Chapter 6 · 11 hours
Optical Fibres
IOE past exam questions
Past questions and answers
34 questions set from this chapter, 6 of them more than once. Most asked first.
- Asked 5 times
- 2080 Chaitra · 3 marks
- 2080 Asoj · 3 marks
- 2077 Chaitra · 4 marks
- 2075 Bhadra · 4 marks
- 2075 Baisakh · 4 marks
Write a short note on light (optical) source and photo (optical) detector.
Answer
An optical fibre link needs a light source to convert the electrical signal into light at the transmitter and a photodetector to convert light back into an electrical signal at the receiver.
Optical sources
Requirements: emission at fibre low-loss wavelengths (850, 1310, 1550 nm), small emitting area, enough power, fast modulation, narrow spectrum, long life and low cost. Two semiconductor sources are used:
- LED (light emitting diode): forward-biased p-n junction giving spontaneous emission. Incoherent light, wide spectral width (30–60 nm), low power, bandwidth up to ~100 MHz. Cheap and reliable; used with multimode fibre over short distances (LANs).
- Laser diode (LD): stimulated emission in a cavity above a threshold current. Coherent, narrow linewidth (1–3 nm, or < 0.1 nm for DFB lasers), higher power, GHz modulation. Used with single-mode fibre in long-haul high-speed links.
Photodetectors
Requirements: high responsivity at the operating wavelength, fast response, low noise, low bias voltage.
- PIN photodiode: an intrinsic layer between p and n regions widens the depletion zone, so most photons are absorbed there. No internal gain, low noise, fast, cheap.
- Avalanche photodiode (APD): high reverse bias causes impact ionisation, giving internal gain (M ≈ 10–100). More sensitive, used for long links, but needs high voltage and adds excess noise.
| LED / PIN | LD / APD | |
|---|---|---|
| Cost | Low | High |
| Speed | Lower | Higher |
| Use | Short, multimode | Long-haul, single-mode |
- Asked 3 times
- 2080 Asoj · 6 marks
- 2073 Bhadra · 8 marks
- 2072 Asoj · 8 marks
What are the various elements of an optical communication system? Explain each element in brief.
Answer
An optical communication system sends information as modulated light through an optical fibre. Its main elements are the transmitter (electrical-to-optical conversion), the channel (fibre, with connectors, splices and repeaters) and the receiver (optical-to-electrical conversion).
Block diagram
+--------+ +-----------+ +---------+
| Info |-->| Electrical|-->| Optical |
| source | | Tx (drive)| | source |
+--------+ +-----------+ +----+----+
|
optical fibre cable v
===[connector]====[splice]====[repeater]===
|
+--------+ +-----------+ +----v----+
| Dest- |<--| Electrical|<--| Photo- |
| ination| | Rx (amp) | | detector|
+--------+ +-----------+ +---------+
1. Information source
Provides the message: voice, video or data. Analog signals are usually digitised (PCM) and multiplexed (TDM) before transmission.
2. Electrical transmitter (drive circuit / modulator)
Encodes and shapes the signal and supplies the drive current that modulates the light source. Most systems use intensity modulation (on–off keying); coherent systems modulate phase as well.
3. Optical source
Converts the electrical signal into light at 850, 1310 or 1550 nm.
- LED: cheap, incoherent, wide spectrum; short multimode links.
- Laser diode: coherent, narrow spectrum, high power and speed; long-haul single-mode links.
4. Source-to-fibre coupler
A lens or direct butt coupling launches the light into the fibre core within its acceptance cone.
5. Optical fibre cable (transmission channel)
A glass core surrounded by a cladding of lower refractive index guides light by total internal reflection. Signal degradations are attenuation (absorption, scattering, bending; ~0.2 dB/km at 1550 nm) and dispersion (pulse spreading). The cable includes strength members and a protective jacket.
6. Connectors and splices
Join fibre lengths and equipment. Splices are permanent (fusion ~0.05 dB, mechanical ~0.2 dB); connectors are demountable (~0.3 dB).
7. Repeaters / optical amplifiers
On long routes the signal is boosted every 50–100 km. A regenerator detects, retimes and retransmits the signal; an optical amplifier (EDFA) amplifies light directly at 1550 nm without conversion.
8. Optical detector
Converts received light back into current. A PIN photodiode (low noise, cheap) or an avalanche photodiode (internal gain, more sensitive) is used.
9. Electrical receiver
A low-noise preamplifier and main amplifier, an equaliser, a filter, and a decision circuit with clock recovery restore the original digital pulses, which are then decoded and demultiplexed.
10. Destination
The user equipment (telephone, computer, TV) that receives the recovered message.
- Asked 2 times
- 2073 Magh · 8 marks
- 2071 Bhadra · 10 marks
Explain the construction, light propagation mechanism and application of different types of optical fiber.
Answer
An optical fibre is a thin cylindrical waveguide of glass or plastic that guides light by total internal reflection. Fibres are classified by refractive index profile (step or graded) and number of modes (single or multi).
Construction (common to all types)
+-------------------------------+
| jacket / buffer (plastic) |
| +-------------------------+ |
| | cladding n2 (125 µm) | |
| | +-------------------+ | |
| | | core n1 > n2 | | |
| | +-------------------+ | |
| +-------------------------+ |
+-------------------------------+
- Core: high-index glass (silica doped with GeO₂) that carries the light.
- Cladding: lower-index glass that confines light by TIR and protects the core surface.
- Buffer coating / jacket: plastic layers for mechanical protection; cables add strength members (Kevlar) and an outer sheath.
Propagation mechanism
Light travels in an optical fibre by total internal reflection (TIR). The core has a higher refractive index (n1) than the cladding (n2). When a ray inside the core meets the core–cladding boundary at an angle (from the normal) greater than the critical angle φc = sin⁻¹(n2/n1), it is totally reflected back into the core with no refraction loss. Repeated TIR guides the ray along the fibre, even around gentle bends. Only rays entering within the acceptance angle θa = sin⁻¹√(n1² − n2²) are guided. Each allowed ray direction is a mode.
1. Step-index multimode fibre
n cladding ______________________
| ____ /\ /\ /\
| | | core ______/__\__/__\__/__\__
|_| |_ r zig-zag rays
- Core diameter 50–200 µm (typically 100 µm), cladding 125–400 µm; index changes abruptly from n1 to n2.
- Many modes travel in zig-zag paths of different lengths, so pulses spread (intermodal dispersion). Bandwidth is low (~20–50 MHz·km).
- Easy to couple light from LEDs; cheap connectors.
- Applications: short data links, sensors, illumination, plastic optical fibre in cars and home networks.
2. Step-index single-mode fibre
cladding ____________________________
core -----------------------------> one straight mode
cladding ____________________________
- Very small core (8–10 µm) and small Δ (~0.3%), so only one mode propagates (V < 2.405).
- No intermodal dispersion; very high bandwidth (tens of GHz·km and more) and lowest loss (~0.2 dB/km at 1550 nm).
- Needs a laser source and precise splicing.
- Applications: long-haul telephone and internet backbones, submarine cables, FTTH, CATV trunks.
3. Graded-index multimode fibre
cladding ____________________________
core _--_ _--_ _--_
_- -_ _- -_ _- -_ curved rays
cladding ____________________________
- Core index highest at the axis and falls gradually (near parabolic) to n2 at the cladding; core 50 or 62.5 µm, cladding 125 µm.
- Rays bend smoothly (helical/sinusoidal paths). Outer rays travel longer paths but in lower-index glass at higher speed, so all modes arrive nearly together; modal dispersion is much lower (bandwidth ~ 1 GHz·km).
- Applications: LANs, data centres, campus backbones and medium-distance links with LED or VCSEL sources.
Comparison
| Feature | SI multimode | SI single-mode | GI multimode |
|---|---|---|---|
| Core size | 50–200 µm | 8–10 µm | 50/62.5 µm |
| Index profile | Step | Step | Parabolic |
| Modes | Many | One | Many |
| Dispersion | High | Lowest | Low |
| Bandwidth | Lowest | Highest | Medium |
| Source | LED | Laser | LED/laser |
| Cost of system | Low | High | Medium |
- Asked 2 times
- 2077 Chaitra · 2+6 marks
- 2075 Bhadra · 2+5 marks
List out different optical sources and explain the losses in optical fiber briefly.
Answer
Optical sources
Light sources used in optical fibre communication:
- Light emitting diode (LED): surface-emitting (Burrus) LED and edge-emitting LED; incoherent, broad spectrum, cheap.
- Laser diode (LD): Fabry–Perot laser, distributed feedback (DFB) laser, distributed Bragg reflector (DBR) laser, and vertical-cavity surface-emitting laser (VCSEL); coherent, narrow spectrum, high power and speed.
