1. Introduction to Optical Fibers
Optical Fiber
A thin, flexible, transparent fiber made of glass (silica) or plastic that transmits light signals over long distances with minimal loss. It works on the principle of total internal reflection.
Optical fibers have revolutionized telecommunications by providing high-bandwidth, low-loss transmission of data as light pulses. A typical optical fiber consists of three parts: the core (where light travels), the cladding (surrounding layer with lower refractive index), and the protective jacket.
Structure of an Optical Fiber
Key Components
- Core: Central region where light propagates. Made of high-purity silica with higher refractive index (n₁). Diameter: 8-62.5 μm
- Cladding: Surrounds the core with lower refractive index (n₂). Causes total internal reflection. Diameter: 125 μm typically
- Buffer/Jacket: Protective outer coating. Provides mechanical strength. Diameter: 250-900 μm
2. Critical Angle
When light travels from a denser medium (higher refractive index n₁) to a rarer medium (lower refractive index n₂), it bends away from the normal. As the angle of incidence increases, the refracted ray bends more and more away from the normal. At a particular angle, called the critical angle, the refracted ray grazes along the interface (angle of refraction = 90°).
Critical Angle (θc)
The angle of incidence in the denser medium at which the angle of refraction in the rarer medium becomes 90°. For angles greater than critical angle, total internal reflection occurs.
Critical Angle and Total Internal Reflection
Derivation of Critical Angle
Using Snell's Law at the critical angle:
n₁ sin θc = n₂ sin 90°
n₁ sin θc = n₂ × 1
sin θc = n₂/n₁
θc = sin⁻¹(n₂/n₁)
Critical Angle Formula (n₁ > n₂)
Example: Critical Angle Calculation
Calculate the critical angle for light traveling from glass core (n₁ = 1.5) to glass cladding (n₂ = 1.45).
Step 1: Apply the critical angle formula
sin θc = n₂/n₁ = 1.45/1.5 = 0.9667
Step 2: Calculate inverse sine
θc = sin⁻¹(0.9667) = 75.16°
Critical Angle θc = 75.16°
3. Total Internal Reflection and Propagation of Light
Total Internal Reflection (TIR)
When light traveling in a denser medium hits the interface with a rarer medium at an angle greater than the critical angle, the light is completely reflected back into the denser medium without any refraction. This is called total internal reflection.
Conditions for Total Internal Reflection
- Light must travel from a denser medium to a rarer medium (n₁ > n₂)
- Angle of incidence must be greater than the critical angle (θ > θc)
In an optical fiber, light enters the core and travels by bouncing off the core-cladding interface through repeated total internal reflections. Since the core has a higher refractive index than the cladding, light striking the interface at angles greater than critical angle is completely reflected back into the core.
Light Propagation in Optical Fiber
Light bounces along the fiber through repeated total internal reflections
4. Acceptance Angle and Acceptance Cone
Acceptance Angle (θa)
The maximum angle (measured from the fiber axis) at which light can enter the fiber and still undergo total internal reflection within the core. Light entering at angles greater than the acceptance angle will not be guided through the fiber.
Acceptance Cone
The cone of light that can enter the fiber and be successfully transmitted. It is formed by rotating the acceptance angle around the fiber axis.
Acceptance Angle and Acceptance Cone
Derivation of Acceptance Angle
Consider light entering the fiber from air (n₀) into core (n₁):
At the air-core interface, applying Snell's law:
n₀ sin θa = n₁ sin θr ... (1)
For TIR at core-cladding interface, the refracted ray must hit at critical angle:
θr + θc = 90° → θr = 90° - θc
Substituting in equation (1):
n₀ sin θa = n₁ sin(90° - θc) = n₁ cos θc
Since sin θc = n₂/n₁, we have cos θc = √(1 - n₂²/n₁²) = √(n₁² - n₂²)/n₁
n₀ sin θa = n₁ × √(n₁² - n₂²)/n₁ = √(n₁² - n₂²)
For air, n₀ = 1:
sin θa = √(n₁² - n₂²)
sin θa = √(n₁² - n₂²) = NA
Acceptance Angle (for fiber in air)
Example: Acceptance Angle
An optical fiber has core refractive index n₁ = 1.50 and cladding refractive index n₂ = 1.46. Calculate the acceptance angle.
Step 1: Calculate sin θa
sin θa = √(n₁² - n₂²) = √(1.50² - 1.46²)
sin θa = √(2.25 - 2.1316) = √0.1184 = 0.344
Step 2: Find acceptance angle
θa = sin⁻¹(0.344) = 20.12°
Acceptance Angle θa = 20.12°
5. Numerical Aperture (NA)
Numerical Aperture (NA)
A dimensionless number that characterizes the light-gathering ability of an optical fiber. It is defined as the sine of the acceptance angle and represents the range of angles over which the fiber can accept light.
