Fiber Optics¶

Exam-ready answer

Optical fiber communication transmits information as intensity-, phase- or frequency-modulated light through a transparent glass or plastic dielectric waveguide. A practical path is input data/voice → encoder and source driver → LED or laser diode → connector/coupler → fiber → PIN or APD photodetector → transimpedance amplifier, filter and decision circuit → recovered output.

The fiber is physically constructed from a core of refractive index \(n_1\), lower-index cladding \(n_2\) and a protective coating, with \(n_1>n_2\). Properly launched rays reach the core-cladding boundary above the critical angle and are guided by total internal reflection. The transmitter converts current to optical power; the receiver converts received photons to photocurrent, approximately \(I_p=RP_r\), where responsivity \(R\) is in A/W and received power \(P_r\) is in W.

For a length \(L\) km, the measurable fiber attenuation is

and a viable link requires received power to remain above receiver sensitivity after fiber, splice and connector losses, with an engineering margin.

Typical applications are telecom backbones, FTTH, submarine cables, LANs, industrial isolation and medical sensing. Limitations are brittle-fiber handling, precise connector/splice alignment, electro-optic conversion cost, inability to carry electrical power and the need for specialized test equipment. Thus fiber is preferred for high-capacity or long-distance links, while copper can remain economical for short low-rate powered connections.

Practice target: 8–9 minutes; draw the complete link, state the guiding principle and give six comparison points plus two limitations.

Exam-ready answer

An optical fiber is a cylindrical dielectric waveguide having a transparent core of refractive index \(n_1\), a lower-index cladding \(n_2\) and an outer protective coating. Refractive index is \(n=c/v\), where \(c\) and \(v\) are light velocities in vacuum and the medium in m/s. At any interface, angles measured from the normal obey Snell's law:

For light going from the denser core to the rarer cladding, \(n_1>n_2\) and the ray bends away from the normal. At the limiting incidence angle \(\theta_c\), the refracted angle is \(90^\circ\); hence

Total internal reflection (TIR) occurs only when (i) light travels from higher to lower refractive index and (ii) its incidence angle at the core-cladding boundary exceeds \(\theta_c\). A source launches light through the end face; rays inside the acceptance cone meet both conditions and undergo repeated TIR, so optical power follows the core even around gentle bends. Rays outside that cone refract into the cladding and are lost. Unlike ordinary refraction, ideal TIR returns essentially all incident power and therefore provides confinement rather than transmission across the boundary.

As a quantitative check, for \(n_1=1.48\) and \(n_2=1.46\), \(\theta_c=\sin^{-1}(1.46/1.48)\approx80.6^\circ\). Thus a guided meridional ray must strike the interface at more than \(80.6^\circ\) to the normal. Guidance enables low-loss telecom, sensing and illumination paths; practical limitations are material/scattering loss, imperfect boundaries and radiation from bends whose radius is too small.

Practice target: 8 minutes; draw the acceptance cone, derive the critical angle and state both TIR conditions before explaining guidance.

Exam-ready answer

All communication fibers have a light-guiding core, lower-index cladding and protective jacket, but they are classified by number of guided modes and radial refractive-index profile. The normalized frequency

where core radius \(a\) and wavelength \(\lambda\) use the same unit, determines modal operation. A step-index fiber is single-mode for \(V<2.405\); a highly multimode step-index fiber supports approximately \(M\simeq V^2/2\) modes, while an ideal graded-index fiber supports about \(V^2/4\).

For example, \(a=4.5\,\mu\text{m}\), \(NA=0.12\) and \(\lambda=1.55\,\mu\text{m}\) give \(V=2\pi(4.5)(0.12)/1.55\approx2.19<2.405\), confirming single-mode operation. Single-mode fiber minimizes pulse spreading but its small core raises source and alignment cost. Step-index MMF is simplest but range-limited by modal dispersion; graded-index MMF costs more to manufacture but gives a useful compromise of easy coupling and higher short-reach bandwidth.

Practice target: 8–9 minutes; reproduce the three profile/ray sketches, the comparison table and the V-number test.

Exam-ready answer

The acceptance angle \(\theta_a\) is the maximum angle, measured from the fiber axis, at which a meridional ray may enter and still undergo total internal reflection. Rotating this limiting ray forms the acceptance cone. The numerical aperture (NA) is the light-gathering measure

where \(n_0\) is the refractive index of the launching medium.