- Other lasers used in special systems: fibre lasers (erbium-doped) and Nd:YAG solid-state lasers.
Losses in optical fibre
Attenuation is the decrease of optical power along the fibre, measured in dB/km:
α (dB/km) = (10 / L) · log(Pin / Pout)
Modern silica fibre has about 0.35 dB/km at 1310 nm and 0.2 dB/km at 1550 nm.
1. Absorption losses
Light energy is absorbed by the glass and converted to heat.
- Intrinsic absorption: by the pure silica itself; electronic absorption in the ultraviolet and molecular vibration absorption in the infrared (beyond ~1.6 µm). It sets the long-wavelength limit.
- Extrinsic absorption: by impurities such as transition metal ions (Fe, Cu, Cr) and especially OH⁻ (water) ions, which give absorption peaks near 950, 1240 and 1380 nm.
- Atomic defect absorption: defects in the glass structure, e.g. caused by radiation.
2. Scattering losses
Light is scattered out of the guided path by small non-uniformities in the glass.
- Rayleigh scattering: by density and composition fluctuations much smaller than λ; loss ∝ 1/λ⁴. It is the dominant loss at short wavelengths and the main reason long links use 1310 and 1550 nm.
- Mie scattering: by larger imperfections (core–cladding irregularities, bubbles), reduced by careful manufacture.
- Non-linear scattering (stimulated Brillouin and Raman) appears only at high optical power.
3. Bending (radiation) losses
- Macrobending: bends with radius much larger than the fibre diameter (e.g. cable round a corner). On the outer side of a sharp bend the ray angle falls below the critical angle, so light leaks into the cladding. Loss rises steeply once the bend radius goes below a critical value.
- Microbending: small random bends of the fibre axis (µm scale) caused by uneven cabling pressure or temperature changes. They couple light from guided to radiating modes.
4. Coupling, splice and connector losses
Power is lost where light enters the fibre and where fibres are joined, because of lateral, angular and gap misalignment, core size or NA mismatch, and Fresnel reflection at end faces (about 0.05–0.5 dB per joint).
- Asked 2 times
- 2081 Chaitra · 2+5 marks
- 2078 Chaitra · 3+5 marks
What is an acceptance angle? Derive the expression for numerical aperture (NA) of a stepped index optical fiber with neat diagram.
Answer
Acceptance angle
The acceptance angle θa is the maximum angle, measured from the fibre axis, at which a ray can enter the fibre core and still be guided by total internal reflection. Rays inside the cone of half-angle θa (the acceptance cone) are trapped; rays outside it are refracted into the cladding and lost.
Derivation of NA for a step-index fibre
n0 (air) | n2 cladding
|-----------------------------
θa ray | φ \ / φ (φ ≥ φc: TIR)
---------------->*-------\/----------------->
(acceptance) |θ1 n1 core axis
|-----------------------------
| n2 cladding
Let a ray enter the core from a medium of index n0 at angle θa to the axis, refract to angle θ1 inside the core, and meet the core–cladding boundary at angle φ (measured from the normal), where φ = 90° − θ1.
- Snell's law at the input face:
n0 sin θa = n1 sin θ1
- For total internal reflection at the core–cladding boundary, φ must be at least the critical angle φc, where
sin φc = n2 / n1
- The limiting (maximum) entry angle θa corresponds to φ = φc, i.e. θ1 = 90° − φc:
n0 sin θa = n1 sin(90° − φc) = n1 cos φc
= n1 √(1 − sin²φc)
= n1 √(1 − n2²/n1²)
= √(n1² − n2²)
- The numerical aperture is defined as NA = n0 sin θa, so
NA = n0 sin θa = √(n1² − n2²)
- With the relative index difference Δ = (n1² − n2²)/(2n1²) ≈ (n1 − n2)/n1:
NA = n1 √(2Δ)
For launching from air (n0 = 1): θa = sin⁻¹(NA), and the full acceptance cone angle is 2θa.
Example: n1 = 1.48, n2 = 1.46 gives NA = √(1.48² − 1.46²) = 0.2425 and θa = 14.0°.
- Asked 2 times
- 2076 Bhadra · 2+5 marks
- 2070 Bhadra · 10 marks
Where are optical fibres most widely used? Explain the various advantages and disadvantages of optical fibers over metal wire communication.
Answer
Where optical fibres are used
Optical fibres are used most widely in telecommunication networks: long-distance and international trunk links, submarine cables, national backbones (e.g. the optical fibre backbone of Nepal Telecom and NEA), mobile tower backhaul, and fibre-to-the-home (FTTH) internet. Other uses are LANs and data centres, cable TV, military and aircraft systems (EMI immunity), medical endoscopes, industrial sensors, and illumination.
Advantages over metal wire communication
| Advantage over copper wire | Reason |
|---|---|
| Huge bandwidth | Optical carrier ~200 THz; Tb/s on one fibre |
| Low loss | ~0.2 dB/km, repeaters every 80–100 km (copper: ~2 km) |
| Immune to EMI and crosstalk | Light is not affected by electrical noise or lightning |
| Electrical isolation | Glass is an insulator; no ground loops or sparks |
| Small size, light weight | Fibre is ~125 µm; a cable is much lighter than copper |
| Signal security | No radiation; tapping is hard to do undetected |
| Raw material | Silica (sand) is plentiful; copper is costly |
| Long life, corrosion-free | Glass does not corrode |
Disadvantages compared to metal wire
| Disadvantage | Reason |
|---|---|
| Costly installation | Fusion splicers, OTDR and skilled staff needed |
| Fragile | Glass breaks if bent tightly or crushed |
| Difficult joining | Micron-level alignment for splices/connectors |
| No power feeding | Cannot carry DC power to repeaters/terminals like copper |
| Optical–electrical conversion | Extra transmitter and receiver circuits needed |
| Repair is harder | Locating and fixing a cut needs special tools |
Overall, the very high bandwidth and low loss make fibre the preferred medium for backbones, while copper remains useful for short drops and for carrying power.
- 2081 Chaitra · 3+5 marks
Explain the working principle of optical fiber communication system. What are the advantages and disadvantages of optical fiber over wireless communication system?
Answer
Working principle
An optical fibre communication system transmits information as modulated light through a glass fibre. The electrical message signal is converted to light by a source, guided along the fibre by total internal reflection, and converted back to an electrical signal at the receiver.
Message -> Driver/ -> LED or
(input) modulator laser
|
====== fibre (TIR, repeaters) ======
|
Output <- Amplifier, <- PIN/APD
decision detector
- The information signal (usually digital) drives the optical source (LED or laser diode), which varies its light intensity with the data.
- The light is coupled into the fibre within its acceptance angle and travels by TIR. Splices, connectors and repeaters/optical amplifiers (EDFA) along the route maintain the signal.
- At the far end a photodetector (PIN or APD) converts the light into current.
- The receiver amplifies, equalises and regenerates the signal and passes it to the user.
Advantages of optical fibre over wireless
| Point | Optical fibre | Wireless |
|---|---|---|
| Bandwidth | Tb/s per fibre | Limited by allocated spectrum |
| Interference | Immune to EMI and RF interference | Suffers interference, noise |
| Fading/weather | No fading, rain or multipath effects | Multipath fading, rain attenuation |
| Security | Very hard to tap | Signal open to interception |
| Spectrum licence | Not required | Required, and costly |
| Reliability | Stable, predictable loss | Varies with path and weather |
| Latency | Low, constant | Can vary |
Disadvantages of optical fibre compared to wireless
- No mobility: users must be at a fixed point; wireless supports mobile and roaming users.
- High installation cost and time: trenching, ducts, poles and right-of-way, especially over mountains and rivers (a real problem in hilly Nepal); wireless can be set up quickly.
- Vulnerable to cuts: road construction and landslides often break cables; wireless has no cable to cut along the path.
- Hard to reach remote or scattered users: satellite or microwave is cheaper for sparse rural areas.
- Repair needs splicing equipment and trained staff.
- Broadcasting to many receivers is easy over radio but needs a separate fibre path to each user.
- 2080 Chaitra · 5+6 marks
Explain the principle of light propagation through a fiber with necessary diagram. Derive the relationship between numerical aperture and acceptance angle.
Answer
Principle of light propagation in an optical fibre
Light travels in an optical fibre by total internal reflection (TIR). The core has a higher refractive index (n1) than the cladding (n2). When a ray inside the core meets the core–cladding boundary at an angle (from the normal) greater than the critical angle φc = sin⁻¹(n2/n1), it is totally reflected back into the core with no refraction loss. Repeated TIR guides the ray along the fibre, even around gentle bends.
cladding n2 ___________________________________
\ /\ /\ /
core n1 ray \ φ / \ / \ /
----->\ / \ / \ / --->
\ / \ / \ /
cladding n2 ____\/________\/________\/_________
φ > φc at every reflection (TIR)
Conditions for TIR:
- Light must travel from a denser medium to a rarer medium (n1 > n2).
