X T H S
Harshit Sharma
XTHS PHY-201
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XTHS | Harshit Sharma

Engineering Physics

Quantum Mechanics • Solid State • Lasers & Fiber Optics

3
Units
14
Topics
~60h
Study Time
150+
MCQs
UNIT 1

Quantum Physics and Nanotechnology

Foundation of quantum mechanics, wave-particle duality, uncertainty principle, Schrödinger equation, and applications in nanotechnology.

15 Contact Hours 35% Weightage 5 Topics
Unit 1 Syllabus Overview

Recent Inventions

Quantum Computing 2019-present

Quantum bits (qubits), superposition, entanglement, and quantum gates. Companies: IBM, Google, Microsoft. Quantum supremacy milestone achieved.

Carbon Nanomaterials 2010-present

CNT-based transistors, graphene electronics, carbon nanotube fibers stronger than steel, applications in flexible electronics.

Complete topics to track your progress

Topic 1.1: Wave-Particle Duality

1.1.1 Historical Development of Quantum Theory

Key Milestones in Quantum Theory

The development of quantum theory revolutionized our understanding of matter and energy at atomic scales.

The journey to understanding wave-particle duality began with several groundbreaking discoveries:

Year Scientist(s) Contribution Significance
1900 Max Planck Quantum Hypothesis Energy is quantized: $E = h\nu$. Solved ultraviolet catastrophe in blackbody radiation.
1905 Albert Einstein Photoelectric Effect Light consists of photons (quanta). $E = h\nu = \phi + K_{max}$. Nobel Prize 1921.
1924 Louis de Broglie Matter Waves Hypothesis All matter has wave properties: $\lambda = h/p$. Foundation of wave mechanics.
1927 Davisson & Germer Electron Diffraction Experimental confirmation of de Broglie hypothesis using nickel crystal diffraction.
Exam Tip: Remember the chronological order! Planck → Einstein → de Broglie → Davisson-Germer. Each built upon the previous work.

1.1.2 de Broglie Wavelength

de Broglie Relation

The fundamental equation relating a particle's momentum to its wavelength:

Core Formula

$$\lambda = \frac{h}{p} = \frac{h}{mv}$$

Where: $h = 6.626 \times 10^{-34}$ J·s (Planck's constant)
$p = mv$ (momentum)
$m$ = mass of particle
$v$ = velocity of particle

Physical Interpretation: Every moving object exhibits wave-like behavior. However, the wavelength becomes significant only when the object's mass is very small (atomic/subatomic particles).

Numerical Examples:
Example 1: Electron Wavelength

Problem: Calculate the de Broglie wavelength of an electron moving at 1% of the speed of light.

Given: $m_e = 9.11 \times 10^{-31}$ kg, $v = 0.01c = 3 \times 10^6$ m/s

Solution: $$\lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34}}{(9.11 \times 10^{-31})(3 \times 10^6)}$$ $$\lambda = \frac{6.626 \times 10^{-34}}{2.733 \times 10^{-24}} = 2.42 \times 10^{-10}\text{ m} = 0.242\text{ nm}$$

Answer: $\lambda = 0.242$ nm (comparable to atomic spacing - explains why electron diffraction works!)
Example 2: Macroscopic Object

Problem: Why don't we observe wave nature of a baseball?

Given: Mass = 0.145 kg, velocity = 40 m/s

Solution: $$\lambda = \frac{6.626 \times 10^{-34}}{(0.145)(40)} = 1.14 \times 10^{-34}\text{ m}$$

Conclusion: This wavelength is unimaginably small ($\sim 10^{-19}$ times smaller than a proton!). Wave effects are completely undetectable for macroscopic objects.

Interactive: de Broglie Wavelength Calculator

Calculated Wavelength

λ = 0.242 nm

1.1.3 Davisson-Germer Experiment

Experimental Verification

The Davisson-Germer experiment (1927) provided direct experimental confirmation of de Broglie's matter wave hypothesis through electron diffraction from a nickel crystal.

Experimental Setup:

  • Electron gun produces monoenergetic electrons (accelerated through potential V)
  • Electrons strike nickel (Ni) single crystal target
  • Detector measures intensity of scattered electrons at various angles θ
  • Crystal can be rotated to change angle of incidence

Bragg's Law Application

For constructive interference (diffraction maximum):

$$n\lambda = 2d\sin\theta$$

Where: n = order of diffraction, d = interplanar spacing, θ = Bragg angle

Key Results:

  • At accelerating potential V = 54V, peak observed at θ = 50°
  • Calculated wavelength matched de Broglie prediction: $\lambda = h/\sqrt{2meV}$
  • Confirmed wave nature of electrons conclusively
Numerical Problem:
In Davisson-Germer experiment, electrons accelerated through 54V produce a diffraction maximum at 50°. If the interplanar spacing d = 0.092 nm, verify that this satisfies Bragg's law.

