Unit-4. LASER and Fiber Optics - Medium Solutions

Part A: Short Answers (1-2 marks)

(1) State Snell's law of refraction

When light passes from one medium to another, the ratio of sine of angle of incidence to sine of angle of refraction is constant (refractive index).

Mathematical Form:

sin i / sin r = μ₂/μ₁ = constant

Where:

i = angle of incidence
r = angle of refraction
μ₁, μ₂ = refractive indices of first and second medium

$Refraction Diagram$ Figure: Light refraction at interface between two media

(2) What is LASER?

LASER = Light Amplification by Stimulated Emission of Radiation

A device that produces intense, coherent, monochromatic, and directional beam of light through stimulated emission.

Key Properties:

Monochromatic: Single wavelength/color
Coherent: Waves in phase (constant phase difference)
Directional: Highly collimated beam (low divergence)
Intense: High power concentration

(3) Define refractive index and give its values for different materials

Refractive Index (μ): Ratio of speed of light in vacuum to speed of light in medium.

μ = c/v = sin i / sin r

Common Values:

Material	Refractive Index (μ)
Vacuum	1.0000
Air	1.0003
Water	1.33
Glass	1.5 - 1.9
Diamond	2.42

Physical Meaning: Higher μ → Light travels slower → More bending

(4) Define critical angle

Critical Angle (θc): Angle of incidence in denser medium for which angle of refraction in rarer medium becomes 90°.

Condition: Light travels from denser to rarer medium

Formula:

sin θc = μ₂/μ₁ = 1/μ (when μ₂ = 1 for air)

θc = sin⁻¹(1/μ)

Figure: Critical angle and total internal reflection

Examples:

Glass (μ=1.5): θc = 41.8°
Water (μ=1.33): θc = 48.8°
Diamond (μ=2.42): θc = 24.4°

(5) State the principle of optical fiber

Principle: Total Internal Reflection (TIR)

Light propagates through optical fiber by undergoing successive total internal reflections at the core-cladding interface.

Conditions for TIR:

Light travels from denser (core) to rarer (cladding) medium
Angle of incidence > critical angle
μ_core > μ_cladding

When i > θc → Total reflection occurs
No refraction, light trapped inside core

Part B: Detailed Answers (2-3 marks)

(1) Refraction of light - definition and examples

Definition: Bending of light when it passes from one transparent medium to another due to change in speed.

$Refraction Examples$ Figure: Refraction in different media

Causes:

Different optical densities
Change in speed of light (v = c/μ)
Light bends towards normal (rarer → denser)
Light bends away from normal (denser → rarer)

Examples:

Swimming Pool: Bottom appears raised due to water-to-air refraction
Atmospheric Refraction: Stars appear at different positions, twinkling
Prism: White light splits into spectrum (dispersion)
Lenses: Converging/diverging light through refraction
Mirage: Hot air acts as rarer medium, causing apparent water on roads

(2) Total Internal Reflection and conditions

Definition: When light traveling from denser to rarer medium is completely reflected back into denser medium (no refraction).

TIR Demonstration Figure: Total internal reflection at critical angle

Essential Conditions:

Medium: Light must travel from optically denser to rarer medium (μ₁ > μ₂)
Angle: Angle of incidence must be greater than critical angle (i > θc)

Applications:

Optical fibers for communication
Periscopes and binoculars (prisms)
Brilliance of diamonds
Mirage and looming effects
Endoscopy in medical field

(3) Difference between LASER light and ordinary light

LASER vs Ordinary Light Figure: Comparison of LASER and ordinary light characteristics

Property	LASER Light	Ordinary Light
Monochromaticity	Single wavelength	Multiple wavelengths (polychromatic)
Coherence	All waves in phase (spatial & temporal)	Random phase (incoherent)
Directionality	Highly directional (0.01° divergence)	Spreads in all directions
Intensity	Very high intensity	Low intensity
Source	Stimulated emission	Spontaneous emission
Power	High power density	Low power density
Wavelength	Fixed (e.g., 632.8 nm for He-Ne)	Broad spectrum (400-700 nm)

Key Advantage: LASER can travel long distances without spreading, making it ideal for communication and precision applications.

(4) Construction and working of optical fiber

Optical Fiber Structure Figure: Cross-section of optical fiber showing core, cladding, and jacket

Construction (Three Layers):

Core (Inner):
- Central light-carrying region
- Made of glass/silica (high μ)
- Diameter: 8-100 μm
- High refractive index (μ₁ ≈ 1.48)
Cladding (Middle):
- Surrounds the core
- Lower refractive index (μ₂ ≈ 1.46)
- Thickness: 125 μm
- Enables TIR at core-cladding boundary
Jacket (Outer):
- Protective polymer coating
- Provides mechanical strength
- Prevents moisture and damage

Working Principle:

Light enters core at specific angle (< acceptance angle)
Strikes core-cladding boundary at angle > critical angle
Undergoes total internal reflection
Multiple reflections carry light through fiber
Light exits at far end with minimal loss

Condition: μ_core > μ_cladding (typically 1.48 > 1.46)

(5) Types of optical fibers

Based on Core Diameter & Refractive Index:

1. Single Mode Fiber (SMF):

Very small core diameter (8-10 μm)
Only one ray path (mode) allowed
Used for long-distance, high-speed transmission
Low signal dispersion
Applications: Telephone, cable TV, internet backbone

2. Multi-Mode Fiber (MMF):

Larger core diameter (50-100 μm)
Multiple ray paths possible
Two subtypes:

a) Step Index Fiber:
- Uniform core refractive index
- Sharp change at core-cladding boundary
- Higher dispersion
- Shorter distances
b) Graded Index Fiber:
- Core refractive index varies (maximum at center, minimum at edge)
- Rays travel at different speeds
- Reduces dispersion
- Better performance than step index

Fiber Types Figure: Types of optical fibers - single mode, step index, and graded index

(6) Acceptance angle and acceptance cone

Acceptance Angle (θₐ): Maximum angle to the fiber axis at which light can enter the core and still propagate through total internal reflection.

