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Module II: Fibre Optics - Question Solutions

Question 1: Critical Angle, Acceptance Angle, and Attenuation

Question
Define the terms: Critical angle, Acceptance angle and Attenuation.
1. Critical Angle (θc)
Definition:
The critical angle is the angle of incidence (measured from the normal) at which the angle of refraction becomes 90°. For angles greater than the critical angle, total internal reflection occurs.
sin θc = n₂/n₁
where n₁ = refractive index of core (denser medium), n₂ = refractive index of cladding (rarer medium)

For light to propagate through an optical fiber by total internal reflection, the angle of incidence at the core-cladding interface must be greater than the critical angle.

2. Acceptance Angle (θa)
Definition:
The acceptance angle (also called maximum acceptance angle) is the maximum angle at which light can enter the fiber and still propagate through it by total internal reflection. Light entering at angles greater than the acceptance angle will not be guided through the fiber.
Acceptance Angle in Optical Fiber
Core (n₁) Cladding (n₂) Cladding (n₂) Incident Ray θa Acceptance Cone
sin θa = √(n₁² - n₂²) / n₀ = NA
where n₀ = refractive index of the medium from which light enters (usually air, n₀ = 1)
3. Attenuation
Definition:
Attenuation is the loss of optical power as light travels through an optical fiber. It represents the reduction in signal strength per unit length and is a critical parameter in fiber optic communication.
α = (10/L) × log₁₀(Pᵢ/Pₒ) dB/km
where α = attenuation coefficient, L = length of fiber, Pᵢ = input power, Pₒ = output power
Causes of Attenuation:
  • Absorption: Material absorption (UV and IR), impurity absorption (OH ions, metal ions)
  • Scattering: Rayleigh scattering (dominant), Mie scattering
  • Bending Losses: Macro-bending and micro-bending losses
  • Connection Losses: Splice losses, connector losses

Question 2: Numerical Aperture Derivation

Question
Derive an expression for numerical aperture of a step index optical fiber. Give its physical significance.
Derivation of Numerical Aperture
Light Propagation in Step Index Fiber
Core (n₁) Cladding (n₂) Cladding (n₂) Air (n₀) θₐ θᵣ θᵢ
Ray diagram showing acceptance angle and TIR in optical fiber

Consider:

  • n₀ = Refractive index of surrounding medium (air = 1)
  • n₁ = Refractive index of core
  • n₂ = Refractive index of cladding
  • θₐ = Acceptance angle (angle of incidence at fiber face)
  • θᵣ = Angle of refraction in core
  • θᵢ = Angle of incidence at core-cladding interface
Step-by-Step Derivation

Step 1: Apply Snell's law at the air-core interface:

n₀ sin θₐ = n₁ sin θᵣ

Step 2: From geometry, at the core-cladding interface:

θᵢ + θᵣ = 90°
Therefore: θᵣ = 90° - θᵢ

Step 3: For total internal reflection, θᵢ must be at least equal to the critical angle θc:

sin θc = n₂/n₁

Step 4: At the maximum acceptance angle, θᵢ = θc (critical condition). Substituting in Step 1:

n₀ sin θₐ = n₁ sin(90° - θc) = n₁ cos θc

Step 5: Using the identity cos θc = √(1 - sin²θc):

cos θc = √(1 - n₂²/n₁²) = √(n₁² - n₂²)/n₁

Step 6: Substituting back:

n₀ sin θₐ = n₁ × √(n₁² - n₂²)/n₁ = √(n₁² - n₂²)

Step 7: For air (n₀ = 1):

sin θₐ = √(n₁² - n₂²)
Final Expression for Numerical Aperture (NA)
NA = sin θₐ = √(n₁² - n₂²)

Alternative forms:

NA = n₁√(2Δ)
where Δ = (n₁ - n₂)/n₁ is the fractional refractive index change
Physical Significance of Numerical Aperture
Importance of NA:
  • Light Gathering Ability: NA indicates the light-gathering capacity of the fiber. Higher NA means more light can enter the fiber.
  • Acceptance Cone: NA defines the acceptance cone - the cone of light that will be guided through the fiber.
  • Coupling Efficiency: Higher NA allows easier coupling of light from sources like LEDs into the fiber.
  • Number of Modes: In multimode fibers, higher NA supports more propagation modes.
  • Bandwidth-Distance Product: Higher NA generally leads to higher dispersion, limiting bandwidth in multimode fibers.
Typical NA Values
  • Single-mode fibers: NA ≈ 0.1 - 0.15
  • Multimode fibers: NA ≈ 0.2 - 0.3
  • Plastic fibers: NA ≈ 0.3 - 0.5

