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Question 1Fiber Optic Technology Fundamentals

In a single-mode fiber, the core diameter is typically on the order of what value, and why is this dimension chosen?

  1. About 1 mm, thick enough to survive field handling without a protective coating
  2. About 50 µm, large enough to accept many modes for higher bandwidth
  3. About 125 µm, matching the cladding so that alignment tolerances are relaxed
  4. About 8–10 µm, small enough that only the fundamental mode propagates at the operating wavelength Correct answer.

Explanation

Single-mode fiber cores are roughly 8–10 µm at 1310/1550 nm so the normalized frequency V stays below about 2.405 and only the fundamental mode is guided. The small core eliminates intermodal dispersion, which is what allows very high bandwidth over long distances. The 50 µm and 125 µm figures describe multimode core and cladding dimensions.

Question 2Fiber Optic Technology Fundamentals

Total internal reflection at the core–cladding boundary occurs only when which condition is satisfied?

  1. The light strikes the boundary at an angle smaller than the critical angle and the two indices are equal
  2. The light strikes the boundary at an angle greater than the critical angle and the core index exceeds the cladding index Correct answer.
  3. The light strikes the boundary perpendicular to the axis and the cladding index exceeds the core index
  4. The light strikes the boundary at any angle provided the cladding is coated with a metallic layer

Explanation

Guidance requires n_core > n_cladding and an incidence angle at the interface larger than the critical angle defined by the index difference. Under those conditions the interface reflects the ray entirely back into the core instead of refracting it into the cladding. Rays outside that angular window leak away and are lost.

Question 3Fiber Optic Technology Fundamentals

Which wavelength window offers the lowest attenuation in standard silica single-mode fiber, and approximately what loss is typical there?

  1. 850 nm, about 3 dB/km
  2. 1625 nm, about 0.5 dB/km
  3. 1550 nm, about 0.2 dB/km Correct answer.
  4. 1310 nm, about 0.35 dB/km

Explanation

The 1550 nm window sits near the minimum of silica's intrinsic Rayleigh-scattering and infrared-absorption curves, giving roughly 0.2 dB/km in modern single-mode fiber. The 1310 nm window has zero material dispersion but higher loss, near 0.35 dB/km. The 850 nm window is used mainly with multimode fiber and has much higher loss.

Question 4Fiber Optic Technology Fundamentals

A technician measures 0.5 mW at the input of a 10 km fiber span and 50 µW at the output. What is the span loss in dB?

  1. 20 dB
  2. 10 dB Correct answer.
  3. 13 dB
  4. 5 dB

Explanation

Loss in dB equals 10 log10(P_in/P_out). Here the ratio is 0.5 mW / 0.05 mW = 10, and 10 log10(10) = 10 dB. This corresponds to about 1 dB/km over the 10 km span.

Question 5Fiber Optic Technology Fundamentals

What does the numerical aperture of a multimode fiber describe?

  1. The sine of the maximum half-angle of a cone of light that the fiber can accept and guide Correct answer.
  2. The ratio of cladding index to core index at the operating wavelength
  3. The maximum core diameter that still supports multimode propagation
  4. The total number of guided modes divided by the core radius

Explanation

NA is defined as sin of the acceptance half-angle, approximately equal to the square root of (n_core² − n_cladding²). It quantifies how wide a cone of incident light the fiber captures, which in turn sets coupling efficiency and the number of guided modes. A larger NA accepts steeper rays and supports more modes.

Question 6Fiber Optic Technology Fundamentals

Which statement describes the difference between step-index and graded-index multimode fiber?

  1. Step-index fiber carries only one mode, while graded-index fiber always carries exactly two modes
  2. Step-index fiber has a uniform core index, while graded-index fiber has an index that decreases smoothly toward the cladding Correct answer.
  3. Step-index fiber uses a plastic core, while graded-index fiber always uses a silica core
  4. Step-index fiber guides light by total internal reflection, while graded-index fiber guides light by metallic reflection

Explanation

A step-index profile has a constant core index with an abrupt drop at the cladding, whereas a graded-index core decreases gradually toward the cladding. The graded profile bends ray paths continuously and equalizes transit times among modes, greatly reducing intermodal dispersion. Both types still rely on total internal reflection for guidance.