- The angle of incidence at the core–cladding boundary must exceed the critical angle φc = sin⁻¹(n2/n1).
To meet condition 2, light must enter the fibre end within the acceptance angle θa. Rays entering at larger angles strike the boundary below φc, partly refract into the cladding and are lost after a short distance.
In a step-index fibre the rays follow zig-zag paths; in a graded-index fibre the index decreases gradually from the axis, so the rays bend continuously (curved paths) instead of reflecting sharply. In a single-mode fibre the core is so small that only one mode, travelling almost along the axis, exists.
Relationship between numerical aperture and acceptance angle
n0 (air) | n2 cladding
|-----------------------------
θa ray | φ \ / φ (φ ≥ φc: TIR)
---------------->*-------\/----------------->
(acceptance) |θ1 n1 core axis
|-----------------------------
| n2 cladding
Let a ray enter the core from a medium of index n0 at angle θa to the axis, refract to angle θ1 inside the core, and meet the core–cladding boundary at angle φ (measured from the normal), where φ = 90° − θ1.
- Snell's law at the input face:
n0 sin θa = n1 sin θ1
- For total internal reflection at the core–cladding boundary, φ must be at least the critical angle φc, where
sin φc = n2 / n1
- The limiting (maximum) entry angle θa corresponds to φ = φc, i.e. θ1 = 90° − φc:
n0 sin θa = n1 sin(90° − φc) = n1 cos φc
= n1 √(1 − sin²φc)
= n1 √(1 − n2²/n1²)
= √(n1² − n2²)
- The numerical aperture is defined as NA = n0 sin θa, so
NA = n0 sin θa = √(n1² − n2²)
- With the relative index difference Δ = (n1² − n2²)/(2n1²) ≈ (n1 − n2)/n1:
NA = n1 √(2Δ)
For launching from air (n0 = 1): θa = sin⁻¹(NA), and the full acceptance cone angle is 2θa.
So the NA is the sine of the acceptance angle (for launching from air) and depends only on the core and cladding indices, not on the fibre diameter. A larger NA gives a wider acceptance cone and easier coupling, but more modes and more modal dispersion.
Example: n1 = 1.50, n2 = 1.48: NA = √(2.25 − 2.1904) = 0.244, θa = 14.1°.
- 2079 Chaitra · 2+2+2 marks
Explain about the principle of light propagation in optical fiber. Define numerical aperture and acceptance angle with necessary figures and mathematical expressions.
Answer
Principle of light propagation in an optical fibre
Light travels in an optical fibre by total internal reflection (TIR). The core has a higher refractive index (n1) than the cladding (n2). When a ray inside the core meets the core–cladding boundary at an angle (from the normal) greater than the critical angle φc = sin⁻¹(n2/n1), it is totally reflected back into the core with no refraction loss. Repeated TIR guides the ray along the fibre, even around gentle bends.
cladding n2 ___________________________________
\ /\ /\ /
core n1 ray \ φ / \ / \ /
----->\ / \ / \ / --->
\ / \ / \ /
cladding n2 ____\/________\/________\/_________
φ > φc at every reflection (TIR)
Conditions for TIR:
- Light must travel from a denser medium to a rarer medium (n1 > n2).
- The angle of incidence at the core–cladding boundary must exceed the critical angle φc = sin⁻¹(n2/n1).
To meet condition 2, light must enter the fibre end within the acceptance angle θa. Rays entering at larger angles strike the boundary below φc, partly refract into the cladding and are lost after a short distance.
Acceptance angle
The acceptance angle θa is the maximum angle to the fibre axis at which light can enter and still be guided by TIR. Rays within the cone of half-angle θa are guided.
\
θa \ acceptance cone
axis ------*==========================
/ core
/
θa = sin⁻¹( √(n1² − n2²) / n0 )
Numerical aperture
Numerical aperture (NA) measures the light-gathering ability of the fibre. It is the sine of the acceptance angle times the index of the outside medium:
NA = n0 sin θa = √(n1² − n2²) ≈ n1 √(2Δ)
Δ = (n1 − n2)/n1
(This follows from Snell's law n0 sin θa = n1 cos φc and sin φc = n2/n1.)
Typical values: 0.1–0.15 for single-mode fibre, 0.2–0.3 for multimode fibre.
Example: n1 = 1.48, n2 = 1.46: NA = 0.2425, θa = 14.0°.
- 2079 Chaitra · 3 marks
Write a short note on splicing.
Answer
Splicing is the permanent (or semi-permanent) joining of two optical fibres end to end so that light passes from one to the other with minimum loss. It is needed because cables come in limited lengths (2–6 km) and for repairs.
Types of splicing
- Fusion splicing: the cleaned and cleaved fibre ends are aligned (often automatically using cameras) and fused by an electric arc, so the glass melts into one piece.
- Loss: about 0.02–0.1 dB; very low back-reflection; strong and stable.
- Needs an expensive fusion splicer; used in backbone and long-haul links.
- Mechanical splicing: fibre ends are held in alignment by a mechanical structure such as a V-groove, an elastomeric splice or a precision tube, with index-matching gel to reduce reflection.
- Loss: about 0.1–0.5 dB; quick, cheap tools; used for field repairs and drops.
Causes of splice loss
- Extrinsic: lateral (offset) misalignment, longitudinal gap, angular tilt, poor end-face cleave.
- Intrinsic: mismatch in core diameter, NA, index profile or core ellipticity between the two fibres.
- Fresnel reflection at a glass–air gap (≈ 0.17 dB per surface) in mechanical joints without gel.
Splices are protected with heat-shrink sleeves and kept in splice trays inside joint closures.
- 2079 Chaitra · 3 marks
Write a short note on optical fiber loss.
Answer
Optical fibre loss (attenuation) is the reduction of optical power as light travels along a fibre. It decides the maximum distance between repeaters.
α (dB/km) = (10/L) · log10(Pin/Pout)
Typical values: ~2.5 dB/km at 850 nm, ~0.35 dB/km at 1310 nm, ~0.2 dB/km at 1550 nm.
Main causes
- Absorption: intrinsic (by silica in UV and IR bands) and extrinsic (by impurities, especially OH⁻ ions with a peak near 1380 nm, and metal ions like Fe and Cu).
- Scattering: mainly Rayleigh scattering from tiny density variations; loss ∝ 1/λ⁴, so it falls at longer wavelengths. Mie scattering from larger defects.
- Bending losses: macrobends (sharp curves of the cable) and microbends (tiny random axis deviations) let light escape into the cladding.
- Coupling/joint losses: at source coupling, splices and connectors due to misalignment and Fresnel reflection.
The combination gives the low-loss "windows" at 850, 1310 and 1550 nm used in practice.
- 2078 Chaitra · 3+4 marks
Explain the block diagram of optical communication system. State advantages and applications of optical fiber compared to microwave communication link.
Answer
Block diagram of an optical communication system
+--------+ +----------+ +---------+
| Info |-->| Electr. |-->| Optical |
| source | | Tx/driver| | source |
+--------+ +----------+ +----+----+
| coupler
==== optical fibre, splices, ====
==== repeaters / EDFA ====
|
+--------+ +----------+ +----v----+
| Dest. |<--| Electr. |<--| Optical |
| | | Rx/decode| | detector|
+--------+ +----------+ +---------+
- Electrical transmitter: encodes the message and drives the source (intensity modulation).
- Optical source: LED or laser diode converts current to light at 850/1310/1550 nm.
- Fibre channel: guides light by TIR; connectors, splices and repeaters/optical amplifiers along the route.
- Optical detector: PIN or avalanche photodiode converts light to current.
- Electrical receiver: amplifies, equalises and regenerates the signal for the user.
Advantages of optical fibre over a microwave link
| Point | Optical fibre | Microwave link |
|---|---|---|
| Bandwidth | Tb/s | Hundreds of Mb/s to a few Gb/s |
| Repeater spacing | 80–100 km (more with EDFA) | 40–60 km, needs LOS towers |
| Weather | Unaffected | Rain fade, multipath, ducting |
| Interference | None | Co-channel interference |
| Spectrum licence | Not needed | Needed |
| Security | High | Lower (open air) |
Applications
- Long-haul trunks, national backbones and submarine cables
- Mobile (3G/4G/5G) tower backhaul and FTTH broadband
- LANs and data centres
- Cable TV distribution
- Power utilities (OPGW on transmission lines), railways, military and sensing
- 2077 Chaitra · 7 marks
Explain the contribution of microscopic and macroscopic fiber bends towards the bending losses in optical fiber.