Let the ray refract at angle \(r\) inside a step-index core of index \(n_1\), with cladding index \(n_2<n_1\). Snell's law at the perpendicular entrance face gives

At the core-cladding wall the incidence angle is \(i=90^\circ-r\). For the limiting guided ray, \(i=\theta_c\) and \(\sin\theta_c=n_2/n_1\). Therefore

Substitution in (1) yields

Thus in air, \(n_0\approx1\) and \(NA=\sin\theta_a=\sqrt{n_1^2-n_2^2}\). With relative index difference \(\Delta=(n_1-n_2)/n_1\) and weak guidance, \(NA\approx n_1\sqrt{2\Delta}\).

For \(n_1=1.50\), \(n_2=1.47\), \(NA=\sqrt{1.50^2-1.47^2}=0.2985\) and \(\theta_a=\sin^{-1}(0.2985)\approx17.4^\circ\) in air. A larger NA eases source coupling and accepts more modes, but generally increases multimode dispersion; a smaller NA offers higher bandwidth at the cost of tighter alignment. NA is therefore specified when choosing sources, connectors and launch optics.

Practice target: 8–9 minutes; mark all three angles on the limiting ray and show both Snell-law substitutions without skipping the geometry.

5-mark answer and 10-mark extension

For 5 marks — write this¶

Attenuation is reduction of optical power with distance. For powers \(P_{in}\) and \(P_{out}\) in the same unit over \(L\) km,

It arises from intrinsic/impurity absorption, microscopic-density Rayleigh scattering, macrobending and microbending, plus discrete connector and splice losses. Silica systems use the \(850\,\text{nm}\) first window for short MMF links, the near-zero material-dispersion region around \(1310\,\text{nm}\), and the lowest-loss region around \(1550\,\text{nm}\), where good fiber is about \(0.2\,\text{dB/km}\).

Dispersion is temporal pulse spreading; overlapping symbols cause ISI and limit the bit-rate-distance product. Modal dispersion is different arrival time among modes and is severe in step-index MMF but reduced in graded-index MMF. Material dispersion comes from wavelength-dependent glass index; waveguide dispersion comes from wavelength-dependent field sharing between core and cladding. Their sum is chromatic dispersion. Polarization-mode dispersion is differential delay between orthogonal polarization modes.

Attenuation reduces pulse amplitude and is overcome by power margin or amplification; dispersion changes pulse shape/time width and is controlled by single-mode or graded-index fiber, narrow-linewidth sources, dispersion compensation or coherent equalization.

Add for a 10-mark variant¶

A physical link is source/driver → connectors and splices → cabled fiber → photodetector/receiver. It must pass two independent tests.

1. Optical power budget. In decibel units,

where powers are in dBm and every loss or engineering margin \(M\) is in dB. The received power must be no lower than receiver sensitivity, while remaining below overload. For \(P_{tx}=0\,\text{dBm}\), \(L=20\,\text{km}\), \(\alpha=0.2\,\text{dB/km}\), five \(0.1\,\text{dB}\) splices, two \(0.5\,\text{dB}\) connectors and \(M=3\,\text{dB}\),

Against \(-18\,\text{dBm}\) sensitivity, the design has \(9.5\,\text{dB}\) spare budget after the stated margin.

2. Rise-time/dispersion budget. Independent broadening contributions add approximately by root-sum-square:

Chromatic broadening may be estimated from \(t_{chromatic}=|D|\,\Delta\lambda L\), with \(D\) in ps/(nm·km), source spectral width \(\Delta\lambda\) in nm and \(L\) in km. For NRZ data the usual criterion is \(t_{sys}\le0.7/B\); for RZ it is about \(0.35/B\), with \(B\) in bit/s and time in seconds. At \(B=1\,\text{Gbit/s}\), \(t_{allow}=0.7\,\text{ns}\). If transmitter, total fiber and receiver contributions are \(0.2\), \(0.4\) and \(0.3\,\text{ns}\), then \(t_{sys}=\sqrt{0.2^2+0.4^2+0.3^2}=0.539\,\text{ns}\), so the timing budget passes.

Hence a link can have adequate received power yet fail through ISI, or adequate bandwidth yet fail sensitivity. Long-haul design must satisfy both budgets and include aging, repair-splice and environmental margin.

Practice target: 9 minutes for the 5-mark definitions and causes, or 18 minutes with both budgets and one numerical pass/fail check.