Question 3: Step Index vs Graded Index Fiber Comparison

Question
Compare Step Index fiber and Graded Index Fiber.
Refractive Index Profiles
Step Index n radius n₂ n₁ Graded Index n radius n₂ n₁
Property Step Index Fiber Graded Index Fiber
Refractive Index Profile Uniform refractive index in core, abrupt change at core-cladding boundary Refractive index decreases gradually from center to cladding (parabolic profile)
Refractive Index Formula n(r) = n₁ for r < a
n(r) = n₂ for r ≥ a		 n(r) = n₁[1 - 2Δ(r/a)ᵅ]^(1/2)
Light Propagation Zig-zag path with TIR at core-cladding interface Sinusoidal/helical path due to continuous refraction
Modal Dispersion High modal dispersion (different path lengths) Low modal dispersion (path lengths equalized)
Bandwidth Lower bandwidth (~20-50 MHz·km) Higher bandwidth (~200-600 MHz·km)
Data Rate Lower data transmission rate Higher data transmission rate
Manufacturing Simpler and cheaper to manufacture Complex and expensive manufacturing
Coupling Efficiency Easier coupling with light sources More difficult coupling
Attenuation Typically higher attenuation Lower attenuation
Applications Short distance communication, LANs, sensors Medium to long distance communication, data networks
Light Path Comparison
Step Index - Zig-zag Path Graded Index - Sinusoidal Path Different path lengths → High dispersion Similar travel times → Low dispersion

Question 4: Numerical Problem - Fiber Parameters Calculation

Question
The refractive indices of core and cladding materials of a step index fibre are 1.48 and 1.45 respectively. Calculate (1) Numerical Aperture, (2) Acceptance angle, (3) Critical angle at core-cladding interface, (4) Fractional refractive indices change.
Given Data:
  • Refractive index of core, n₁ = 1.48
  • Refractive index of cladding, n₂ = 1.45
Part (1): Numerical Aperture (NA)

Using the formula:

NA = √(n₁² - n₂²)

Substituting values:

NA = √(1.48² - 1.45²)
NA = √(2.1904 - 2.1025)
NA = √0.0879
NA = 0.296 ≈ 0.30
Part (2): Acceptance Angle (θₐ)

Using:

sin θₐ = NA = 0.296
θₐ = sin⁻¹(0.296)
θₐ = 17.22° ≈ 17.2°
Part (3): Critical Angle (θc)

Using the formula:

sin θc = n₂/n₁
sin θc = 1.45/1.48 = 0.9797
θc = sin⁻¹(0.9797)
θc = 78.45° ≈ 78.5°
Part (4): Fractional Refractive Index Change (Δ)

Using the formula:

Δ = (n₁ - n₂)/n₁
Δ = (1.48 - 1.45)/1.48
Δ = 0.03/1.48
Δ = 0.0203 ≈ 2.03%

Question 5: Acceptance Angle Numerical

Question
An optical fiber has an acceptance angle of 30° and core refractive index of 1.5. Calculate the refractive index of cladding.
Given Data:
  • Acceptance angle, θₐ = 30°
  • Refractive index of core, n₁ = 1.5
  • Assuming air medium, n₀ = 1
Step 1: Calculate Numerical Aperture
NA = sin θₐ = sin 30° = 0.5
Step 2: Use NA formula to find n₂

We know:

NA = √(n₁² - n₂²)

Squaring both sides:

NA² = n₁² - n₂²
n₂² = n₁² - NA²
n₂² = (1.5)² - (0.5)²
n₂² = 2.25 - 0.25 = 2.00
n₂ = √2.00
n₂ = 1.414

Question 6: Antireflection Coating

Question
What is antireflection coating? What should be the refractive index and minimum thickness of the coating?
Definition

An antireflection coating (AR coating) is a thin transparent film applied to optical surfaces to reduce reflection and increase light transmission. It works on the principle of destructive interference between light reflected from the top and bottom surfaces of the coating.

Antireflection Coating Principle
Air (n₀ = 1) AR Coating (nf) Substrate (ns) Incident R₁ R₂ t R₁ and R₂ interfere destructively when path difference = λ/2
Conditions for Effective AR Coating
1. Refractive Index Condition
  • For complete destructive interference, the amplitudes of both reflected rays should be equal
  • This occurs when the coating refractive index is the geometric mean of air and substrate
nf = √(n₀ × ns)
where nf = refractive index of film, n₀ = refractive index of air (≈1), ns = refractive index of substrate

Example: For glass with ns = 1.5:

nf = √(1 × 1.5) = √1.5 ≈ 1.22
MgF₂ (n = 1.38) is commonly used as it's close to this ideal value
2. Minimum Thickness Condition
  • For destructive interference, the optical path difference must equal λ/2
  • The light travels through the coating twice (down and up)
  • Total optical path = 2 × nf × t
Derivation of Minimum Thickness

For destructive interference (minimum reflection):

2 × nf × t = (2m + 1) × λ/2
where m = 0, 1, 2, ... (order of interference)

For minimum thickness, m = 0:

2 × nf × t = λ/2
Minimum Thickness Formula
t = λ/(4nf)
This is called a "quarter-wave" coating
Practical Example
For green light (λ = 550 nm) with MgF₂ coating (nf = 1.38):
t = 550/(4 × 1.38) = 550/5.52 ≈ 99.6 nm ≈ 100 nm
Applications of AR Coatings
  • Camera lenses and optical instruments
  • Eyeglasses and sunglasses
  • Solar cells (to maximize light absorption)
  • Optical fiber connectors
  • Display screens (monitors, smartphones)
  • Laser optics
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