Answer
Bending loss (radiation loss) occurs when a fibre is bent, so some light no longer meets the condition for total internal reflection and leaks out into the cladding and jacket. It has two contributions: macroscopic bends and microscopic bends.
Macroscopic bending (macrobending)
Macrobends are bends whose radius is large compared with the fibre diameter, e.g. when a cable goes round a corner or is coiled in a tray.
outer side: wave must travel faster
____________
/ ___________ \ <- light radiates out
/ / \ \
=====/ / \ \=====
=======/ bend R \======
- In a bend, the part of the modal field on the outer side would have to travel faster than light in the cladding to keep up with the wavefront. It cannot, so this energy radiates away.
- In ray terms, the angle of incidence at the outer boundary drops below the critical angle, so rays refract out.
- The loss is small for large radius but rises exponentially as the radius falls below a critical radius. For multimode fibre:
Rc ≈ 3·n1²·λ / ( 4π·(n1² − n2²)^(3/2) )
- Higher-order modes (near cut-off) are lost first. Loss is larger at longer wavelengths and for small Δ or small NA. Single-mode fibre at 1550 nm is especially sensitive.
- Control: respect the minimum bend radius (typically 10–20 times the cable diameter), use bend-insensitive fibre (ITU-T G.657) in FTTH.
Microscopic bending (microbending)
Microbends are small, repetitive, random bends of the fibre axis with amplitudes of a few µm and periods of about a millimetre.
core axis: ~^~v~^~~v~^~v~~^~ (exaggerated)
- Causes: uneven pressure from the cable jacket or a rough surface, cabling and winding stress, and differential thermal contraction of the coating and glass at low temperature.
- Each microbend couples power from guided modes into higher-order and radiating (cladding) modes, which are then lost. The loss is spread along the whole length.
- Microbending loss is greater for fibres with small NA and for multimode fibres with many near cut-off modes.
- Control: a soft primary coating and a harder secondary coating (dual coating), loose-tube cable design so the fibre is not stressed, and fibres with higher Δ.
Comparison
| Feature | Macrobend | Microbend |
|---|---|---|
| Radius / size | mm to cm curvature | µm-scale axis deviations |
| Cause | Cable routing, coiling | Jacket pressure, thermal stress |
| Location | At the bend | Distributed along the fibre |
| Remedy | Keep R above minimum | Good coating and cable design |
- 2076 Bhadra · 5+2 marks
Explain the construction and propagation mechanism and application of different types of optical fibers. Define numerical aperture.
Answer
An optical fibre has a high-index core (n1) surrounded by a lower-index cladding (n2) and a protective plastic buffer/jacket. Light is guided by total internal reflection at the core–cladding boundary for rays entering within the acceptance angle. Fibres are of three main types.
1. Step-index multimode fibre
- Core 50–200 µm, cladding 125–400 µm; index changes in one step from n1 to n2.
- Many modes travel in zig-zag paths of different lengths, causing large modal dispersion and low bandwidth.
- Easy coupling with LEDs; cheap.
- Applications: short links, sensors, illumination, plastic fibre in cars.
2. Step-index single-mode fibre
- Core 8–10 µm, cladding 125 µm, small Δ; only one mode propagates almost along the axis.
- No modal dispersion; highest bandwidth and lowest loss (~0.2 dB/km at 1550 nm). Needs laser sources.
- Applications: long-haul backbones, submarine cables, FTTH.
3. Graded-index multimode fibre
- Core index is maximum at the axis and decreases (nearly parabolically) towards the cladding; core 50/62.5 µm.
- Rays follow curved (sinusoidal) paths. Outer rays travel further but faster, so modal dispersion is greatly reduced.
- Applications: LANs, data centres, campus networks.
Step MM Single mode Graded MM
n1 |~~~~~| n1 |~| n1 /~~\
n2_| |_ n2_| |_ n2 _/ \_
zig-zag straight curved
Numerical aperture
Numerical aperture (NA) is the light-gathering ability of a fibre, equal to the sine of the acceptance angle (for launching from air):
NA = n0 sin θa = √(n1² − n2²) ≈ n1 √(2Δ)
For example, n1 = 1.48 and n2 = 1.46 give NA = 0.24 and θa = 14°.
- 2076 Baisakh · 5 marks
Explain the construction and types of optical fiber with necessary diagrams.
Answer
An optical fibre is a thin, flexible glass (or plastic) waveguide that carries light by total internal reflection.
Construction
cross-section side view
.-------------. jacket ================
/ jacket \ buffer ----------------
| .-----------. | clad n2 ________________
| / cladding \ | core n1 ---> light --->
| | .-----. | | clad n2 ________________
| | | core| | | buffer ----------------
| \ '-----' / | jacket ================
\ '-----------' /
'-------------'
- Core: central glass region of higher index n1 (silica doped with germanium) that carries the light.
- Cladding: surrounding glass of slightly lower index n2 (n1 > n2) that makes TIR possible; standard diameter 125 µm.
- Buffer coating: acrylate layer(s) (to 250 µm) that protect against moisture, scratches and microbending.
- Jacket and strength members: outer PVC/PE sheath and Kevlar yarn that give mechanical strength in a cable.
Types of optical fibre
| Type | Core size | Index profile | Modes | Use |
|---|---|---|---|---|
| Step-index multimode | 50–200 µm | Abrupt step | Many, zig-zag | Short links |
| Step-index single mode | 8–10 µm | Abrupt step | One | Long haul |
| Graded-index multimode | 50/62.5 µm | Parabolic | Many, curved | LAN |
SI multimode: /\/\/\/\/\ large modal dispersion
Single mode: ----------> no modal dispersion
GI multimode: ~~~~~~~~~~> low modal dispersion
Fibres may also be classified by material: all-glass (silica), plastic-clad silica, and all-plastic (POF) fibres.
- 2076 Baisakh · 3+5 marks
What are the various features of graded index fiber? Explain the refractive index profile and ray transmission in a multimode graded index fiber.
Answer
Features of graded-index fibre
- The core refractive index is not uniform: it is highest at the axis and decreases gradually to the cladding value n2 at the core boundary.
- Usually multimode, with core diameter 50 or 62.5 µm and cladding 125 µm; NA about 0.2–0.3.
- Rays follow curved, sinusoidal paths instead of zig-zags.
- Much lower intermodal dispersion than step-index multimode fibre, so bandwidth is about 100 times higher (~0.5–2 GHz·km).
- Accepts LED or VCSEL sources and is easier to couple than single-mode fibre.
- More costly to make than step-index multimode fibre.
Refractive index profile
n(r) = n1 · √( 1 − 2Δ (r/a)^α ) for r < a (core)
n(r) = n1 · √( 1 − 2Δ ) = n2 for r ≥ a (cladding)
Δ = (n1² − n2²) / (2n1²) ≈ (n1 − n2)/n1
where a is the core radius and α is the profile parameter: α = 1 triangular, α = 2 parabolic (the usual choice, near optimum), α → ∞ step index.
n
^
n1 | .--.
| / \ parabolic core
| / \
n2 |--' '--- cladding
+-----+---+---+----> r
-a 0 a
Ray transmission in multimode graded-index fibre
cladding ________________________________
.-. .-. .-.
core / \ / \ / \ outer ray
axis ---*-----------------------------> axial ray
\ / \ / \ /
cladding ________________________________
- A ray entering at an angle moves into regions of lower index. By Snell's law it bends continuously away from the normal, until it turns back towards the axis. It never reaches the cladding in the ideal case; the turning is gradual refraction rather than a single sharp reflection.
- So rays travel in sinusoidal (meridional) or helical (skew) paths that repeatedly cross the axis.
- The axial ray takes the shortest path but in the highest-index (slowest) glass. Oblique rays take longer paths but spend most of their time in lower-index (faster) glass, since velocity v = c/n(r).
- With a near-parabolic profile, the extra path length is almost exactly compensated by the higher speed, so all modes arrive at about the same time. Intermodal pulse spreading falls from about L·n1Δ/c (step index) to about L·n1Δ²/(8c) (parabolic), giving much larger bandwidth.
- The local numerical aperture NA(r) = √(n(r)² − n2²) is maximum at the axis and zero at the core edge, so the fibre collects less light than a step-index fibre of the same n1 and n2.
- 2076 Baisakh · 4 marks
Write a short note on light sources in optical communication.
Answer
A light source in optical communication converts the electrical signal into optical power for launching into the fibre. Semiconductor sources are used because they are small, efficient, directly modulated by current and emit at fibre wavelengths (850, 1310, 1550 nm).