Figure: Acceptance angle and acceptance cone in optical fiber

Formula:

sin θₐ = √(μ₁² - μ₂²) = NA (Numerical Aperture)

For air: sin θₐ = √(μ_core² - μ_cladding²)

Acceptance Cone: 3D cone of light that can enter and propagate through fiber

Cone angle = 2θₐ
Larger NA → Larger acceptance cone → More light collection

Significance:

Determines light-gathering capability
Higher NA → Easier coupling but more dispersion
Typical values: 0.1 to 0.5

(7) Numerical Aperture (NA)

Definition: Measure of light-gathering ability of an optical fiber. It's the sine of acceptance angle.

Formula:

NA = sin θₐ = √(μ₁² - μ₂²) = √(n_core² - n_cladding²)

Alternative: NA = μ₁√(2Δ)
where Δ = (μ₁ - μ₂)/μ₁ (relative refractive index difference)

Physical Meaning:

Higher NA → Better light collection
Lower NA → Less dispersion, better for long distance
Typical range: 0.1 - 0.5

Trade-off:

Large NA: Good coupling, high dispersion
Small NA: Poor coupling, low dispersion

(8) Advantages of optical fiber communication

Technical Advantages:

Wide Bandwidth: Can carry millions of channels simultaneously (THz bandwidth)
Low Attenuation: Signal loss < 0.2 dB/km (can transmit 100+ km without amplification)
No Interference: Immune to electromagnetic interference (EMI) and radio frequency interference (RFI)
No Cross-talk: Multiple fibers in cable don't interfere with each other
Security: Difficult to tap, no radiation leakage

Physical Advantages:

Light Weight: Glass fiber is lighter than copper cables
Small Size: Compact, easy to install
Flexibility: Can bend around corners
Corrosion Resistant: Not affected by moisture or chemicals

Economic Advantages:

Cost-Effective: Lower installation and maintenance costs for long distances
Abundant Material: Silica (sand) is readily available

(9) Applications of optical fiber

1. Communication:

Telephone networks (local and long-distance)
Internet backbone and broadband
Cable television (CATV)
Video conferencing
Military and secure communications

2. Medical (Endoscopy):

Internal body examination (stomach, intestines)
Minimally invasive surgery
Laser surgery delivery
Medical imaging

3. Sensing:

Temperature sensors
Pressure sensors
Rotation sensors (gyroscopes)
Structural health monitoring

4. Industrial:

Computer networking (LAN, data centers)
Automotive (car lighting, sensors)
Lighting and decoration
Remote sensing

5. Defense:

Secure communications
Missile guidance
Surveillance systems

Part C: Numerical Problems (3 marks)

(1) Calculate critical angle for glass-air interface (μ = 1.5)

Given: μ = 1.5 (glass), air μ = 1

Formula:

sin θc = 1/μ = 1/1.5 = 0.6667

θc = sin⁻¹(0.6667) = 41.81° ≈ 42°

Answer: Critical angle = 41.81°

(2) Wavelength of light in glass (λ₀ = 6000 Å, μ = 1.5)

Given: λ₀ = 6000 Å (in vacuum/air), μ = 1.5

Concept: Wavelength decreases in denser medium

Formula:

λ = λ₀/μ = 6000/1.5 = 4000 Å

Answer: λ = 4000 Å

Note: Frequency remains constant, only wavelength and speed change

(3) Numerical Aperture (μ_core = 1.50, μ_cladding = 1.48)

Given: μ₁ = 1.50, μ₂ = 1.48

Formula:

NA = √(μ₁² - μ₂²)
NA = √(1.50² - 1.48²)
NA = √(2.25 - 2.1904)
NA = √0.0596 = 0.244

Answer: NA = 0.244

(4) Acceptance angle for NA = 0.39

Given: NA = 0.39

Formula:

sin θₐ = NA = 0.39
θₐ = sin⁻¹(0.39) = 22.95° ≈ 23°

Answer: Acceptance angle = 22.95°

Acceptance cone angle = 2θₐ = 45.9°

(5) NA and acceptance angle (μ_core = 1.53, μ_cladding = 1.50)

Given: μ₁ = 1.53, μ₂ = 1.50

Step 1: Calculate NA

NA = √(μ₁² - μ₂²)
NA = √(1.53² - 1.50²)
NA = √(2.3409 - 2.25)
NA = √0.0909 = 0.3015

Step 2: Calculate acceptance angle

sin θₐ = NA = 0.3015
θₐ = sin⁻¹(0.3015) = 17.54°

Answers:

NA = 0.30
Acceptance angle = 17.54°

Quick Reference

Important Formulas

Snell's Law: μ₁ sin i = μ₂ sin r
Refractive Index: μ = c/v = sin i / sin r
Critical Angle: sin θc = μ₂/μ₁ = 1/μ (for air)
Numerical Aperture: NA = √(μ₁² - μ₂²) = sin θₐ
Acceptance Angle: θₐ = sin⁻¹(NA)
Wavelength in medium: λ = λ₀/μ
Relative RI difference: Δ = (μ₁ - μ₂)/μ₁
Alternative NA: NA = μ₁√(2Δ)

Key Constants

Speed of light (c): 3 × 10⁸ m/s
1 Angstrom (Å): 10⁻¹⁰ m
1 nanometer (nm): 10⁻⁹ m
1 micrometer (μm): 10⁻⁶ m
Visible light: 400-700 nm

Medium Solutions - Unit 4