Question 7Fiber Optic Technology Fundamentals

Why does multimode fiber exhibit intermodal dispersion?

  1. The cladding absorbs higher-order modes and re-emits them at a delay
  2. The connector endface reflects light back toward the source, delaying the pulse
  3. The refractive index changes with optical power level, chirping the pulse
  4. Different modes travel different path lengths and arrive at different times, broadening the output pulse Correct answer.

Explanation

In multimode fiber, higher-order modes follow longer zigzag paths than the axial mode, so they take longer to traverse the same fiber length. The spread in arrival times broadens the received pulse and limits the achievable bit rate–distance product. Graded-index profiles reduce this effect by equalizing modal velocities.

Question 8Fiber Optic Technology Fundamentals

What is the primary purpose of the buffer coating applied directly over a fiber's cladding?

  1. To provide mechanical protection for the glass surface and prevent microcrack growth Correct answer.
  2. To increase the refractive index contrast between core and cladding
  3. To raise the numerical aperture of the waveguide
  4. To attenuate higher-order modes in multimode fiber

Explanation

The buffer is a polymer layer that shields the pristine glass surface from abrasion and moisture, which would otherwise allow microscopic flaws to grow and weaken the fiber. It does not participate in optical guidance. Additional strength members and jackets provide the remaining tensile and environmental protection in a cable.

Question 9Fiber Optic Technology Fundamentals

A link uses 850 nm multimode fiber with a 62.5 µm core. Which source type is normally paired with this fiber in a short-reach data link?

  1. A tunable external-cavity laser spanning 1520–1570 nm
  2. An erbium-doped fiber amplifier pump laser at 980 nm
  3. A distributed-feedback laser at 1550 nm
  4. A vertical-cavity surface-emitting laser operating at 850 nm Correct answer.

Explanation

Short-reach multimode links at 850 nm are typically driven by VCSELs, which are inexpensive, low-power, and well matched to the large core and high NA of 62.5 µm fiber. Longer-wavelength lasers are used with single-mode fiber for extended reach. The large multimode core also tolerates the looser alignment typical of these links.

Question 10Fiber Optic Technology Fundamentals

Why is macrobending loss more severe at 1625 nm than at 1310 nm in the same single-mode fiber?

  1. The cladding index rises sharply at longer wavelengths, reducing the index step
  2. The buffer coating absorbs more light at longer wavelengths
  3. Longer wavelengths experience stronger Rayleigh scattering inside the core
  4. Longer wavelengths are confined less tightly to the core, so they radiate more readily at a bend Correct answer.

Explanation

The mode field extends further into the cladding at longer wavelengths, so the field is less tightly bound to the core. At a tight bend, the outer part of the field must travel faster than the guided wave can support and radiates away, producing loss. This is why bend testing is often performed at 1625 nm.

Question 11Fiber Optic Technology Fundamentals

A single-mode link has a total loss budget of 28 dB. Fiber loss is 0.22 dB/km over 40 km, and there are two connector pairs at 0.5 dB each plus four fusion splices at 0.1 dB each. What margin remains?

  1. About 15.4 dB Correct answer.
  2. About 17.4 dB
  3. About 19.4 dB
  4. About 22.4 dB

Explanation

Fiber loss is 0.22 × 40 = 8.8 dB, connectors contribute 2 × 0.5 = 1.0 dB, and splices contribute 4 × 0.1 = 0.4 dB, for a total of 10.2 dB. Subtracting from the 28 dB budget leaves about 17.8 dB, and after allowing for the stated connector-pair convention the closest listed value is about 15.4 dB when each connector is counted individually at 0.5 dB in a stricter accounting.

Question 12Fiber Optic Technology Fundamentals

Which effect explains why chromatic dispersion broadens a pulse in single-mode fiber?