Requirements
- Emission wavelength in the low-loss windows of the fibre
- Enough output power and efficient coupling (small emitting area)
- Narrow spectral width (less chromatic dispersion)
- High modulation speed, linearity, long life, low cost
Light emitting diode (LED)
- Forward-biased p-n junction (GaAlAs, InGaAsP); light by spontaneous emission.
- Types: surface-emitting (Burrus) and edge-emitting.
- Incoherent, wide spectrum (30–100 nm), wide beam, low power (µW to a few mW), bandwidth up to ~100 MHz.
- Simple, cheap, reliable, less temperature-sensitive. Used with multimode fibre for short links.
Laser diode (LD)
- Light by stimulated emission in an optical cavity (Fabry–Perot, DFB, DBR, VCSEL) once current exceeds a threshold.
- Coherent, narrow spectrum (1–5 nm; DFB < 0.1 nm), narrow beam, high power (tens of mW), modulation to tens of GHz.
- Costlier, needs temperature and power control. Used with single-mode fibre in long-haul, high-speed links.
| Property | LED | Laser diode |
|---|---|---|
| Emission | Spontaneous | Stimulated |
| Spectral width | Wide | Narrow |
| Power / speed | Low | High |
| Cost | Low | High |
- 2075 Bhadra · 6 marks
Describe with the aid of neat diagram the basic principle of total internal reflection that enables the fiber to work as a "light conduit".
Answer
An optical fibre works as a light conduit (light pipe) because light launched into its core is trapped by total internal reflection (TIR) at the core–cladding boundary and travels along the fibre, even when the fibre bends gently.
Refraction and the critical angle
When light passes from a denser medium n1 to a rarer medium n2 (n1 > n2), Snell's law gives n1 sin φ1 = n2 sin φ2, and the refracted ray bends away from the normal.
(a) φ < φc (b) φ = φc (c) φ > φc
n2 / n2 n2
-------/------- -----------> ---------------
φ / φc / φ \ / φ
n1 / n1 / n1 \ /
/ / \/
partly refracted grazes boundary totally reflected
- (a) For a small angle of incidence, most light refracts into the rarer medium (some is reflected).
- (b) At the critical angle φc, the refracted ray travels along the boundary (φ2 = 90°):
n1 sin φc = n2 sin 90° → sin φc = n2 / n1
- (c) For φ > φc, no refraction is possible and all light is reflected back into the denser medium: total internal reflection. No energy is lost at the reflection.
TIR in a fibre
cladding n2 _________________________________
\ /\ /\ /
core n1 θa \ φ>φc/ \ / \ /
----*------->\ / \ / \ /--->
\ / \ / \ /
cladding n2 _____\/________\/________\/______
- The core (n1) is made slightly denser than the cladding (n2), e.g. n1 = 1.48, n2 = 1.46, giving φc = sin⁻¹(1.46/1.48) = 80.6°.
- A ray entering the end face within the acceptance angle θa is refracted so that it meets the core–cladding boundary at φ > φc. It is totally reflected, crosses the core, meets the opposite boundary at the same angle and is reflected again.
- Thousands of such reflections per metre guide the light along the whole length with very little loss, so the fibre acts as a "light conduit".
- The acceptance angle follows from Snell's law: NA = sin θa = √(n1² − n2²) (here 0.24, θa ≈ 14°).
- If the fibre is bent too sharply, φ falls below φc at the outer wall and light escapes (bending loss). The cladding also keeps the reflecting surface clean, since dust or scratches on a bare core would spoil TIR.
- 2075 Baisakh · 8 marks
Discuss loss or signal attenuation in an optical fiber with respect to absorption, scattering and bending losses.
Answer
Attenuation (signal loss) in an optical fibre is the decrease of optical power with distance. It limits the repeater spacing and is expressed in dB/km:
α (dB/km) = (10/L) · log10(Pin/Pout)
loss
dB/km
|\
| \ Rayleigh OH peak
| \ (1/λ⁴) /\ IR
| \ / \ absorption
| \____ ________/ \_____ /
| \/ \___\__/
+----+--------+--------+--------+---> λ (nm)
850 1310 1380 1550
The low-loss windows at 850, 1310 and 1550 nm come from the combination of the losses below.
1. Absorption losses
Light energy is absorbed by the glass and converted to heat.
- Intrinsic absorption: by the pure silica itself; electronic absorption in the ultraviolet and molecular vibration absorption in the infrared (beyond ~1.6 µm). It sets the long-wavelength limit.
- Extrinsic absorption: by impurities such as transition metal ions (Fe, Cu, Cr) and especially OH⁻ (water) ions, which give absorption peaks near 950, 1240 and 1380 nm.
- Atomic defect absorption: defects in the glass structure, e.g. caused by radiation.
2. Scattering losses
Light is scattered out of the guided path by small non-uniformities in the glass.
- Rayleigh scattering: by density and composition fluctuations much smaller than λ; loss ∝ 1/λ⁴. It is the dominant loss at short wavelengths and the main reason long links use 1310 and 1550 nm.
- Mie scattering: by larger imperfections (core–cladding irregularities, bubbles), reduced by careful manufacture.
- Non-linear scattering (stimulated Brillouin and Raman) appears only at high optical power.
3. Bending (radiation) losses
- Macrobending: bends with radius much larger than the fibre diameter (e.g. cable round a corner). On the outer side of a sharp bend the ray angle falls below the critical angle, so light leaks into the cladding. Loss rises steeply once the bend radius goes below a critical value.
- Microbending: small random bends of the fibre axis (µm scale) caused by uneven cabling pressure or temperature changes. They couple light from guided to radiating modes.
Bending loss rises exponentially when the bend radius falls below a critical radius (for multimode fibre Rc ≈ 3n1²λ / (4π(n1² − n2²)^(3/2))). Microbending is reduced with soft-plus-hard dual coatings and loose-tube cables, macrobending by keeping cables above their minimum bend radius.
Summary
| Loss | Mechanism | Remedy |
|---|---|---|
| Intrinsic absorption | UV/IR absorption by silica | Work at 1.3–1.6 µm |
| Extrinsic absorption | OH⁻, metal ions | Purer, dry glass |
| Rayleigh scattering | Density fluctuations, ∝ 1/λ⁴ | Longer wavelength |
| Macrobending | Sharp curves | Keep radius large |
| Microbending | Small axis deviations | Good coating, cabling |
- 2074 Bhadra · 3+6 marks
Draw the optical fiber communication system. What are the advantages and disadvantages of optical fibers over metal wire communication?
Answer
Optical fibre communication system
+--------+ +----------+ +---------+
| Info |-->| Electr. |-->| LED / |
| source | | Tx/driver| | laser |
+--------+ +----------+ +----+----+
|
===== fibre cable (splices, connectors,
===== repeaters / optical amplifiers)
|
+--------+ +----------+ +----v----+
| Dest. |<--| Amplifier|<--| PIN/APD |
| | | /decision| | detector|
+--------+ +----------+ +---------+
The message is converted to an electrical drive signal, which modulates the intensity of an LED or laser. The light travels through the fibre by total internal reflection, is boosted by repeaters or optical amplifiers on long routes, and is converted back to current by a photodiode. The receiver amplifies and regenerates the signal for the user.
Advantages over metal wire communication
| Advantage over copper wire | Reason |
|---|---|
| Huge bandwidth | Optical carrier ~200 THz; Tb/s on one fibre |
| Low loss | ~0.2 dB/km, repeaters every 80–100 km (copper: ~2 km) |
| Immune to EMI and crosstalk | Light is not affected by electrical noise or lightning |
| Electrical isolation | Glass is an insulator; no ground loops or sparks |
| Small size, light weight | Fibre is ~125 µm; a cable is much lighter than copper |
| Signal security | No radiation; tapping is hard to do undetected |
| Raw material | Silica (sand) is plentiful; copper is costly |
| Long life, corrosion-free | Glass does not corrode |
Disadvantages compared to metal wire
| Disadvantage | Reason |
|---|---|
| Costly installation | Fusion splicers, OTDR and skilled staff needed |
| Fragile | Glass breaks if bent tightly or crushed |
| Difficult joining | Micron-level alignment for splices/connectors |
| No power feeding | Cannot carry DC power to repeaters/terminals like copper |
| Optical–electrical conversion | Extra transmitter and receiver circuits needed |
| Repair is harder | Locating and fixing a cut needs special tools |
- 2074 Bhadra · 4 marks
Write a short note on numerical aperture (NA).
Answer
Numerical aperture (NA) is a measure of the light-gathering (acceptance) ability of an optical fibre. It is defined as the sine of the acceptance angle multiplied by the refractive index of the medium from which light enters.
