ETA FOT CERTIFICATION PRACTICE EXAM (2026
UPDATE)
Section 1 – Fiber Optics Fundamentals (Questions 1–20)
1. What is the typical core diameter of single-mode fiber?
A) 50 µm
B) 62.5 µm
C) 9 µm
D) 100 µm
Answer: C) 9 µm
Rationale: Single-mode fiber has a small core (8–10 µm) allowing only one mode
of light propagation, typically 9 µm for ITU-T G.652.
2. Which wavelength is not commonly used in multimode fiber systems?
A) 850 nm
B) 1300 nm
C) 1550 nm
D) 1310 nm
Answer: C) 1550 nm
Rationale: 1550 nm is used in single-mode for long distances. Multimode
primarily uses 850 nm and 1300/1310 nm.
,3. The numerical aperture (NA) of a fiber describes its:
A) Bandwidth
B) Light-gathering ability
C) Attenuation coefficient
D) Tensile strength
Answer: B) Light-gathering ability
Rationale: NA = sin(θ_max) defines the acceptance angle for light entering the
fiber.
4. Which glass impurity primarily causes absorption loss at 850 nm?
A) OH⁻ ions (water peak)
B) Iron
C) Germanium
D) Phosphorus
Answer: A) OH⁻ ions (water peak)
Rationale: Hydroxyl ions cause absorption peaks around 1383 nm, but also
contribute to loss at 850 nm and 950 nm in some fibers.
5. Modal dispersion is a problem mainly in:
A) Single-mode fiber
B) Multimode fiber
C) Plastic optical fiber only
D) Polarization-maintaining fiber
Answer: B) Multimode fiber
Rationale: Different modes travel different path lengths, causing pulse spreading.
Single-mode fiber has negligible modal dispersion.
,6. What is the typical chromatic dispersion value for standard single-mode fiber at
1550 nm?
A) 0 ps/(nm·km)
B) ~17 ps/(nm·km)
C) ~85 ps/(nm·km)
D) ~3 ps/(nm·km)
Answer: B) ~17 ps/(nm·km)
Rationale: G.652 fiber has zero dispersion near 1310 nm; at 1550 nm dispersion is
~17 ps/(nm·km).
7. The fiber bending loss increases exponentially when bend radius is less than:
A) 50 mm
B) 30 mm
C) 15 mm (for single-mode)
D) 100 mm
Answer: C) 15 mm (for single-mode)
Rationale: Standard single-mode fibers become highly lossy below ~15 mm
radius. Bend-insensitive fibers perform better.
8. Which fiber type has the highest bandwidth?
A) Step-index multimode
B) Graded-index multimode
C) Single-mode
D) Plastic optical fiber
Answer: C) Single-mode
Rationale: Single-mode eliminates modal dispersion, allowing extremely high
bandwidth (Tbps·km).
, 9. Attenuation in optical fiber is usually expressed in:
A) dB/km
B) dB/m
C) %/km
D) mW/km
Answer: A) dB/km
Rationale: Standard unit for fiber loss per unit length.
10. Which wavelength has the lowest intrinsic loss in silica fiber?
A) 850 nm
B) 1310 nm
C) 1550 nm
D) 1625 nm
Answer: C) 1550 nm
Rationale: Theoretical minimum attenuation for silica is ~0.16 dB/km near 1550
nm.
11. The main cause of Rayleigh scattering in fiber is:
A) Core-cladding boundary irregularities
B) Doping fluctuations
C) Macrobends
D) Connector misalignment
Answer: B) Doping fluctuations (and density variations)
Rationale: Rayleigh scattering is caused by microscopic refractive index variations,
proportional to 1/λ⁴.
UPDATE)
Section 1 – Fiber Optics Fundamentals (Questions 1–20)
1. What is the typical core diameter of single-mode fiber?
A) 50 µm
B) 62.5 µm
C) 9 µm
D) 100 µm
Answer: C) 9 µm
Rationale: Single-mode fiber has a small core (8–10 µm) allowing only one mode
of light propagation, typically 9 µm for ITU-T G.652.
2. Which wavelength is not commonly used in multimode fiber systems?
A) 850 nm
B) 1300 nm
C) 1550 nm
D) 1310 nm
Answer: C) 1550 nm
Rationale: 1550 nm is used in single-mode for long distances. Multimode
primarily uses 850 nm and 1300/1310 nm.
,3. The numerical aperture (NA) of a fiber describes its:
A) Bandwidth
B) Light-gathering ability
C) Attenuation coefficient
D) Tensile strength
Answer: B) Light-gathering ability
Rationale: NA = sin(θ_max) defines the acceptance angle for light entering the
fiber.
4. Which glass impurity primarily causes absorption loss at 850 nm?
A) OH⁻ ions (water peak)
B) Iron
C) Germanium
D) Phosphorus
Answer: A) OH⁻ ions (water peak)
Rationale: Hydroxyl ions cause absorption peaks around 1383 nm, but also
contribute to loss at 850 nm and 950 nm in some fibers.
5. Modal dispersion is a problem mainly in:
A) Single-mode fiber
B) Multimode fiber
C) Plastic optical fiber only
D) Polarization-maintaining fiber
Answer: B) Multimode fiber
Rationale: Different modes travel different path lengths, causing pulse spreading.
Single-mode fiber has negligible modal dispersion.
,6. What is the typical chromatic dispersion value for standard single-mode fiber at
1550 nm?
A) 0 ps/(nm·km)
B) ~17 ps/(nm·km)
C) ~85 ps/(nm·km)
D) ~3 ps/(nm·km)
Answer: B) ~17 ps/(nm·km)
Rationale: G.652 fiber has zero dispersion near 1310 nm; at 1550 nm dispersion is
~17 ps/(nm·km).
7. The fiber bending loss increases exponentially when bend radius is less than:
A) 50 mm
B) 30 mm
C) 15 mm (for single-mode)
D) 100 mm
Answer: C) 15 mm (for single-mode)
Rationale: Standard single-mode fibers become highly lossy below ~15 mm
radius. Bend-insensitive fibers perform better.
8. Which fiber type has the highest bandwidth?
A) Step-index multimode
B) Graded-index multimode
C) Single-mode
D) Plastic optical fiber
Answer: C) Single-mode
Rationale: Single-mode eliminates modal dispersion, allowing extremely high
bandwidth (Tbps·km).
, 9. Attenuation in optical fiber is usually expressed in:
A) dB/km
B) dB/m
C) %/km
D) mW/km
Answer: A) dB/km
Rationale: Standard unit for fiber loss per unit length.
10. Which wavelength has the lowest intrinsic loss in silica fiber?
A) 850 nm
B) 1310 nm
C) 1550 nm
D) 1625 nm
Answer: C) 1550 nm
Rationale: Theoretical minimum attenuation for silica is ~0.16 dB/km near 1550
nm.
11. The main cause of Rayleigh scattering in fiber is:
A) Core-cladding boundary irregularities
B) Doping fluctuations
C) Macrobends
D) Connector misalignment
Answer: B) Doping fluctuations (and density variations)
Rationale: Rayleigh scattering is caused by microscopic refractive index variations,
proportional to 1/λ⁴.