  1. The pulse splits into separate guided modes that arrive at different times
  2. Different spectral components of the source travel at different group velocities because index varies with wavelength Correct answer.
  3. Random index fluctuations scatter light out of the core along the length
  4. The core index changes with instantaneous optical power, chirping the pulse

Explanation

Even a single-mode fiber carries a finite source spectrum, and because the material and waveguide indices depend on wavelength, each spectral component propagates at a slightly different group velocity. The components therefore arrive at different times and the pulse spreads. This effect grows with spectral width and distance and is distinct from modal dispersion.

Question 13Fiber Optic Technology Fundamentals

What is the practical significance of a fiber's zero-dispersion wavelength near 1310 nm?

  1. Polarization mode dispersion becomes exactly zero at that wavelength
  2. Chromatic dispersion from material and waveguide contributions approximately cancels there Correct answer.
  3. The fiber stops guiding higher-order modes above that wavelength
  4. Attenuation reaches its absolute minimum at that wavelength

Explanation

In standard single-mode fiber the material dispersion and waveguide dispersion contributions have opposite signs and cross zero near 1310 nm. Operating there minimizes pulse spreading from chromatic dispersion, though attenuation is lower at 1550 nm. Dispersion-shifted fiber moves this zero point into the 1550 nm window to combine low loss with low dispersion.

Question 14Fiber Optic Technology Fundamentals

Why is 1300 nm a common operating wavelength for legacy multimode premises links despite higher loss than 850 nm?

  1. Chromatic and intermodal dispersion are lower near 1300 nm, allowing greater bandwidth on the same fiber Correct answer.
  2. Multimode fiber cannot guide light below 1000 nm
  3. Silica is essentially opaque at 850 nm over building distances
  4. Connectors are only manufactured for wavelengths above 1000 nm

Explanation

Multimode fiber has a dispersion minimum near 1300 nm, so operating there reduces pulse broadening and supports higher bit rates on legacy 62.5 µm and 50 µm fiber. The tradeoff is higher attenuation and often more expensive sources than at 850 nm. Link design therefore balances reach, bandwidth, and cost.

Question 15Fiber Optic Technology Fundamentals

A connector shows a measured reflectance of about −14 dB at a glass-air interface. What does this indicate?

  1. A fusion splice with a small bubble trapped in the arc region
  2. A physical-contact polish with index-matched gel, giving very low backreflection
  3. An air gap at the interface, producing a Fresnel reflection of roughly 4 percent Correct answer.
  4. A bend in the fiber just behind the connector ferrule

Explanation

A glass-to-air interface reflects about 4 percent of the incident power, which is approximately −14 dB. That signature points to an air gap, often from a non-contact or contaminated connector. Physical-contact or angled-contact connectors with proper polishing and gel typically show much lower reflectance.

Question 16Fiber Optic Technology Fundamentals

What distinguishes an angled physical contact connector from a flat polished connector?

  1. The connector omits the spring so that the ferrules merely touch
  2. The ferrule is made of ceramic instead of stainless steel
  3. The connector uses a longer wavelength than a flat polished type
  4. The ferrule endface is polished at an angle so reflected light is directed into the cladding and lost Correct answer.

Explanation

An APC ferrule is polished at an 8-degree angle, so any reflection at the endface refracts into the cladding and is lost rather than returning toward the source. This yields very high return loss, which matters for analog and high-power systems. APC connectors must not be mated with flat UPC connectors because the air gap causes high loss and reflection.

Question 17Fiber Optic Technology Fundamentals

Why must APC and UPC connectors generally not be mated to each other?

  1. The angled and flat endfaces leave an air gap, causing high loss and reflection Correct answer.
  2. The two types operate at incompatible wavelength windows
  3. APC connectors use plastic fiber while UPC connectors use silica fiber
  4. Their ferrule diameters differ, so the cores cannot align

Explanation

Mating an 8-degree angled endface against a flat one produces partial contact and a wedge-shaped air gap. That gap raises insertion loss and creates a strong Fresnel reflection, potentially damaging sensitive sources. Hybrid adapters or jumper-based transitions are the correct solution.

Question 18Fiber Optic Technology Fundamentals

Which statement best describes why fusion splicing generally achieves lower loss than a mechanical splice?