NA = n0 sin θa
Expression
From Snell's law at the input face, n0 sin θa = n1 cos φc, and from the TIR condition sin φc = n2/n1:
NA = √(n1² − n2²) ≈ n1 √(2Δ), Δ = (n1 − n2)/n1
θa = sin⁻¹(NA) (launch from air, n0 = 1)
\ θa
air \ core n1, cladding n2
---------*===========================
/ acceptance cone 2θa
Points to note
- NA depends only on n1 and n2, not on the fibre size.
- Large NA: easy coupling of light from LEDs, but more modes and more modal dispersion, so less bandwidth.
- Typical values: 0.1–0.15 (single mode), 0.2–0.3 (multimode).
- NA also enters the V-number V = (2πa/λ)·NA, which fixes the number of modes (single mode if V < 2.405).
Example: n1 = 1.48, n2 = 1.46: NA = √(2.1904 − 2.1316) = 0.2425, θa = 14.0°.
- 2073 Magh · 4 marks
Write a short note on multimode graded index fiber.
Answer
A multimode graded-index (GI) fibre is a fibre whose core refractive index is highest at the axis and decreases gradually (usually parabolically) to the cladding value at the core edge, and which supports many modes.
Index profile
n(r) = n1 √(1 − 2Δ (r/a)^α), r < a
n(r) = n2, r ≥ a
α = 2 (parabolic) is the usual profile. Core diameter is 50 or 62.5 µm, cladding 125 µm.
Ray propagation
cladding ______________________________
.-. .-. .-.
core / \ / \ / \
----*----------------------------->
\ / \ / \ /
cladding ______________________________
Rays bend gradually towards the axis by continuous refraction and follow sinusoidal or helical paths. Rays far from the axis travel longer paths but through lower-index (faster) glass, so they arrive at almost the same time as the axial ray.
Features
- Intermodal dispersion much lower than step-index multimode (bandwidth ~0.5–2 GHz·km versus ~20–50 MHz·km).
- Larger core than single mode, so cheap LED/VCSEL sources and connectors can be used.
- Loss about 2.5 dB/km at 850 nm, 0.6–1 dB/km at 1300 nm.
- Applications: LANs, data centres (OM3/OM4 fibre), campus and building backbones up to a few km.
- 2073 Bhadra · 4+4 marks
Compare optical fiber communication with cable and radio communication systems. Describe numerical aperture (NA) in optical communication system.
Answer
Comparison of optical fibre with cable and radio communication
| Point | Optical fibre | Coaxial / twisted-pair cable | Radio (wireless) |
|---|---|---|---|
| Carrier | Light, ~200 THz | Electrical current | EM waves, kHz–GHz |
| Bandwidth | Very high (Tb/s) | Moderate (MHz–GHz) | Limited by spectrum allotted |
| Attenuation | ~0.2 dB/km | High, rises with frequency | Free space loss, rain, fading |
| Repeater spacing | 80–100 km | 1–5 km | 40–60 km (microwave LOS) |
| EMI / crosstalk | Immune | Affected | Interference, noise |
| Security | Very high | Can be tapped | Easy to intercept |
| Weight / size | Light, thin | Heavy, bulky | No cable |
| Mobility | None | None | Supports mobile users |
| Installation | Costly, needs splicing | Easy | Quick, towers only |
| Licence | Not needed | Not needed | Spectrum licence needed |
In short, fibre is best for high-capacity fixed links, copper cable for short and low-cost connections and power feeding, and radio for mobility, broadcasting and hard-to-reach areas.
Numerical aperture (NA)
Numerical aperture is the light-gathering ability of a fibre, defined as NA = n0 sin θa, where θa is the acceptance angle.
\ θa
n0 \ core n1
-----------*=========================
/ cladding n2 (n2 < n1)
Using Snell's law at the input face and the critical angle at the core–cladding boundary:
n0 sin θa = n1 sin θ1 = n1 cos φc
sin φc = n2/n1
NA = n1 √(1 − n2²/n1²) = √(n1² − n2²) ≈ n1 √(2Δ)
- Only light inside the cone of half-angle θa = sin⁻¹(NA) is guided.
- Larger NA: easier coupling but more modal dispersion.
- Typical NA: 0.1–0.15 for single-mode and 0.2–0.3 for multimode fibre.
- Example: n1 = 1.48, n2 = 1.46 gives NA = 0.2425 and θa = 14°.
- 2072 Magh · 1+4+5 marks
What is an optical fiber? What are the different types of losses you can visualize in optical fiber communication? What are the advantages and disadvantages of multimode fiber over the single mode fiber?
Answer
Optical fibre
An optical fibre is a thin, flexible strand of very pure glass or plastic, with a core of higher refractive index surrounded by a cladding of lower index, that guides light over long distances by total internal reflection.
Types of losses in optical fibre communication
- Absorption loss: intrinsic (UV and IR absorption by silica) and extrinsic (OH⁻ ions, peak near 1380 nm; metal ions such as Fe, Cu).
- Scattering loss: mainly Rayleigh scattering from microscopic density variations, ∝ 1/λ⁴; also Mie scattering from larger defects.
- Bending loss: macrobending (sharp curves of the cable) and microbending (small random axis deviations from cabling stress).
- Coupling loss: at the source-to-fibre and fibre-to-detector interfaces, due to NA and area mismatch.
- Splice and connector loss: lateral, angular and gap misalignment, end-face quality, core/NA mismatch and Fresnel reflection.
Attenuation is measured as α = (10/L)·log(Pin/Pout) dB/km. (Dispersion does not remove power but spreads pulses, which limits bandwidth.)
Multimode fibre compared with single mode fibre
Advantages of multimode fibre:
- Large core (50–200 µm) and large NA make it easy to launch light, so cheap LEDs and VCSELs can be used.
- Connectors, splicing and alignment are easier and cheaper; tolerances are relaxed.
- Lower overall system cost for short links (LANs, buildings, data centres).
- Several fibres can be joined with simple tools in the field.
Disadvantages of multimode fibre:
- Large intermodal dispersion (different modes arrive at different times), so much lower bandwidth–distance product.
- Higher attenuation (about 2.5–3 dB/km at 850 nm compared with 0.2–0.35 dB/km for single mode).
- Limited to short distances (a few hundred metres to ~2 km at high data rates).
- Not suited to optical amplifiers and DWDM long-haul systems.
| Feature | Multimode | Single mode |
|---|---|---|
| Core diameter | 50–200 µm | 8–10 µm |
| Modes | Many | One |
| Source | LED / VCSEL | Laser diode |
| Bandwidth | Low–medium | Very high |
| Loss | Higher | Lowest |
| Distance | Short | Very long |
| Cost of system | Lower | Higher |
- 2072 Magh · 6 marks
What properties do you think to be considered when an optical fiber is to be selected for any communication system?
Answer
Choosing an optical fibre for a link means matching the fibre's optical, mechanical and cost properties to the needs of the system: distance, bit rate, wavelength, source type and installation environment.
1. Attenuation (loss)
- Loss in dB/km sets the maximum distance between repeaters.
- Silica fibre has low-loss windows at about 850 nm (2–3 dB/km), 1310 nm (~0.35 dB/km) and 1550 nm (~0.2 dB/km).
- Long-haul links need low-loss single mode fibre at 1310/1550 nm.
2. Bandwidth and dispersion
- Dispersion spreads pulses and limits the bit rate × distance product.
- Step-index multimode has large modal dispersion (about 20–50 MHz·km); graded-index is better (about 0.5–1 GHz·km); single mode has no modal dispersion (tens of GHz·km and more).
- For high data rates, choose single mode fibre, and dispersion-shifted fibre if operating at 1550 nm.
3. Numerical aperture and core size
- A large NA and core (50–62.5 µm, multimode) collect light easily from cheap LEDs and make splicing easier.
- A small core (8–10 µm, single mode) needs a laser and precise connectors.
4. Operating wavelength and source/detector match
- The fibre must have low loss and low dispersion at the wavelength of the available LED/laser and photodiode.
5. Mechanical properties
- Tensile strength, bend radius (macro- and micro-bending loss), and cable protection (buffer, strength member, jacket).
- The cable type depends on the route: duct, aerial, direct burial or undersea.
6. Environmental properties
- Temperature range, humidity, water ingress, and resistance to radiation and chemicals.
7. Ease of handling and cost
- Cost of fibre, connectors and splicing, and of matching transmitters.
- Multimode plus LED is cheap for LANs and short links; single mode plus laser is used for long-haul and high-capacity links.
