Difficulty: Medium
A 20 km fibre has attenuation of 0.25 dB/km. What is its fibre loss?
A 5 dB
B 0.5 dB
C 10 dB
D 20 dB
Answer: A. 5 dB. Multiply length by attenuation coefficient: 20 × 0.25 = 5 dB.
Fibre loss is 5 dB, two connectors lose 0.5 dB each, and four splices lose 0.1 dB each. Total?
A 5.4 dB
B 6.4 dB
C 6.0 dB
D 7.4 dB
Answer: B. 6.4 dB. Add 5 + 1.0 + 0.4 dB.
Transmit power is 0 dBm and receiver sensitivity is -10 dBm. What is the power budget?
A 10 dB
B 0 dB
C -10 dB
D 20 dB
Answer: A. 10 dB. The difference between launch power and sensitivity is 10 dB.
Launching at -3 dBm through 7 dB loss gives which received power?
A -10 dBm
B 4 dBm
C -7 dBm
D -21 dBm
Answer: A. -10 dBm. Subtract the link loss in dB from the transmitted dBm level.
A 12 dB budget and 9 dB installed loss leave what margin?
A 2 dB
B 9 dB
C 3 dB
D 21 dB
Answer: C. 3 dB. Available margin is budget minus actual loss.
Approximately what fraction of optical power remains after 3 dB loss?
A One-half
B One-quarter
C One-tenth
D Two-thirds
Answer: A. One-half. A 3 dB decrease is close to a factor of two in power.
Four WDM channels each carry 2.5 Gb/s. What is the aggregate gross rate?
A 2.5 Gb/s
B 5 Gb/s
C 40 Gb/s
D 10 Gb/s
Answer: D. 10 Gb/s. Four independent channels each carrying 2.5 Gb/s total 10 Gb/s.
A source emits 2 mW. Its power is approximately what in dBm?
A 0 dBm
B +3 dBm
C -3 dBm
D +10 dBm
Answer: B. +3 dBm. 10 log10(2 mW/1 mW) is approximately +3 dBm.
With refractive index 1.5, the optical phase speed in glass is about what?
A 3 × 10^8 m/s
B 1 × 10^8 m/s
C 1.5 × 10^8 m/s
D 2 × 10^8 m/s
Answer: D. 2 × 10^8 m/s. Speed is c/n, approximately 3 × 10^8 / 1.5 m/s.
Light travels at 2 × 10^8 m/s in fibre. What is the one-way delay over 10 km?
A 5 microseconds
B 100 microseconds
C 50 microseconds
D 500 microseconds
Answer: C. 50 microseconds. Delay is 10,000 metres divided by 2 × 10^8 metres per second.
For a step-index fibre in air with n-core 1.5 and n-cladding 1.4, NA is about what?
A 0.10
B 0.54
C 0.75
D 1.40
Answer: B. 0.54. NA is sqrt(1.5² – 1.4²) = sqrt(0.29), about 0.54.
A fibre has NA 0.5 in air. What is its acceptance half-angle?
A 15°
B 30°
C 45°
D 60°
Answer: B. 30°. The sine of the acceptance half-angle equals NA in air.
Different wavelengths travel at different group speeds. Which effect broadens the pulse?
A Modal gain
B Fresnel loss
C Polarization filtering
D Chromatic dispersion
Answer: D. Chromatic dispersion. Spectral components arrive at different times.
A dispersion coefficient is 17 ps/(nm·km), linewidth is 0.1 nm, and length is 10 km. Approximate broadening?
A 1.7 ps
B 170 ps
C 17 ps
D 1700 ps
Answer: C. 17 ps. Multiply 17 by 0.1 by 10.
Different polarization states propagate with different delays. What is this impairment?
A Polarization-mode dispersion
B Rayleigh scattering
C Connector attenuation
D Laser chirp
Answer: A. Polarization-mode dispersion. PMD separates polarization components in time.
Which transmitter change generally reduces broadening caused by chromatic dispersion?
A Narrower optical spectrum
B Wider optical spectrum
C More connector loss
D Larger cladding
Answer: A. Narrower optical spectrum. A narrower wavelength spread reduces group-delay spread for a fixed fibre.
Why can index-matching material reduce reflection at a connection?
A More fibre modes
B Higher transmitter power
C Smaller index step
D Shorter wavelength only
Answer: C. Smaller index step. Reducing the refractive-index discontinuity lowers Fresnel reflection.
An OTDR sees a return after 100 microseconds. If speed is 2 × 10^8 m/s, how distant is the event?
A 5 km
B 20 km
C 10 km
D 100 km
Answer: C. 10 km. Distance is speed times round-trip time divided by two.
Which event commonly produces a reflective spike on an OTDR trace?
A Ideal fusion splice
B Mated connector
C Uniform fibre segment
D Buffer coating
Answer: B. Mated connector. An interface at a connector can reflect a portion of the test pulse.
A launch cable is added before OTDR testing. Which near-end issue does it help overcome?
A Rayleigh limit
B Chromatic dispersion
C Dead zone
D Laser linewidth
Answer: C. Dead zone. The extra fibre separates the first tested connector from the instrument’s dead zone.
Why average OTDR splice loss from both directions?
A Double bit rate
B Remove all noise
C Increase power budget
D Backscatter mismatch
Answer: D. Backscatter mismatch. Different fibre backscatter coefficients can bias one-way readings.
An OTDR uses a wider test pulse to reach farther. What tradeoff follows?
A Zero backscatter
B Poorer event resolution
C Lower range
D No dead zone
Answer: B. Poorer event resolution. A longer pulse increases returned energy but makes nearby events harder to separate.
What does optical signal-to-noise ratio compare?
A Signal to optical noise
B Length to diameter
C Voltage to current
D Clock to bit rate
Answer: A. Signal to optical noise. OSNR is optical signal power relative to optical noise power in a stated bandwidth.
One bit error occurs in one million transmitted bits. What is the measured BER?
A 10^-3
B 10^-9
C 10^6
D 10^-6
Answer: D. 10^-6. BER is errors divided by the number of transmitted bits.
An ideal 1:4 splitter receives 1 mW. How much reaches each output?
A 0.1 mW
B 0.5 mW
C 4 mW
D 0.25 mW
Answer: D. 0.25 mW. Ignoring excess loss, the input is divided evenly among four outputs.