NA = sin θa = √(n₁² - n₂²)
Numerical Aperture
The NA can also be expressed in terms of the relative refractive index difference (Δ):
Δ = (n₁ - n₂)/n₁
Relative Refractive Index Difference
NA = n₁√(2Δ)
NA in terms of Δ
Derivation of NA = n₁√(2Δ)
Starting with NA² = n₁² - n₂² = (n₁ + n₂)(n₁ - n₂)
Since n₁ ≈ n₂ (small difference), we can approximate n₁ + n₂ ≈ 2n₁
NA² ≈ 2n₁(n₁ - n₂) = 2n₁² × (n₁ - n₂)/n₁ = 2n₁²Δ
NA = n₁√(2Δ)
Significance of Numerical Aperture
- Light Gathering: Higher NA means fiber can collect more light (larger acceptance cone)
- Coupling Efficiency: NA determines how easily light from a source can be coupled into the fiber
- Modal Dispersion: Higher NA allows more modes → more dispersion → limits bandwidth
- Typical Values: Single-mode fibers: 0.1-0.15, Multimode fibers: 0.2-0.5
Example: Numerical Aperture and Δ
A step-index fiber has core refractive index 1.52 and Δ = 0.0197. Calculate (a) Cladding refractive index (b) Numerical aperture (c) Acceptance angle
Step 1: Calculate cladding refractive index
Δ = (n₁ - n₂)/n₁ → n₂ = n₁(1 - Δ)
n₂ = 1.52(1 - 0.0197) = 1.52 × 0.9803 = 1.49
Step 2: Calculate Numerical Aperture
NA = n₁√(2Δ) = 1.52 × √(2 × 0.0197)
NA = 1.52 × √0.0394 = 1.52 × 0.1985 = 0.302
Step 3: Calculate Acceptance Angle
θa = sin⁻¹(NA) = sin⁻¹(0.302) = 17.58°
n₂ = 1.49, NA = 0.302, θa = 17.58°
6. Types of Optical Fibers
6.1 Based on Mode of Propagation
Single Mode Fiber (SMF)
- Very small core diameter (8-10 μm)
- Only one mode (ray path) propagates
- Low dispersion → High bandwidth
- Used for long-distance communication
- Requires laser source for launching
- Difficult to couple light
- NA: 0.08 - 0.15
Multimode Fiber (MMF)
- Larger core diameter (50-62.5 μm)
- Multiple modes propagate simultaneously
- Higher dispersion → Lower bandwidth
- Used for short-distance communication
- Can use LED sources
- Easy to couple light
- NA: 0.2 - 0.5
Single Mode vs Multimode Fiber
6.2 Based on Refractive Index Profile
Step Index Fiber
- Refractive index changes abruptly at core-cladding boundary
- Core has uniform refractive index n₁
- Cladding has uniform refractive index n₂
- Light travels in zigzag path (sharp reflections)
- Higher intermodal dispersion
- Simple to manufacture
Graded Index Fiber
- Refractive index decreases gradually from center to cladding
- Parabolic profile: n(r) = n₁[1 - 2Δ(r/a)²]^½
- Light travels in curved/sinusoidal path
- Lower intermodal dispersion
- Rays near axis travel slower but shorter path
- More complex to manufacture
Step Index vs Graded Index Profile
| Property | Step Index SMF | Step Index MMF | Graded Index MMF |
|---|---|---|---|
| Core Diameter | 8-10 μm | 50-200 μm | 50-62.5 μm |
| Cladding Diameter | 125 μm | 125-400 μm | 125 μm |
| NA | 0.08-0.15 | 0.2-0.5 | 0.2-0.3 |
| Bandwidth | Very High (>10 GHz·km) | Low (~50 MHz·km) | Medium (~500 MHz·km) |
| Dispersion | Very Low | High | Medium |
| Application | Long-haul telecom | Short links, LANs | Medium distance LANs |
7. Attenuation in Optical Fibers
Attenuation
The reduction in power (intensity) of the optical signal as it propagates through the fiber. It is measured in decibels per kilometer (dB/km).
α = (10/L) × log₁₀(P_in/P_out) dB/km
Attenuation Coefficient
Alternatively, expressed as:
P_out = P_in × 10^(-αL/10)
Power after distance L
Where: α = attenuation coefficient (dB/km), L = fiber length (km), P_in = input power, P_out = output power
Factors Affecting Attenuation
Absorption Losses
- Intrinsic Absorption: Due to the silica material itself absorbing light at certain wavelengths (UV and IR regions)
- Extrinsic Absorption: Due to impurities like OH ions (water), metal ions. OH absorption peak at 1380 nm
Scattering Losses
- Rayleigh Scattering: Due to microscopic variations in glass density. Dominant at shorter wavelengths. Proportional to 1/λ⁴
- Mie Scattering: Due to larger imperfections, bubbles, or core-cladding irregularities
Bending Losses
- Macrobending: Large-scale bends in fiber cause light to exceed critical angle and escape
- Microbending: Small random bends due to pressure, manufacturing defects cause loss
Attenuation vs Wavelength in Silica Fiber
Modern fibers have minimum loss (~0.2 dB/km) at 1550 nm wavelength
Optical Transmission Windows
- 1st Window (850 nm): Used with multimode fibers, ~3 dB/km loss
- 2nd Window (1310 nm): Zero dispersion wavelength, ~0.5 dB/km loss
- 3rd Window (1550 nm): Minimum attenuation (~0.2 dB/km), preferred for long-haul
Example: Attenuation Calculation
An optical fiber has an attenuation of 0.5 dB/km at 1310 nm. If 2 mW power is launched into a 50 km fiber, what is the output power?
