with its numerator gives (one − Rx)² − (one − R)²x = (one − x)(one − R²x), which is nonnegative for the stipulated passive domain.
Thus Tmax ≤ one, and every other phase gives T ≤ Tmax. The inequality verifies passivity of the transmitted-power result without requiring the explicit reflected-field formula. With x = one, equality is possible at resonance. With α positive and L positive, the resonance transmission is below one for R < one.
Do not substitute R = one into a lossless resonance expression and interpret the resulting zero-over-zero form as a working perfectly transmitting device. R = one was excluded from the convergent driven model: the interfaces then do not admit the assumed input coupling. Limits taken while parameters approach an excluded boundary require their own interpretation. Domain statements belong alongside equations, especially when an attractive special case appears.
CHAPTER 11 OF 31 · Work the passive examples
11. Keep stored field and transmitted power distinct
Near resonance, repeated contributions can build a large internal circulating field relative to one first-pass contribution. That field is a state of the driven section. It is not the same quantity as net transmitted power divided by incident power.
In the lossless symmetric case, the total outgoing power divides between transmission and reflection. At resonance, the reflected contributions combine so that all incident power is transmitted within this ideal model. Away from resonance, less is transmitted and more is reflected. When x is below one, some power is dissipated in the section and a lossless transmission-plus-reflection equality no longer describes the whole balance.
The field series is therefore consistent with energy conservation, but only when the amplitudes and reference-plane normalisation are handled consistently. Adding an internal circulating power to output power as if they were independent sources double-counts passages of the same driven field. A clear path ledger prevents that error before a more detailed cavity-energy analysis is attempted.
CHAPTER 12 OF 31 · Work the passive examples
12. Why the logarithm of transmission is not simply linear in length
For propagation alone, T = exp(−αL) gives lnT = −αL. With coherent interface feedback, lnT also contains a constant interface term and the logarithm of the phase-dependent denominator. Since δ changes with L, that extra term is generally not linear.
Using the two lossy examples, a naive coefficient −ln[T(1.0)/T(0.5)] divided by 0.5 metres is negative, because the measured transmission rises. The true model input α is positive, 0.1 per metre. The negative fitted number exposes the inadequacy of the propagation-only interpretation for this two-point record.
It does not show that the passive bulk medium amplifies. Nor does discarding the resonant point establish the correct coefficient automatically. A justified inference must model or independently constrain the interface and phase contributions, or use an observation regime where their averaging is established. The logarithm is a mathematical transformation, not a device that removes unmodelled physical factors.
CHAPTER 13 OF 31 · Average and invert
13. Distinguish a fixed phase from a phase average
A coherent measurement can have an unknown but fixed phase. That situation is not equivalent to averaging uniformly over every phase. Lack of phase knowledge is a statement about information; physical phase averaging is a statement about the measurement process.
For a genuinely uniform phase average with R and x fixed, the mean transmission is Tavg = (one − R)²x/(one − R²x²). One way to derive it is to expand the field series and observe that cross terms with different phase powers average to zero. The remaining powers form a geometric series.
The same expression follows by adding separately resolved or mutually incoherent transmitted-path powers under the corresponding assumptions. That agreement has a reason: the phase-sensitive cross terms no longer contribute. It should not be invoked merely because phase was omitted from a notebook. Establish the source coherence, observation time or averaging procedure that makes the mean appropriate for the question.
CHAPTER 14 OF 31 · Average and invert
14. Do not replace the mean denominator with a denominator of the mean
A tempting shortcut notes that the average cosine over a full phase cycle is zero, then inserts zero into the coherent denominator. That gives (one − R)²x/(one + R²x²), which is not the correct phase average.
The reason is that the average of a reciprocal is generally not the reciprocal of an average. Transmission is a nonlinear function of cosδ. For R = one-half and x = one, the incorrect shortcut gives 0.2, while the correct mean is one-third.
This provides an original transfer lesson beyond waveguides: averaging an input before a nonlinear transformation can produce a different result from averaging the output after transformation. In an actual spectrum, the weighting may also be nonuniform and R, α or β may vary across the measurement band. A uniform-phase formula is then a conditional model rather than a universal correction for any broad-band measurement.
CHAPTER 15 OF 31 · Average and invert
15. Calculate fringe visibility on a defined scan
Suppose a phase scan changes δ through a complete cycle while R and x remain effectively fixed. Define visibility V = (Tmax − Tmin)/(Tmax + Tmin). Substituting the two extrema gives V = 2p/(one + p²), where p = Rx.
Visibility is dimensionless and lies between zero and one in the stipulated domain. It measures the contrast of the phase-dependent transmission record, not absolute throughput. A highly contrasting fringe can still have a low peak transmission when propagation loss is substantial.