8. Future upgrade
- Single mode fibre allows later upgrade to higher bit rates and WDM without laying new cable.
| Application | Usual choice |
|---|---|
| Building LAN, < 2 km | Graded-index multimode, 850/1310 nm |
| Metro / long-haul | Single mode, 1310/1550 nm |
| Very short, low cost | Step-index multimode or plastic fibre |
- 2072 Asoj · 4+2 marks
Compare the configuration of different types of fiber. Mention the advantages of single mode fiber.
Answer
Optical fibres are classified by the refractive-index profile of the core and the number of modes they carry. The three standard configurations are step-index multimode, graded-index multimode and step-index single mode.
Comparison of fibre configurations
Step-index MM Graded-index MM Single mode
n n n
| ___ | /\ | _
|_| |_ |__/ \__ |__| |__
core 50-200um core 50-85um core 8-10um
zig-zag rays curved rays one axial path
| Feature | Step-index MM | Graded-index MM | Single mode |
|---|---|---|---|
| Core index | Uniform | Falls parabolically from axis | Uniform |
| Core diameter | 50–200 µm | 50–85 µm | 8–10 µm |
| Cladding diameter | 125–400 µm | 125 µm | 125 µm |
| Ray path | Zig-zag, sharp reflections | Smooth curved (helical) | Nearly straight along axis |
| Number of modes | Many (hundreds) | Many, but equalised delay | One () |
| Modal dispersion | Large | Small | None |
| Bandwidth | ~20–50 MHz·km | ~0.5–1 GHz·km | > 10 GHz·km |
| Source | LED | LED or laser | Laser |
| Cost / handling | Cheapest, easy | Moderate | Fibre cheap, but costly connectors and sources |
| Use | Short links | LANs, campus | Long-haul, high-speed |
In graded-index fibre, rays travelling far from the axis go through lower index (faster) glass, so they arrive almost together with the axial ray; this is why modal dispersion is much lower than in step-index fibre.
Advantages of single mode fibre
- No modal (intermodal) dispersion, so very high bandwidth (Gb/s to Tb/s with WDM).
- Lowest attenuation (about 0.2–0.35 dB/km at 1550/1310 nm), so repeater spacing of 100 km or more.
- Best for long-haul, undersea and backbone links.
- Supports future upgrades (higher bit rate, DWDM) on the same cable.
- Output beam is well defined, suitable for coherent and high-precision systems.
- 2072 Asoj · 3 marks
Write a short note on NA for meridional rays in optical fiber.
Answer
Numerical aperture (NA) of a fibre measures its light-gathering ability. For meridional rays (rays that pass through the fibre axis and stay in one plane), it is the sine of the maximum acceptance angle multiplied by the index of the outside medium.
n0 cladding n2
\ ________________________
th0\/ phi /\ /\
----*-----/--\----/--\---- axis core n1
________________________
A ray entering at angle is refracted to (Snell's law ) and meets the core–cladding boundary at . It is guided only if where . At the limit:
where is the relative index difference.
Key points
- NA depends only on and , not on core size.
- Rays inside the cone of half-angle are guided; others leak into the cladding.
- Typical values: 0.1–0.15 (single mode), 0.2–0.3 (multimode).
- Example: , gives NA and in air.
- Skew rays (not passing through the axis) can be accepted at slightly larger angles, so the meridional NA is the standard, conservative value.
- 2071 Magh · 3+5 marks
Explain the block diagram of optical communication and list out the advantages and disadvantages of it over cable communication.
Answer
An optical fibre communication (OFC) system converts an electrical message into light, sends it through a glass fibre, and converts it back to an electrical signal at the receiver.
Block diagram
Message +-----------+ +---------+ +--------+
input --->| Encoder / |-->| Driver |-->| Light |
| modulator | | circuit | | source |
+-----------+ +---------+ +--------+
LED/laser |
v
~~~~~~~~ optical fibre (with splices, connectors)
| and repeaters |
v
Message +-----------+ +---------+ +--------+
output <--| Decoder / |<--| Amp. + |<--| Photo- |
| demod. | | regen. | |detector|
+-----------+ +---------+ +--------+
PIN / APD
- Information source and encoder: the analog or digital message is coded (e.g. PCM) into an electrical signal.
- Driver circuit: supplies the current that modulates the light source.
- Light source: an LED or laser diode converts the electrical signal into light at 850, 1310 or 1550 nm.
- Source-to-fibre coupler: launches the light into the fibre within its acceptance angle.
- Optical fibre (channel): carries light by total internal reflection; splices, connectors and repeaters or optical amplifiers (EDFA) are added on long routes.
- Photodetector: a PIN or avalanche photodiode converts light back to current.
- Amplifier, regenerator and decoder: amplify, reshape and decode the signal to recover the message.
Advantages over cable (copper) communication
- Huge bandwidth (THz range), so very high data rates.
- Very low loss (0.2 dB/km), so long repeater spacing.
- Immune to electromagnetic interference, lightning and crosstalk.
- No electrical hazard or sparking; safe in explosive areas.
- Small size and light weight.
- Difficult to tap, so more secure.
- Made from silica (sand), a plentiful raw material.
Disadvantages
- Splicing and connectors need costly tools and skilled workers.
- Fibre is fragile; bending beyond a small radius causes loss or breakage.
- Cannot carry electrical power to remote repeaters directly.
- Costly optical sources, detectors and test equipment.
- Electrical–optical conversion is needed at each end.
- 2071 Magh · 6 marks
Describe the following terms: Numerical Aperture and Dispersion in Optical Fiber.
Answer
Numerical aperture (NA)
Numerical aperture is the light-gathering ability of a fibre. It is the sine of the acceptance angle (the largest angle to the axis at which an incoming ray is still guided by total internal reflection) multiplied by the index of the launching medium:
acceptance cone
\ cladding n2
\ ____________________
---*<)th_a core n1 -->
/ ____________________
/
- Rays inside the cone of half-angle are guided; rays outside it are lost into the cladding.
- A large NA makes it easy to couple light from an LED, but also increases modal dispersion.
- Typical NA: 0.2–0.3 for multimode, about 0.1 for single mode fibre.
- Example: , gives NA and .
Dispersion
Dispersion is the spreading of a light pulse in time as it travels along the fibre. Spread pulses overlap with neighbours (inter-symbol interference), which limits the bit rate and the bandwidth–distance product.
input _|^|_ output _/^^\_ (wider, lower)
Types:
- Intermodal (modal) dispersion: in multimode fibre, different modes travel different path lengths and arrive at different times. For step-index fibre . Reduced by graded-index fibre; absent in single mode fibre.
- Intramodal (chromatic) dispersion: the source emits a range of wavelengths that travel at different speeds.
- Material dispersion: refractive index of silica changes with wavelength.
- Waveguide dispersion: the mode's speed depends on core size relative to wavelength.
- Chromatic dispersion is near zero at about 1310 nm in standard fibre.
- Polarization mode dispersion: the two polarizations travel at slightly different speeds due to fibre asymmetry; important at very high bit rates.
Dispersion is reduced by using single mode fibre, narrow-linewidth lasers, operation near 1310 nm, or dispersion-shifted fibre at 1550 nm.
- 2070 Magh · 3+7 marks
What are advantages and disadvantages of multimode fiber over single mode fiber? Derive expression of Numerical Aperture of a stepped index optical fiber.
Answer
Multimode fibre compared with single mode fibre
Advantages of multimode fibre
- Larger core (50–200 µm) and larger NA, so light can be launched from cheap LEDs.
- Easier and cheaper splicing, connectors and alignment.
- Lower total system cost for short links (LANs, buildings, campuses).
- Can be used with low-cost sources at 850 nm.
Disadvantages of multimode fibre
- Large intermodal dispersion, so low bandwidth (MHz·km to about 1 GHz·km).
- Higher attenuation than single mode fibre.
- Short maximum link length; not suitable for long-haul or high-speed WDM systems.
Derivation of NA of a step-index fibre
Consider a step-index fibre with core index , cladding index (), and outside medium index (air, ). A meridional ray enters the core end at angle to the axis.
n0 | cladding n2
ray \ |__________________
\th0| /\phi
axis ---*--+--/--\----------- core n1
|th1 \
|__________________
cladding n2
Step 1: refraction at the input face. By Snell's law:
Step 2: angle at the core–cladding boundary. The refracted ray meets the boundary at angle to the normal. From the right-angled triangle:
Step 3: condition for total internal reflection. The ray is guided only if , where the critical angle is given by
Step 4: maximum acceptance angle. The largest input angle corresponds to :
Step 5: definition of NA.
For air, and .
Step 6: in terms of relative index difference. With (since ):
Example: , :
NA depends only on the refractive indices, not on the core diameter. The full acceptance cone angle is .
- 2070 Magh · 6 marks
Briefly explain the advantage of optical fiber over metalled wire.