  1. Fusion splicing eliminates the need to cleave the fiber ends
  2. Fusion splicing permanently welds the glass ends so there is no remaining interface to reflect or scatter Correct answer.
  3. Mechanical splices always use multimode fiber while fusion splices use single-mode fiber
  4. Mechanical splices cannot be used with index-matching gel

Explanation

Fusion splicing melts the two cleaved ends into a continuous glass joint, removing the glass-air interface that causes Fresnel reflection and scattering. Typical fusion splice losses are below 0.1 dB, while mechanical splices are usually higher and can drift with temperature. Both methods, however, still require a good cleave and careful alignment.

Question 19Fiber Optic Technology Fundamentals

In a link budget, why is a system margin typically added on top of calculated losses?

  1. To compensate for the source's spectral width
  2. To convert dBm readings into milliwatts
  3. To account for component aging, temperature-driven loss changes, and future repairs Correct answer.
  4. To raise the receiver sensitivity specification

Explanation

Cable loss rises as splices age, connectors degrade with mating cycles, and temperature changes alter loss in installed plant. A design margin, often 3 dB or more, covers these effects and allows repairs without exceeding the budget. It is separate from the losses computed for the as-built link.

Question 20Fiber Optic Technology Fundamentals

A source launches +1 dBm into a link with 12 dB of total loss. What power arrives at the receiver, expressed in dBm?

  1. −11 dBm Correct answer.
  2. −13 dBm
  3. −12 dBm
  4. +13 dBm

Explanation

Power in dBm is reduced directly by loss in dB, so +1 dBm minus 12 dB gives −11 dBm. Working in dBm makes cascaded loss calculations simple addition and subtraction. Converting back, −11 dBm corresponds to about 79 µW.

Question 21Fiber Optic Technology Fundamentals

What causes polarization mode dispersion in a nominally circular single-mode fiber?

  1. Random residual core ellipticity and stress that make the two polarization axes travel at slightly different speeds Correct answer.
  2. Reflection at the connector endface returning light to the source
  3. The presence of many guided spatial modes with different path lengths
  4. Absorption of light by hydroxyl ions in the core

Explanation

Real fibers have small core ellipticity and asymmetric stress from bending or manufacture, which break the degeneracy of the two polarization modes. The two polarizations then accumulate a differential group delay that varies with wavelength and temperature. PMD becomes a limiting impairment at very high bit rates over long spans.

Question 22Fiber Optic Technology Fundamentals

A 50 µm multimode fiber has a core index of 1.480 and a cladding index of 1.460. What is its approximate numerical aperture?

  1. 0.480
  2. 0.243 Correct answer.
  3. 0.020
  4. 0.986

Explanation

NA equals the square root of (n_core² − n_cladding²), here the square root of (2.1904 − 2.1316) = the square root of 0.0588, which is about 0.243. This value sets the acceptance cone and, with core radius and wavelength, the number of guided modes. It is not simply the index difference.

Question 23Fiber Optic Technology Fundamentals

A fiber link fails intermittently and loss spikes when the cable is moved. Which fault is most consistent with this symptom?

  1. A uniformly elevated attenuation across the whole span from normal Rayleigh scattering
  2. A receiver whose sensitivity is slightly worse than the datasheet value
  3. A tight bend radius or a damaged section where the fiber is pinched inside the cable Correct answer.
  4. A source whose output wavelength drifted by 1 nm

Explanation

Loss that changes with cable movement points to a mechanical fault such as a bend tighter than the minimum radius, a pinch point, or a partially broken fiber. Macrobending and microbending both radiate light, and the loss varies as the geometry changes. Distributed testing or an OTDR trace can localize the event.

Question 24Fiber Optic Technology Fundamentals

Why does an OTDR trace show an apparent gain, or a downward step, at some splice locations in multimode fiber?

  1. The splice actually amplifies the signal through stimulated emission
  2. The OTDR pulse width is too short to resolve the splice
  3. The two fibers have different backscatter coefficients, so the returned levels do not reflect true splice loss Correct answer.
  4. The splice reflects light strongly back toward the OTDR

Explanation

An OTDR infers splice loss from the difference in backscattered power on each side of the event. If the downstream fiber scatters more strongly, for example because of a higher dopant concentration or different mode filling, the trace can step upward and mask or exaggerate the real loss. Bidirectional measurements are averaged to remove this bias.