The scan assumption deserves attention. Changing frequency may also change reflectivity, attenuation, modal content and the propagation constant. If those variations are not negligible, the simple fixed-R, fixed-x visibility interpretation requires revision. The mathematical definition of contrast remains usable, but its identification with 2p/(one + p²) belongs to this model and scan contract.
CHAPTER 16 OF 31 · Average and invert
16. Invert visibility without identifying too much
For positive V below one, the physical solution is p = V/[one + √(one − V²)]. This stable form is equivalent to the smaller root of the quadratic relation. At V = zero, use p = zero directly.
Visibility identifies the product Rx within the model, not R and x separately. For p = 0.2, R = 0.4 with x = 0.5 and R = 0.5 with x = 0.4 both give V = 0.4/1.04, about 0.384615. Their interface reflectivities and propagation losses differ.
The two cases do not have the same absolute transmission levels. Their visibility agrees because the same round-trip field-magnitude product controls the contrast. This is an important inverse distinction: a normalised shape statistic can constrain one combination of parameters while discarding information needed to separate them. State the identified combination before reporting a separately named material or interface property.
Suppose visibility has supplied p and an independently calibrated peak gives C = Tmax(one − p)². In the model C = (one − R)²x. Dividing by p gives C/p = one/R − two + R when p is positive.
This additional relationship can determine the physical R below one, after which x = p/R. The solution must still satisfy x ≤ one. If it does not, the supplied observations and assumptions are inconsistent with this passive symmetric model. Calibration is essential: an unknown multiplicative detector gain would enter C and preserve an additional ambiguity.
For the first p = 0.2 example, R = 0.4 and x = 0.5 give C = 0.18. Then C/p = 0.9 and R solves R² − 2.9R + one = zero. Its physical root is 0.4; the other root is 2.5 and violates the interface domain. Substitution returns x = 0.5 and checks the original quantities.
CHAPTER 18 OF 31 · Average and invert
18. Check the alternative equal-visibility record
For R = 0.5 and x = 0.4, the same p = 0.2 gives C = 0.1. The peak is C/(one − p)² = 0.15625 and the minimum is C/(one + p)², about 0.0694444. The visibility remains about 0.384615.
For R = 0.4 and x = 0.5, the peak is 0.28125 and the minimum is 0.125. The second record is brighter at both extrema while having the same normalised contrast. This pair shows exactly what an absolute calibration adds to the visibility-only account.
If both records were divided by their own peaks, that information would be removed. A normalised plot can be helpful for comparing shape, provided its discarded scale is not later treated as known. Retain the original calibrated power ratios and the normalisation rule alongside any contrast summary. This keeps the inverse problem reproducible rather than relying on visual similarity between curves.
CHAPTER 19 OF 31 · Average and invert
19. Relate a frequency scan to propagation phase carefully
With a fixed reflection-phase contribution, δ = 2β(ω)L plus that contribution. Its local frequency derivative contains dβ/dω, the reciprocal group velocity of the stipulated uniform mode. A fringe-spacing interpretation therefore concerns the change in phase with frequency, not simply the value of β at one frequency.
If interface reflection phases also vary with frequency, their derivatives add to the round-trip phase response. A measured two-interface phase slope can then differ from the propagation-only group delay. The waveguide delay article gives the uniform-mode foundation and its cutoff uncertainty; the present model adds a feedback contribution at the boundaries.
Do not infer a whole-device delay from L divided by a phase velocity without checking the observable and reference planes. Nor should the constant-β numbers in our length comparison be interpreted as a complete frequency-dispersion model. They compare two supplied lengths at one stipulated propagation state. The frequency question requires additional relationships that the length table did not supply.
CHAPTER 20 OF 31 · Test the observation model
20. State the coherence and observation limits
The field-series result assumes the repeated contributions overlap coherently in the measured output state. A short pulse can produce distinguishable returns rather than one steady monochromatic amplitude. A source with limited coherence can reduce interference between sufficiently separated paths.
A detector that sums powers over modes or frequencies can also report an average different from the single-mode, single-frequency T. To model that reading, define its weighting and the coherence relationships among the contributions. A scalar phase average should not be attached to every detector merely because it reports a power value.
The existing aperture article considers how a spatially restricted measurement responds to modal fields. That is another observation effect, distinct from the longitudinal round-trip feedback treated here. Both lessons begin with the same useful question: which fields reach the observed state, and are they combined as coherent amplitudes or as separate powers? The answer determines the correct calculation.
CHAPTER 21 OF 31 · Test the observation model
21. Use the cavity source for the geometric-series principle
University of New Mexico Diels Laboratory teaching notes on the Fabry–Perot interferometer develop repeated reflected fields and their geometric-series transmission. They provide the authoritative optical setting for the field-summation principle used here.
