Difficulty: Hard
A step-index fibre has normalized frequency V = 2.2. Which modal condition applies?
A Many modes
B No guided mode
C Single mode
D Cladding mode only
Answer: C. Single mode. For a conventional step-index fibre, V below about 2.405 supports only the fundamental mode.
For fixed core radius and numerical aperture, halving wavelength does what to V?
A Halves V
B Doubles V
C Leaves V unchanged
D Reduces V fourfold
Answer: B. Doubles V. Normalized frequency is proportional to 1/λ when other parameters are fixed.
A link launches at +3 dBm; sensitivity is -20 dBm. Fixed losses total 2 dB and reserve is 3 dB. At 0.25 dB/km, what span fits?
A 48 km
B 72 km
C 64 km
D 92 km
Answer: B. 72 km. The 23 dB power budget leaves 18 dB for fibre, so 18/0.25 = 72 km.
An optical amplifier raises input from -18 dBm to -3 dBm. What is its gain?
A 3 dB
B 18 dB
C 15 dB
D 21 dB
Answer: C. 15 dB. Gain is output minus input power in dB.
At 25 Gb/s, what is one bit period?
A 4 ps
B 25 ps
C 40 ps
D 250 ps
Answer: C. 40 ps. The bit period is 1/(25 × 10^9) seconds, or 40 ps.
A fibre has D = 17 ps/(nm·km), source width 0.2 nm, and length 80 km. What is broadening?
A 27.2 ps
B 272 ps
C 136 ps
D 2720 ps
Answer: B. 272 ps. Approximate spread is D × spectral width × length.
A 272 ps pulse spread occurs in a 25 Gb/s link. Which impairment is likely?
A Mode locking
B Clock multiplication
C Lower attenuation
D Inter-symbol interference
Answer: D. Inter-symbol interference. The spread greatly exceeds the 40 ps bit period, causing adjacent symbols to overlap.
A link has +800 ps/nm dispersion and a compensator provides -760 ps/nm. What remains?
A -40 ps/nm
B +760 ps/nm
C +1560 ps/nm
D +40 ps/nm
Answer: D. +40 ps/nm. The signed dispersion contributions add algebraically.
Fibre length increases fourfold while PMD coefficient stays constant. RMS differential delay changes by what factor?
A Twofold
B Fourfold
C Eightfold
D Unchanged
Answer: A. Twofold. Statistical PMD delay scales approximately with the square root of length.
Which WDM condition can make four-wave mixing especially troublesome?
A Near phase matching
B Large guard bands
C Low launch power
D Wide spacing only
Answer: A. Near phase matching. Nonlinear mixing builds efficiently when phase mismatch is small.
In self-phase modulation, which signal property changes the fibre’s refractive index?
A Connector colour
B Optical intensity
C Cable length only
D Receiver voltage
Answer: B. Optical intensity. The Kerr effect makes refractive index depend on optical intensity.
What does amplifier spontaneous emission primarily do to an amplified link?
A Degrades OSNR
B Removes dispersion
C Raises receiver sensitivity
D Eliminates reflections
Answer: A. Degrades OSNR. ASE adds optical noise alongside the amplified signal.
Four equal WDM channels each launch at 0 dBm. What is their approximate total power?
A 0 dBm
B +3 dBm
C +6 dBm
D +12 dBm
Answer: C. +6 dBm. Four channels at 1 mW each sum to 4 mW, about +6 dBm.
An ideal 1:8 splitter divides power equally. What is its splitting loss per output?
A About 3 dB
B About 6 dB
C About 9 dB
D About 18 dB
Answer: C. About 9 dB. The ideal split loss is 10 log10(8), approximately 9 dB.
Many WDM channels drive an EDFA beyond its operating range. What effect can appear?
A Negative attenuation
B Gain saturation
C Lower photon energy
D Zero ASE
Answer: B. Gain saturation. Excess total input power can saturate the amplifier gain.
Why place an optical isolator beside a laser or amplifier?
A Split wavelengths
B Measure dispersion
C Raise launch power
D Suppress backward light
Answer: D. Suppress backward light. An isolator reduces harmful reflections returning toward the source.
An OTDR reports an apparent loss in one direction and gain in reverse. What may explain it?
A Backscatter mismatch
B Actual optical amplification
C Negative fibre length
D A new wavelength
Answer: A. Backscatter mismatch. Different backscatter coefficients across an event can create a directional measurement artifact.
What does a coherent optical receiver mix with the incoming field?
A Clock cable
B Reference fibre
C Power splitter
D Local oscillator
Answer: D. Local oscillator. A local optical oscillator enables recovery of field information.
A direct-detection receiver measures optical intensity. What does it generally discard?
A Signal power
B Bit timing
C Wavelength
D Optical phase
Answer: D. Optical phase. Direct detection responds to optical power and does not recover the carrier phase.
Why can an avalanche photodiode detect weak signals, and what tradeoff follows?
A Internal gain adds noise
B No bias needed
C No dark current
D Unlimited bandwidth
Answer: A. Internal gain adds noise. Avalanche multiplication provides gain but also introduces excess noise and high-voltage requirements.
How can forward error correction improve a noisy optical link?
A Increase fibre diameter
B Add redundant bits
C Remove all noise
D Change connector type
Answer: B. Add redundant bits. Redundant coding bits let the receiver detect and correct some errors.
Near 1550 nm, a 100 GHz channel spacing corresponds to about what wavelength spacing?
A 0.8 nm
B 0.08 nm
C 1.55 nm
D 8 nm
Answer: A. 0.8 nm. Using Δλ ≈ λ²Δf/c gives about 0.8 nm.
Laser frequency chirp combined with fibre dispersion can cause what?
A Lower connector loss
B More splitter outputs
C Zero PMD
D Extra pulse spreading
Answer: D. Extra pulse spreading. Frequency changes across a pulse travel at different group delays in dispersive fibre.
Why can EDFA saturation cause unequal powers across WDM channels?
A Every channel uses copper
B Wavelengths become identical
C Spectral gain varies
D Fibre core shrinks
Answer: C. Spectral gain varies. Saturation can change the amplifier’s gain spectrum across channels.
Which component can counter wavelength-dependent gain across an EDFA band?
A Gain-flattening filter
B Optical circulator
C Launch cable
D Photodiode
Answer: A. Gain-flattening filter. A gain-flattening filter compensates for EDFA gain variation over wavelength.