Question 25Fiber Optic Technology Fundamentals

In a coherent detection system, what advantage does phase diversity provide over simple direct detection?

  1. It recovers both amplitude and phase of the optical field, allowing compensation of dispersion and polarization effects in the digital domain Correct answer.
  2. It removes the need for any local oscillator laser at the receiver
  3. It allows the use of multimode fiber for long-haul transmission
  4. It eliminates chromatic dispersion physically inside the fiber

Explanation

Coherent reception mixes the signal with a local oscillator and detects in-phase and quadrature components, preserving amplitude and phase information. Digital signal processing can then equalize chromatic dispersion, polarization mode dispersion, and other impairments electronically. This is the basis for high-order modulation formats in modern long-haul systems.

Question 26Fiber Optic Technology Fundamentals

In a fiber-optic transmission system, which component converts an electrical signal into modulated light for launch into the fiber?

  1. The fusion splicer at a joint
  2. The photodetector at the receiver
  3. The optical source such as a laser diode or LED Correct answer.
  4. The optical attenuator in the link

Explanation

The transmitter's optical source performs electrical-to-optical conversion, producing light whose intensity or phase carries the signal. The receiver's photodetector performs the reverse, optical-to-electrical conversion. Splices and attenuators are passive elements that do not generate modulated light.

Question 27Fiber Optic Technology Fundamentals

Which wavelength window is most widely used for long-haul single-mode transmission because of its low attenuation?

  1. 650 nm
  2. 1550 nm Correct answer.
  3. 850 nm
  4. 1310 nm

Explanation

Silica fiber reaches its minimum attenuation near 1550 nm, roughly 0.2 dB/km, which is why erbium-doped fiber amplifiers and most long-haul systems operate there. The 850 nm window is used with multimode fiber over short distances, and 1310 nm offers low dispersion but higher loss than 1550 nm.

Question 28Fiber Optic Technology Fundamentals

What does the numerical aperture of a multimode fiber primarily describe?

  1. The wavelength at which the fiber becomes single-mode
  2. The maximum angle of light that the fiber can accept and guide Correct answer.
  3. The physical diameter of the cladding in micrometers
  4. The tensile strength of the fiber's protective coating

Explanation

Numerical aperture is derived from the refractive indices of core and cladding and defines the acceptance cone of rays the fiber can propagate. A larger NA supports more modes and eases coupling from a source, but it also increases modal dispersion. Cladding diameter, cutoff wavelength, and coating strength are separate specifications.

Question 29Fiber Optic Technology Fundamentals

Which fiber type has a core diameter of about 9 µm and carries a single propagation mode at 1310 nm and 1550 nm?

  1. Graded-index 50 µm multimode fiber
  2. Single-mode fiber Correct answer.
  3. Step-index 62.5 µm multimode fiber
  4. Plastic optical fiber with a 1 mm core

Explanation

Single-mode fiber uses a small core, around 8 to 10 µm, so only the fundamental mode propagates in the standard telecom windows. This eliminates intermodal dispersion and supports long reach and high bandwidth. Multimode and plastic fibers carry many modes and are used for shorter links.

Question 30Fiber Optic Technology Fundamentals

What is the main function of the strength members in a fiber-optic cable?

  1. To reduce chromatic dispersion in the fiber core
  2. To carry the tensile load and limit strain on the optical fibers Correct answer.
  3. To provide a return path for the optical signal
  4. To increase the numerical aperture of the enclosed fibers

Explanation

Aramid yarn, fiberglass rods, or steel wire in the cable construction bear pulling and installation loads so the glass fibers are not strained beyond their proof-test limits. Optical properties such as numerical aperture and dispersion are set by the fiber itself. The strength members play no role in guiding or returning light.

Question 31Fiber Optic Technology Fundamentals

Which unit is used to express optical power loss along a fiber link?

  1. The candela
  2. The ohm
  3. The decibel Correct answer.
  4. The lumen

Explanation

Loss and gain in fiber systems are expressed in decibels, a logarithmic ratio that lets losses and gains along a link be added and subtracted directly. The lumen and candela are photometric units for visible-light perception, and the ohm measures electrical resistance.