Our power attenuation convention, fictional lengths, numerical records, visibility inverse examples and passivity check are derived from the explicitly stated symmetric model. The notes are not evidence that a particular fabricated waveguide has the supplied interface coefficients. Real waveguide junctions need their own modal and scattering description before the idealised two-interface result can be applied.
Read the university Fabry–Perot field-summation notes, checked on 24 October 2026. Keep field amplitudes, power ratios and propagation conventions visible when comparing formulas from different sources. A factor of two or a phase sign can reflect a different definition, but it should be reconciled explicitly rather than ignored.
CHAPTER 22 OF 31 · Test the observation model
22. Try a changed-reflectivity problem
Take a lossless section with R = one-third and the same β = π radians per metre and zero combined reflection phase. Calculate the transmitted power ratios at half a metre and one metre. Then calculate the uniform-phase mean and the fringe visibility.
Next, suppose only the visibility is given. Explain which parameter combination it identifies if x is unknown. Would the visibility alone establish that the section is lossless? Construct another allowed R and x pair giving the same product if possible, and state which additional calibrated observation would help distinguish them.
Work from the definitions before using the next chapter. This changes the interface reflectivity and asks about both forward and inverse conclusions. The exercise is complete only when the numerical answers are accompanied by a statement of what the observed contrast can identify. A remembered resonance value of one is not enough to answer the loss-identification question.
CHAPTER 23 OF 31 · Test the observation model
23. Check the changed-record solution
For x = one, the half-metre antiresonance gives Tmin = [(two-thirds)/(four-thirds)]² = one-quarter. At one metre, resonance gives Tmax = one. The mean transmission is (one − R)/(one + R) = one-half, while the visibility is 2R/(one + R²) = 0.6.
If x is unknown, V = 0.6 identifies p = Rx = one-third, rather than losslessness. Another allowed pair is R = two-thirds and x = one-half. It has the same p and therefore the same visibility, but its absolute peak and minimum differ from the lossless example.
For that alternative, C = (one-third)² times one-half = one-eighteenth. The peak is C/(two-thirds)² = one-eighth. A calibrated peak therefore distinguishes it from the unit peak of the lossless record within the symmetric model. Check the passive domain and reference-plane normalisation before turning this example into a general procedure for a real junction.
CHAPTER 24 OF 31 · Test the observation model
24. Plan evidence around the unresolved quantity
If the question is bulk attenuation, record whether the measured transmission includes interface feedback and whether phase was resolved or averaged. If the question is reflectivity, ask which independent information constrains propagation loss. If the question is fringe contrast, retain both the normalised contrast and the calibrated absolute levels.
A useful record states input and output reference planes, mode definition, frequency, source coherence, detector response and the assumed interface symmetry. It distinguishes supplied coefficients from quantities inferred using the same data. A fitted parameter is not an independent check of the model that produced it.
For an initial learning exercise, reproduce the two-length ledger and the passivity inequality. For a stronger inference task, reconstruct both equal-visibility parameter pairs and identify the extra observation that separates them. This sequence makes the observation model visible before asking for a material property that the observation may not uniquely determine.
CHAPTER 25 OF 31 · Test the observation model
25. Repair, stabilise and extend the explanation
If your passive model gives transmission above one, check amplitude versus power conventions, the round-trip multiplier and the port normalisation. If the single-pass and coherent columns disagree, identify whether later fields reinforce or cancel rather than treating their difference as unexplained gain.
To stabilise the method, derive the geometric series and check the R = zero and x = one limits. Then verify the phase mean by averaging powers after field summation, not by replacing the cosine inside a nonlinear denominator. Reconstruct the changed-reflectivity example without copying the original numbers.
For extension, allow unequal interfaces or frequency-dependent reflection phases, but write the revised field multiplier before calculating. The transferable habit is to preserve the full path and observation contract. A longer-section transmission increase is then a checkable consequence of coherent feedback, while the bulk attenuation coefficient retains its own definition and sign.
CHAPTER 26 OF 31 · Test the observation model
26. Recover propagation loss only after separating its factor
Suppose an independently justified analysis gives x = one-half for a section of length two metres. With our power convention, α = −lnx/L = ln(two)/two, about 0.346574 per metre. That inference uses the separated one-way propagation factor, not the total two-interface transmission.
If the same x were assigned to a one-metre section, the inferred coefficient would be ln(two), twice as large. A dimensionless retention factor belongs to a stated path length; it is not itself an attenuation coefficient. Changing the unit from metres to centimetres changes the numerical coefficient by the corresponding reciprocal-length conversion, while leaving αL and x unchanged.