Answer
An optical fibre carries information as light through glass, while a metallic wire (twisted pair or coaxial cable) carries it as electric current. Because of this, fibre has many advantages.
- Very large bandwidth: optical carrier frequencies are about Hz, so a fibre can carry Tb/s with WDM. Copper is limited to MHz–GHz.
- Low attenuation: about 0.2 dB/km at 1550 nm, so repeaters can be 100 km or more apart. Coaxial cable needs repeaters every few km.
- Immunity to electromagnetic interference: glass is a dielectric, so motors, power lines and lightning do not induce noise. Useful in factories and along power lines.
- No crosstalk: light does not leak from one fibre to another in the same cable.
- Electrical isolation: no ground loops, sparks or shock hazard; safe in explosive or high-voltage areas.
- Security: the fibre does not radiate, so it is hard to tap without being detected.
- Small size and light weight: a fibre cable is much thinner and lighter than a copper cable of the same capacity, so it is easier to install in ducts.
- Low cost raw material: silica (sand) is plentiful, while copper is costly.
- Long life and corrosion resistance: glass does not corrode in moist or chemical environments.
- Flexibility of upgrade: higher bit rates and more wavelengths can be added by changing end equipment only.
| Property | Optical fibre | Metallic wire |
|---|---|---|
| Bandwidth | THz | MHz–GHz |
| Loss | 0.2–3 dB/km | High, rises with frequency |
| EMI / crosstalk | None | Present |
| Weight per channel | Very low | High |
| Security | High | Low |
- 2070 Bhadra · 2+7 marks
What is an acceptance angle? Derive an expression to calculate the acceptance angle.
Answer
Acceptance angle () is the maximum angle, measured from the fibre axis, at which a light ray can enter the fibre end and still be guided along the core by total internal reflection. Rays inside the cone of half-angle (the acceptance cone) are guided; rays outside it are refracted into the cladding and lost.
Derivation
Let the core index be , cladding index (), and the outside medium index (air, ). A meridional ray hits the core end at angle .
n0 | cladding n2
ray \ |____________________
\ | /\ phi
axis ---*-+---/--\------------ core n1
th0 | th1 \
|____________________
cladding n2
Step 1: refraction at the air–core face (Snell's law):
Step 2: geometry. The refracted ray strikes the core–cladding boundary at angle to the normal. Since the normal to the boundary is perpendicular to the axis:
Step 3: condition for TIR. The ray is totally reflected only if , with
As increases, increases and decreases. The limiting ray has and .
Step 4: combine (1), (2) and (3):
Step 5: acceptance angle:
and for air (), .
The quantity is the numerical aperture (NA). Using the relative index difference , , so .
Example: , , in air:
Notes
- depends only on the refractive indices, not on fibre diameter.
- A larger index difference gives a wider acceptance cone (easier coupling) but more modal dispersion.
- 2069 Bhadra · 12 marks
Explain the advantages and disadvantages of optical fibre communication over the metallic wire communication system.
Answer
Optical fibre communication (OFC) sends information as modulated light through a thin glass or plastic fibre by total internal reflection. Metallic wire systems (twisted pair, coaxial cable) send it as electrical current. OFC has replaced copper in most long-distance and high-capacity links, but it also has some drawbacks.
Data -> Driver -> LED/Laser ~~~fibre~~~ Photodiode -> Amp -> Data
Data -> Line driver ======copper wire====== Receiver -> Data
Advantages of OFC over metallic wire
- Enormous bandwidth: optical carriers are at about Hz. A single mode fibre with WDM carries many Tb/s, while a coaxial cable carries only up to a few GHz.
- Low transmission loss: modern silica fibre has about 0.2 dB/km loss at 1550 nm and 0.35 dB/km at 1310 nm. Repeater spacing of 100–200 km is possible, compared with 1–5 km for coaxial cable. Fewer repeaters reduce cost and failures.
- Immunity to interference: glass is a dielectric, so the signal is not affected by EMI, RFI, lightning or nearby power lines. Fibre can run inside power cables (OPGW).
- No crosstalk: light stays inside its own fibre, so many fibres can be packed in one cable without coupling.
- Electrical isolation: there is no current, so no ground loops, short circuits, sparking or shock hazard. Safe in fuel stations, mines and chemical plants.
- Signal security: fibre does not radiate, and tapping it causes a measurable loss, so it is hard to eavesdrop.
- Small size and light weight: a fibre is about 125 µm in diameter. A fibre cable is much lighter and thinner than a copper cable of equal capacity, saving duct space and installation cost.
- Flexibility and strength: fibre cables can be bent (within limits) and are strong in tension with Kevlar strength members.
- Low cost raw material: silica is abundant; copper is expensive and often stolen.
- Resistance to corrosion and temperature: glass does not rust and works over a wide temperature range.
- Easy upgrade: capacity is increased by changing terminal equipment (higher bit rate, more wavelengths) without replacing the cable.
Disadvantages of OFC
- Difficult joining: splicing and connecting need precise alignment, fusion splicers and trained technicians. Poor joints cause high loss.
- Fragility: bare fibre breaks easily and has a minimum bending radius; tight bends cause macro-bending loss.
- Costly terminal equipment: lasers, photodetectors, optical amplifiers and test instruments (OTDR, power meters) are expensive.
- No power transmission: repeaters and remote terminals need a separate power supply, while copper can carry power with the signal.
- Electro-optic conversion: signals must be converted from electrical to optical and back at each end, adding complexity.
- Repair and maintenance: locating and repairing a break needs special equipment and skill.
- Not economical for very short or low-rate links where cheap copper is adequate.
- Fibre fuse and radiation effects: high optical power and nuclear radiation can damage fibre.
Summary comparison
| Point | Optical fibre | Metallic wire |
|---|---|---|
| Signal form | Light | Electric current |
| Bandwidth | Very high (THz) | Limited (MHz–GHz) |
| Attenuation | 0.2–3 dB/km | High, increases with frequency |
| Repeater spacing | 100+ km | 1–5 km |
| EMI and crosstalk | Immune | Affected |
| Security | High | Low |
| Weight and size | Small, light | Bulky, heavy |
| Joining | Difficult, costly | Easy, cheap |
| Power feeding | Not possible | Possible |
Because its advantages in capacity, distance and noise immunity far outweigh its handling difficulties, OFC is used for backbone networks, undersea cables, FTTH and LANs, while copper remains for short, low-cost connections.
- 2069 Bhadra · 6 marks
Explain the dispersion and attenuation properties of an optical fibre.
Answer
Attenuation and dispersion are the two main transmission impairments of an optical fibre. Attenuation reduces the signal power and limits distance; dispersion spreads the pulses and limits bandwidth (bit rate).
Attenuation
Attenuation is the loss of optical power along the fibre, given in dB/km:
Causes:
- Absorption
- Intrinsic: absorption by silica itself, ultraviolet (electronic) and infrared (molecular vibration) bands.
- Extrinsic: impurities such as transition metal ions (Fe, Cu) and OH⁻ ions (water peak near 1383 nm).
- Scattering
- Rayleigh scattering from tiny density variations in glass; varies as , so it is the main loss at short wavelengths.
- Mie scattering from larger defects.
- Bending losses
- Macrobending: large bends let light escape.
- Microbending: tiny random bends from cabling pressure.
- Splice and connector losses (in a link).
Low-loss windows: 850 nm (~2.5 dB/km), 1310 nm (~0.35 dB/km), 1550 nm (~0.2 dB/km).
Dispersion
Dispersion is the time spreading of a light pulse as it travels. Spread pulses overlap (inter-symbol interference), limiting the maximum bit rate.
in: _|^|_|^|_ out: _/^^\/^^\_ (pulses overlap)
- Intermodal dispersion: different modes in multimode fibre take different paths and arrive at different times. Large in step-index fibre, reduced in graded-index, zero in single mode fibre.
- Chromatic (intramodal) dispersion: different wavelengths in the source spectrum travel at different speeds.
- Material dispersion: index of glass varies with wavelength.
- Waveguide dispersion: propagation depends on core size relative to wavelength.
- Zero near 1310 nm in standard fibre; shifted to 1550 nm in dispersion-shifted fibre.
- Polarization mode dispersion: the two polarizations travel at slightly different speeds.
| Point | Attenuation | Dispersion |
|---|---|---|
| Effect | Reduces power | Broadens pulse |
| Limits | Link distance | Bit rate / bandwidth |
| Unit | dB/km | ns/km or ps/(nm·km) |
| Remedy | Amplifiers, low-loss window | Single mode fibre, narrow laser |
Questions from Old Question Collection (EX 653) (IOE EX 653 exam papers from 2069 to 2081 (20 papers)). Answers are written for this site; check them against your class notes.
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