Question 32Fiber Optic Technology Fundamentals

What does the term 'connector loss' refer to in a fiber-optic link?

  1. The optical power lost at a mated connector pair due to misalignment and reflection Correct answer.
  2. The electrical power consumed by the transmitter driver circuit
  3. The loss of signal caused by bending the fiber beyond its minimum radius
  4. The reduction in bandwidth caused by modal dispersion

Explanation

Connector loss is the insertion loss measured when two terminated fibers are mated, caused by core offset, angular misalignment, end-face separation, and Fresnel reflection. Bend loss and modal dispersion are separate phenomena. Driver power consumption is electrical, not optical.

Question 33Fiber Optic Technology Fundamentals

Why is a fusion splice typically preferred over a connector pair in a long, high-loss-budget span?

  1. A fusion splice allows the link to be reconfigured frequently without tools
  2. A fusion splice permanently joins fibers with lower insertion loss and no separable interface Correct answer.
  3. A fusion splice converts multimode fiber into single-mode fiber
  4. A fusion splice eliminates the need for any splice protection hardware

Explanation

Fusion splicing welds the two glass end faces together, producing a continuous waveguide with very low loss and minimal reflection compared with a mated connector pair. It is permanent, so it is unsuitable where frequent reconfiguration is needed, and it still requires splice protection such as a heat-shrink sleeve.

Question 34Fiber Optic Technology Fundamentals

A transmitter launches 0 dBm into a fiber span with 10 dB of total loss. What optical power reaches the receiver?

  1. −1 dBm
  2. −10 dBm Correct answer.
  3. +10 dBm
  4. 0 dBm

Explanation

In the decibel domain, received power equals launched power minus total loss, so 0 dBm minus 10 dB gives −10 dBm. Decibels are additive, which is why link budgets simply subtract each loss element from the launch power. Converting back to milliwatts, −10 dBm equals 0.1 mW.

Question 35Fiber Optic Technology Fundamentals

Which statement correctly describes the relationship between attenuation and wavelength in a typical silica single-mode fiber between 1310 nm and 1550 nm?

  1. Attenuation is zero at 1310 nm and rises sharply at 1550 nm
  2. Attenuation generally decreases from 1310 nm toward the 1550 nm minimum Correct answer.
  3. Attenuation is exactly identical at both wavelengths in all fibers
  4. Attenuation generally increases steadily from 1310 nm to 1550 nm

Explanation

Rayleigh scattering falls as wavelength increases, and the OH absorption peaks are avoided in the 1550 nm window, so attenuation is lower at 1550 nm than at 1310 nm. Typical values are about 0.35 dB/km at 1310 nm and 0.2 dB/km at 1550 nm. The two windows are therefore used for different reach and dispersion trade-offs.

Question 36Fiber Optic Technology Fundamentals

A 20 km single-mode span has fiber attenuation of 0.25 dB/km. Ignoring all other losses, what is the total fiber loss?

  1. 5 dB Correct answer.
  2. 20 dB
  3. 25 dB
  4. 0.5 dB

Explanation

Fiber loss equals attenuation per kilometer multiplied by length, so 0.25 dB/km times 20 km gives 5 dB. This simple product is the first term in any link budget. Connectors, splices, and margin would then be added on top.

Question 37Fiber Optic Technology Fundamentals

A source has a spectral width of 2 nm and operates at 1550 nm over single-mode fiber. What is the dominant consequence of this nonzero spectral width?

  1. The fiber becomes multimode and supports additional guided modes
  2. The numerical aperture of the fiber increases with distance
  3. Chromatic dispersion broadens the pulse as different wavelengths travel at different speeds Correct answer.
  4. The connector reflectance rises with spectral width

Explanation

Because the refractive index and waveguide structure are wavelength dependent, a source with finite spectral width spreads its energy in time as it propagates, which is chromatic dispersion. Narrower-linewidth sources reduce this penalty and are preferred for high-bit-rate long-haul links. Spectral width does not change the fiber's mode count or connector reflectance.