A mistaken amplitude convention would create another factor of two. An amplitude retention of one-half corresponds to a power retention of one-quarter, not one-half. The correct logarithm follows from the observable’s definition. These checks are useful before interpreting a fitted loss as a material property, especially when instrument software and a theoretical source use different conventions.
CHAPTER 27 OF 31 · Test the observation model
27. Preserve the observation plane when comparing devices
Our T is defined between specified input and output reference planes. Moving either plane can include additional propagation, connector effects or detector acceptance. A change in reported transmission can therefore occur without changing the uniform section whose α is being discussed.
For example, suppose a fixed extra path contributes a power factor g after the output plane. A detector beyond that path reports gT rather than T if its calibration otherwise matches. A common g cancels in a ratio of two lengths, but it remains in absolute peak levels used to separate R and x. If g changes between measurements, even that ratio no longer cancels it.
This is why the visibility-only inverse and the calibrated-level inverse have different requirements. A common positive scale leaves visibility unchanged, while the C quantity built from an absolute peak changes with that scale. State which comparison is robust to a common multiplier and which depends on knowing it. Do not treat a cancellation valid for one statistic as a universal property of the whole measurement.
CHAPTER 28 OF 31 · Test the observation model
28. Examine when a two-interface model becomes inadequate
A bend, discontinuity or asymmetric junction can couple the incident field into several propagating or evanescent components. Our single scalar round-trip factor then omits relationships that may matter to the output. An evanescent contribution near an interface is also different from treating the whole uniform section as the propagating mode used in the numerical examples.
The appropriate extension may use mode-dependent amplitudes and a scattering matrix, with power and phase conventions defined for each observed channel. That statement does not supply the missing coefficients or authorise a scalar fit to stand in for them. It identifies the boundary of the simple model.
For a learner, the useful first question is whether the same mode and observation state can be followed through every counted path. If not, draw the additional channels before selecting a formula. The present result remains valuable as a complete, checkable example of coherent feedback. Its limitations help explain which extra information a more realistic device analysis would need.
CHAPTER 29 OF 31 · Test the observation model
29. Use a forward check after every inverse calculation
An inverse calculation should end by rebuilding the observations from the inferred parameters. In the example with p = 0.2 and C = 0.18, the physical R = 0.4 and x = 0.5 return Rx = 0.2 and (one − R)²x = 0.18. They then return the expected peak 0.28125 and minimum 0.125.
This substitution checks more than the quadratic arithmetic. It verifies that the selected root respects the passive domain and reproduces both independent response quantities. A root outside the domain is not rescued by satisfying one algebraic rearrangement. A parameter pair that reproduces contrast but misses the calibrated level has solved only part of the supplied problem.
The same habit applies when observations have uncertainty: test the reconstructed quantities against their admissible ranges and preserve any shared calibration uncertainty. An apparently precise parameter estimate is useful only if its response is consistent with the data and the model contract that justified the inversion.
Does the longer section amplify the input?
No. Both original examples are passive and their transmitted power ratios stay at or below one. The longer section has a more favourable phase relationship between repeated transmitted contributions. Positive propagation loss can coexist with that higher total transmission.
Can I add the powers of all transmitted passes?
Only under an observation contract that removes the relevant coherent cross terms, such as justified incoherent or phase-averaged contributions. For overlapping coherent fields, sum amplitudes first and calculate the squared magnitude afterward. Unknown phase alone does not establish physical averaging.
Does a resonance peak always have transmission one?
Only in the stipulated lossless symmetric case with the defined power-normalised ports. Propagation loss, unequal interfaces, mode conversion and other omitted effects can reduce or alter the peak. Keep the conditions attached to the familiar unit-transmission limit.
Does fringe visibility identify attenuation?
Not by itself in the two-parameter model. It identifies Rx, combining interface reflectivity and one-way power retention. A justified independent reflectivity, calibrated absolute level or other informative observation is needed to separate them within the stated model.
CHAPTER 31 OF 31 · Questions and onward routes
31. Continue through the relevant wave routes
For mode, cutoff and interface foundations, return to Waveguides. For propagation delay under a defined uniform-mode model, read bounding group delay when cutoff is uncertain. For a different spatial observation effect, use equal modal weights and different aperture readings.
For the two-beam interference foundation, read optical interferometry, fringe counting and phase unwrapping. For resolved echo magnitudes rather than coherent steady transmission, compare boundary reflection and bulk attenuation in echo ratios. Their observation equations answer different questions.
For phase change with frequency, read ring-resonator spacing and group index, keeping its ring path distinct from this two-interface section. For uncertainty shared by calibrated optical readings, use cascaded optical losses and shared calibration. These routes help identify which extra relationship a transmission inference needs.
Continue through Learning That Transfers into Life and How X Works. A useful explanation keeps propagation, reflection, coherence and the measured power ratio separate until the field-path model connects them.