Optical fibres, light sources, detectors and photonic components

Use ray and wave pictures within their scope

In a simple step-index fibre, a higher-index core is surrounded by lower-index cladding. Total internal reflection provides a useful ray picture for suitable angles. A fuller wave description uses guided modes and their field distributions. Not every launched direction becomes a guided mode, and light can be lost through absorption, scattering, bends and imperfect connections.

Multiple modes can have different propagation delays. Chromatic dispersion concerns wavelength-dependent delay, and waveguide properties also influence dispersion. A single-mode fibre avoids intermodal dispersion but does not eliminate every form of pulse spreading. Broadening can make successive symbols overlap even when average received optical power appears adequate.

Ask which rays can remain in the core

Refractive index n=c/v compares the vacuum speed of light c with phase speed v in the specified medium: how fast a fixed phase of the wave advances. At an ideal transparent boundary, the ray angles obey n₁sinθ₁=n₂sinθ₂, with both angles measured from the normal, the perpendicular to the boundary. Here sine is the opposite-side fraction of a right triangle. For n₁>n₂, the transmitted angle grows until it reaches 90°. Beyond that critical incidence, sinθ₁>n₂/n₁ would require sinθ₂>1, so there is no propagating transmitted ray in this ideal picture: total internal reflection occurs.

Supply a hypothetical core index 1.50 and cladding index 1.20. Their ratio is 0.80, giving critical angle arcsin(0.80)≈53.13°; arcsin asks which angle has that sine. A 60° internal incidence is above it. The sketch follows such a ray between parallel boundaries. It is a bounded ray model of guidance, not a fibre cross-section design or a claim that bending, scattering and absorption disappear. The existing wave/mode description remains necessary beyond this simplified picture; phase speed alone also does not determine pulse delay in a dispersive link.

Follow conversion without skipping its limits

A light source converts supplied energy into an optical field. LEDs and lasers differ in emission mechanisms and spectral/coherence properties; source suitability depends on coupling, modulation and receiver needs. Information can be imposed by varying the source or using a separate modulator. Multiplexers combine selected optical channels and filters separate them.

A photodetector converts absorbed light into an electrical signal. Responsivity Rλ relates photocurrent to incident optical power at a specified wavelength and condition, with units A/W. Dark current, noise, response speed and saturation also matter. A responsivity number is not an error-rate specification, and electrical power supply to an amplifier should not be confused with optical signal power.

Follow energy into light and absorbed light into charge

A semiconductor LED supplies electrons and holes to an emitting region. A hole is an unoccupied valence-band state; an excited electron can fill it and lose energy. In a radiative recombination event, that energy leaves as a photon. Spontaneous emission means that the event is not triggered by an already present photon in the emitted optical mode. Nonradiative processes can instead transfer energy to the material. Not every supplied electron produces light that escapes and couples into the fibre.

In stimulated emission, an existing optical field can trigger a matching emission event, adding energy to that field. A pumped or electrically supplied gain medium must be in an appropriate excited state so that stimulated emission can outweigh absorption. A laser oscillator also returns part of the light through optical feedback, such as a resonator. Initially the gain must overcome round-trip losses in a supported mode; in a steady lasing state, saturated gain balances those losses, including the useful output leaving the resonator. Feedback does not supply the missing energy. This mechanism can concentrate emission into selected modes, but it does not imply a perfectly single frequency, an infinitely narrow beam or universal superiority to an LED.

A junction photodetector follows the reverse energy direction. An absorbed photon of suitable energy can excite an electron into a mobile conduction-band state, leaving a hole. The junction field moves electrons and holes in opposite directions; carriers generated nearby can also diffuse toward the collecting region. Recombination before collection, reflection and wavelength-dependent absorption can reduce the collected charge. Electrodes connect that charge motion to the measured photocurrent. Thus the stated A/W responsivity summarises absorption and collection under specified conditions; it is not a guarantee that every arriving photon gives the same useful signal. Collection time and the readout circuit also limit rapid changes.

Keep logarithmic loss separate from time spreading

For an assumed attenuation α in dB/km and length Lkm, fibre loss isαL decibels. Add stated connector or splice losses in decibels, then convert total loss A into power ratio 10^(−A/10). Powers themselves are added in linear units. Loss depends on wavelength, condition and route; a classroom coefficient is not a universal silica property.

Dispersion needs its own model or measurement. A power budget alone supplies no pulse width, timing margin or detector bandwidth. Regeneration, optical amplification and equalization can address different problems and bring additional noise or constraints. Component performance must therefore be assessed with the source spectrum, path, receiver and intended signalling together.

One hypothetical wavelength and a declared power account

Supply launch power 2 mW, fibre coefficient .4 dB/km at the chosen hypothetical wavelength and 5 km length. Two assumed .5 dB connectors add 1 dB, so total loss is 3 dB. Received power is 2×10^(−3/10)=1.002374 mW. A supplied detector responsivity .5 A/W predicts signal photocurrent 0.501187 mA before dark current and readout effects.

The plot separates fibre-only power from the same fibre plus fixed connector loss. Even at zero fibre length the latter remains below 2 mW. The numbers are original assumptions for a compatible hypothetical source/fibre/detector; they are not a component selection or an actual material data table.

Keep the original received optical power 1.002374 mW, but stipulate responsivity 0.25 A/W for a changed wavelength/collection condition. The signal-current prediction becomes about 0.250594 mA, half the earlier value. Holding incident power fixed did not hold collected current fixed. The original 3 dB path loss also cannot simply be carried to a new wavelength unless its attenuation and connection assumptions remain supported.

Try the reasoning

1. Does 3 dB loss mean exactly half the power?

   Show the answer and reasoning

   It is approximately half: 10^(−3/10)=.501187. Exactly half corresponds to about 3.0103 dB.

2. Does a single-mode link have no dispersion?

   Show the answer and reasoning

   No. It avoids intermodal delay but wavelength and waveguide effects can still broaden signals.

3. Can constant average received power prove unchanged high-speed performance?

   Show the answer and reasoning

   No. Pulse spreading, timing, noise and detector bandwidth can change while the mean power remains similar.

4. Keep the supplied 1.50 core index and 60° internal incidence, but change the cladding index to 1.40. Does the same ray still satisfy total internal reflection?

   Show the answer and reasoning

   No. Total internal reflection requires sinθ₁>n₂/n₁. Here sin60°≈0.866 is below 1.40/1.50≈0.933. A propagating refracted ray is allowed, so the old guidance claim for that ray no longer holds. This does not say all its power disappears: reflection can still occur, but it is not the stated total-reflection case. Reassess guidance rather than carrying forward the old loss account automatically.

5. A laser resonator is left unchanged while the supplied excitation is removed. Can the mirrors maintain the original steady output? Does a detector with unchanged incident optical power necessarily keep its current if the wavelength changes?

   Show the answer and reasoning

   The mirrors return light but provide no replacement for energy lost to output, absorption and other losses. Without appropriate excitation and gain, the original steady lasing claim cannot continue. At the detector, a new wavelength can change absorption and carrier collection, hence responsivity; the earlier current cannot be inferred from equal power alone. Source generation, propagation and detector conversion require their own supported conditions.

What to carry forward

Check optical power, temporal spreading and conversion limits as separate parts of the same link.