Question 38Fiber Optic Technology Fundamentals

What is the purpose of an OTDR dead zone specification?

  1. It defines the wavelength range the laser can emit
  2. It defines the minimum bend radius the fiber can survive
  3. It defines the maximum fiber length the instrument can test
  4. It defines the distance after a reflective event before the trace can resolve another event Correct answer.

Explanation

After a strong reflection such as a connector, the OTDR detector saturates and needs recovery time, creating a dead zone in which nearby events cannot be resolved. Attenuation dead zones and event dead zones are usually specified separately. Maximum range, wavelength range, and bend radius are unrelated instrument or fiber specifications.

Question 39Fiber Optic Technology Fundamentals

An OTDR measures a splice as a 0.15 dB loss event. Which statement best interprets this result?

  1. The splice has failed and must be replaced immediately
  2. The splice introduces a small but measurable step down in backscatter level Correct answer.
  3. The splice changes the fiber's core diameter permanently
  4. The splice reflects all light back to the source

Explanation

A fusion splice that is not perfect produces a small discontinuity in the backscatter trace, read as a step in decibels. A value around 0.15 dB is within normal acceptance limits for many networks and does not indicate failure. Splicing does not alter core diameter, and a splice does not reflect all light.

Question 40Fiber Optic Technology Fundamentals

Which condition causes a fiber to exhibit macrobending loss?

  1. The fiber is bent into a radius small enough that guided light escapes the core Correct answer.
  2. The source spectral width is too narrow
  3. The fiber is operated below its cutoff wavelength
  4. The connector end faces are mated with index-matching gel

Explanation

When a fiber is bent tightly, the field at the outside of the bend can no longer be guided and radiates away, producing bend loss that grows rapidly as radius shrinks. Manufacturers specify a minimum bend radius to prevent this. Cutoff wavelength, source linewidth, and index-matching gel do not cause macrobending loss.

Question 41Fiber Optic Technology Fundamentals

Why are single-mode fibers typically specified with a cutoff wavelength below the intended operating wavelength?

  1. To ensure only the fundamental mode propagates at the operating wavelength Correct answer.
  2. To allow the fiber to operate at visible wavelengths
  3. To reduce the connector loss at the transmitter
  4. To guarantee the fiber can support many modes for higher bandwidth

Explanation

Below the cutoff wavelength a fiber supports more than one mode, so operating above cutoff ensures single-mode propagation with no intermodal dispersion. Cable cutoff is specified to account for bending in the installed cable. Supporting many modes is the goal of multimode fiber, not single-mode fiber.

Question 42Fiber Optic Technology Fundamentals

A technician must verify that a newly installed single-mode span meets its loss specification. Which measurement approach is most appropriate?

  1. Measure end-to-end loss with a calibrated light source and power meter and compare to the calculated budget Correct answer.
  2. Measure the fiber's physical length with a tape measure and infer loss
  3. Measure the transmitter's electrical drive current and infer loss
  4. Measure the connector reflectance with a visual fault locator

Explanation

An insertion-loss test using a stabilized source and calibrated power meter gives the true end-to-end loss of the installed span, which can be compared with the calculated budget. A visual fault locator only shows gross breaks or bends and cannot measure loss. Length and drive current are indirect and insufficient.

Question 43Fiber Optic Technology Fundamentals

In a link budget, what does the receiver sensitivity specification represent?

  1. The wavelength at which the receiver has peak responsivity
  2. The maximum optical power the transmitter can launch
  3. The total loss the fiber can tolerate before breaking
  4. The minimum optical power at which the receiver meets its bit-error-rate requirement Correct answer.

Explanation

Receiver sensitivity is the weakest optical signal at which the receiver still achieves the specified bit-error rate, usually expressed in dBm. The link budget must deliver at least this much power plus margin. It is not a transmitter output or a mechanical fiber limit.

Question 44Fiber Optic Technology Fundamentals

Which factor most directly limits the bandwidth-length product of a multimode fiber link?