Step 1: Calculate total attenuation
Total Loss = α × L = 0.5 × 50 = 25 dB
Step 2: Convert dB loss to power ratio
P_out/P_in = 10^(-Loss/10) = 10^(-25/10) = 10^(-2.5)
P_out/P_in = 0.00316
Step 3: Calculate output power
P_out = P_in × 0.00316 = 2 mW × 0.00316 = 0.00632 mW
P_out = 6.32 μW
Output Power = 6.32 μW
8. Fiber Optic Communication System
A fiber optic communication system converts electrical signals to optical signals, transmits them through optical fiber, and converts them back to electrical signals at the receiver.
Basic Fiber Optic Communication System
Transmitter
- Input Signal: Electrical signal (voice, data, video)
- Modulator/Driver: Converts electrical signal to modulate light intensity
- Light Source: LED (for short distance) or Laser Diode (for long distance) converts electrical signal to optical signal
Transmission Medium (Optical Fiber)
- Carries light signal from transmitter to receiver
- Signal experiences attenuation and dispersion
- For long distances, optical amplifiers (EDFA) or repeaters may be used
Receiver
- Photodetector: PIN diode or APD converts optical signal back to electrical signal
- Amplifier: Amplifies weak electrical signal
- Signal Processing: Demodulates and reconstructs original signal
| Component | Options | Characteristics |
|---|---|---|
| Light Source | LED | Low cost, moderate speed, wide spectrum, multimode |
| Laser Diode | High cost, high speed, narrow spectrum, single mode | |
| Photodetector | PIN Diode | Simple, reliable, moderate sensitivity |
| APD | Internal gain, higher sensitivity, requires bias | |
| Amplifier | EDFA | Erbium-Doped Fiber Amplifier for 1550 nm band |
9. Advantages of Optical Fiber Communication
Bandwidth & Speed
- Extremely high bandwidth (THz range)
- Data rates of Tbps possible
- Much higher than copper cables
- Supports multiple wavelengths (WDM)
Low Loss
- Attenuation as low as 0.2 dB/km
- Signals travel longer without amplification
- Fewer repeaters needed
- Lower operational cost
Immunity
- No electromagnetic interference (EMI)
- No radio frequency interference (RFI)
- No crosstalk between fibers
- Safe in explosive environments
Security
- Very difficult to tap
- No electromagnetic emissions
- Ideal for secure communications
- Military and banking applications
Physical Properties
- Small size and light weight
- Flexible and durable
- Made from abundant silica
- No electrical hazard
Cost-Effective
- Lower material cost than copper
- Higher capacity per cable
- Longer lifespan
- Lower maintenance
| Parameter | Copper Cable | Optical Fiber |
|---|---|---|
| Bandwidth | ~100 MHz·km | >10 GHz·km (single mode) |
| Attenuation | High (dB/100m) | Very low (0.2 dB/km) |
| EMI Susceptibility | High | None |
| Security | Easy to tap | Very secure |
| Weight | Heavy | Very light |
| Distance | <100 m (high speed) | >100 km |
10. Practice Numerical Problems
Example: Complete Fiber Analysis
A step-index fiber has a core refractive index of 1.48 and a numerical aperture of 0.20. Calculate: (a) Cladding refractive index (b) Critical angle (c) Acceptance angle (d) Relative refractive index difference Δ
Step 1: Calculate cladding refractive index from NA
NA = √(n₁² - n₂²)
0.20 = √(1.48² - n₂²)
0.04 = 2.1904 - n₂²
n₂² = 2.1504 → n₂ = 1.466
Step 2: Calculate critical angle
sin θc = n₂/n₁ = 1.466/1.48 = 0.9905
θc = sin⁻¹(0.9905) = 82.1°
Step 3: Calculate acceptance angle
sin θa = NA = 0.20
θa = sin⁻¹(0.20) = 11.54°
Step 4: Calculate relative refractive index difference
Δ = (n₁ - n₂)/n₁ = (1.48 - 1.466)/1.48
Δ = 0.014/1.48 = 0.00946 ≈ 0.95%
n₂ = 1.466, θc = 82.1°, θa = 11.54°, Δ = 0.95%
Summary: Key Formulas
θc = sin⁻¹(n₂/n₁)
Critical Angle
NA = sin θa = √(n₁² - n₂²)
Numerical Aperture
NA = n₁√(2Δ)
NA in terms of Δ
Δ = (n₁ - n₂)/n₁
Relative RI Difference
α = (10/L)log₁₀(P_in/P_out)
Attenuation (dB/km)
P_out = P_in × 10^(-αL/10)
Output Power