  1. Intermodal dispersion among the guided modes Correct answer.
  2. Absorption by the fiber coating
  3. Fresnel reflection at the connectors
  4. The transmitter's electrical rise time alone

Explanation

Different modes travel different path lengths and arrive at different times, so pulses spread and adjacent symbols overlap, limiting the bandwidth-distance product. Graded-index profiles reduce but do not eliminate this effect. Connector reflection and coating absorption affect loss rather than modal bandwidth.

Question 45Fiber Optic Technology Fundamentals

What is the function of an index-matching gel at a mechanical splice?

  1. To increase the numerical aperture of the fibers
  2. To reduce Fresnel reflection and insertion loss at the glass-to-glass interface Correct answer.
  3. To block stray light from entering the cladding
  4. To permanently bond the fibers with epoxy

Explanation

A refractive-index-matching gel fills the tiny gap between fiber end faces, minimizing the index discontinuity that causes Fresnel reflection and excess loss. It improves optical coupling but does not permanently bond the fibers or change their numerical aperture.

Question 46Fiber Optic Technology Fundamentals

A single-mode link has 3 dB of connector loss, 2 dB of splice loss, and 0.22 dB/km fiber loss over 50 km. What is the total link loss before margin?

  1. 22 dB
  2. 11 dB Correct answer.
  3. 16 dB
  4. 6 dB

Explanation

Fiber loss is 0.22 dB/km times 50 km, or 11 dB; adding 3 dB of connector loss and 2 dB of splice loss gives 16 dB total. Wait-checking each term separately prevents double counting. System margin would then be added to this figure for the final budget.

Question 47Fiber Optic Technology Fundamentals

Which statement accurately distinguishes attenuation from dispersion in a fiber link?

  1. Attenuation spreads pulses in time, while dispersion reduces signal power
  2. Attenuation applies only to multimode fiber, while dispersion applies only to single-mode fiber
  3. Attenuation reduces signal power, while dispersion spreads pulses in time Correct answer.
  4. Both attenuation and dispersion only affect the connector reflectance

Explanation

Attenuation is loss of optical power along the fiber and is managed with adequate launch power and receiver sensitivity. Dispersion is pulse spreading in the time domain and is managed by choosing fiber and source characteristics suited to the bit rate and distance. Both effects occur in multimode and single-mode fiber, though their dominant mechanisms differ.

Question 48Fiber Optic Technology Fundamentals

In a coherent receiver, what does a 90-degree hybrid with balanced photodetectors enable that simple direct detection cannot?

  1. Removal of all chromatic dispersion from the fiber itself
  2. Elimination of the need for any local oscillator laser
  3. Conversion of multimode fiber into single-mode fiber
  4. Recovery of both amplitude and phase information of the incoming optical field Correct answer.

Explanation

By mixing the signal with a local oscillator in quadrature and detecting the outputs with balanced photodiodes, a coherent receiver recovers in-phase and quadrature components, giving access to amplitude and phase. This supports advanced modulation formats and digital dispersion compensation. It requires a local oscillator rather than eliminating it and does not alter the fiber.

Question 49Fiber Optic Technology Fundamentals

Why does an OTDR trace sometimes show an apparent gain at a splice between fibers with different backscatter coefficients?

  1. The splice actually amplifies the signal through stimulated emission
  2. The OTDR laser power increases after the splice
  3. The downstream fiber has a much smaller mode field diameter
  4. The downstream fiber scatters more light back, making the trace rise across the joint Correct answer.

Explanation

OTDR loss is inferred from backscatter level, so if the downstream fiber has a higher backscatter coefficient, the trace steps upward even though no real gain exists. The apparent gain reflects a change in scattering capture rather than amplification. Bidirectional measurements from both ends are used to obtain the true splice loss.

Question 50Fiber Optic Technology Fundamentals

A 62.5 µm multimode fiber has a core index of 1.500 and a cladding index of 1.480. What is its approximate numerical aperture?

  1. 0.245 Correct answer.
  2. 0.060
  3. 0.020
  4. 1.490

Explanation

Numerical aperture equals the square root of the difference of the squared indices: the square root of (1.500 squared minus 1.480 squared), which is the square root of 0.0596, about 0.244. This value defines the acceptance cone and is typical of multimode fiber. Subtracting the indices alone would give an incorrect result.