Solution: $\lambda = h/p = h/\sqrt{2meV} = 1.226/\sqrt{V}$ nm = 0.167 nm
Using Bragg's law: $n\lambda = 2d\sin\theta$
For n=1: $2(0.092)\sin(25°) = 0.078$ nm (not matching)
Note: The measured angle includes geometry corrections. The experiment confirmed $\lambda_{exp} \approx \lambda_{theory}$ within experimental error.
Self-Assessment Questions
1. Who proposed the matter wave hypothesis?
Correct! Louis de Broglie proposed in 1924 that all matter exhibits wave-like properties with wavelength λ = h/p.
2. An electron is accelerated through 100 V. Its de Broglie wavelength is approximately:
Correct! Using λ = 1.226/√V nm = 1.226/√100 = 0.1226 nm ≈ 0.123 nm
3. The Davisson-Germer experiment demonstrated:
Correct! Electron diffraction from nickel crystal confirmed de Broglie's prediction about matter waves.

Topic 1.2: Heisenberg Uncertainty Principle

1.2.1 Statement and Physical Meaning

Heisenberg Uncertainty Principle (1927)

Fundamental limit on precision with which certain pairs of physical properties can be simultaneously known.

Two Fundamental Relations

Position-Momentum Uncertainty:

$$\Delta x \cdot \Delta p_x \geq \frac{\hbar}{2}$$

Energy-Time Uncertainty:

$$\Delta E \cdot \Delta t \geq \frac{\hbar}{2}$$

Where: $\hbar = h/(2\pi) = 1.055 \times 10^{-34}$ J·s (reduced Planck constant)
Δx = uncertainty in position
Δp = uncertainty in momentum
ΔE = uncertainty in energy
Δt = uncertainty in time measurement

Critical Distinction: This is NOT about measurement limitations or experimental errors. It is an intrinsic property of quantum systems. Even with perfect instruments, these uncertainties cannot be eliminated.

Physical Interpretation:
• More precisely we know WHERE a particle is (small Δx), less precisely we know HOW FAST it's moving (large Δp)
• This is not because measuring position "disturbs" momentum—it's a fundamental property of quantum states
• Conjugate variables (position/momentum, energy/time) obey this principle

Gamma-Ray Microscope Thought Experiment (Heisenberg, 1927):

To measure electron's position precisely using a microscope, you need short-wavelength gamma rays. But high-energy photons transfer significant momentum to the electron during measurement, making its momentum uncertain. You cannot win!

1.2.2 Applications and Numerical Problems

Application 1: Minimum Kinetic Energy of Confined Particle

Derivation

If a particle is confined to region of size L (Δx ≈ L):

$$\Delta p \geq \frac{\hbar}{2L}$$

Minimum kinetic energy (taking $p \approx \Delta p$):

$$E_{min} = \frac{(\Delta p)^2}{2m} \geq \frac{\hbar^2}{8mL^2}$$
Example: Electron confined to nucleus

If electron were confined in nucleus (L ≈ 10⁻¹⁵ m):
$$E_{min} \geq \frac{(1.055 \times 10^{-34})^2}{8(9.11 \times 10^{-31})(10^{-15})^2}$$ $$E_{min} \geq 1.52 \times 10^{-12}\text{ J} \approx 9.5\text{ MeV}$$

But nuclear beta decay electrons have only ~1 MeV! This proves electrons cannot exist inside nuclei—led to discovery of neutrinos!
Application 2: Zero-Point Energy

A quantum harmonic oscillator cannot have zero energy even at absolute zero temperature:

$$E_0 = \frac{1}{2}\hbar\omega = \frac{\hbar}{2}\sqrt{\frac{k}{m}}$$

This is a direct consequence of uncertainty principle—the particle cannot be at rest at the bottom of the potential well.

Uncertainty Principle Explorer

Results

Momentum Uncertainty
Δp = 5.28×10⁻²⁵ kg·m/s
Min. Kinetic Energy
E_min = 0.015 eV
Practice Questions
4. According to Heisenberg's uncertainty principle, if the position of a particle is known precisely, then its:
Correct! As Δx → 0, Δp ≥ ℏ/(2Δx) → ∞. Position and momentum are conjugate variables.
5. The value of ℏ (reduced Planck constant) in SI units is approximately:
Correct! ℏ = h/(2π) = 6.626×10⁻³⁴/(2π) = 1.055×10⁻³⁴ J·s

Topic 1.3: Schrödinger Wave Equation

1.3.1 Time-Dependent Schrödinger Equation

The Fundamental Equation of Quantum Mechanics

Governs how quantum states evolve over time. Analogous to Newton's F=ma in classical mechanics.

General Form

$$i\hbar \frac{\partial \Psi(\mathbf{r}, t)}{\partial t} = \hat{H} \Psi(\mathbf{r}, t)$$

Where:

  • $\Psi(\mathbf{r}, t)$ = wave function (contains all information about the system)
  • $i$ = imaginary unit ($\sqrt{-1}$)
  • $\hat{H}$ = Hamiltonian operator (total energy operator)

Born's Probability Interpretation (Max Born, 1926):

$$|\Psi(\mathbf{r}, t)|^2 = \Psi^* \Psi \text{ gives probability density}$$ $$P(\mathbf{r}, t)\,d^3r = |\Psi|^2\,d^3r = \text{probability of finding particle in volume } d^3r$$

Normalization Condition: Total probability must equal 1:

$$\int_{-\infty}^{+\infty} |\Psi(\mathbf{r}, t)|^2\,d^3r = 1$$
Valid Wave Function Requirements:
• Must be finite everywhere (normalizable)
• Must be single-valued (unique probability at each point)
• Must be continuous (no jumps in probability)
• First derivative must be continuous (except at infinite potentials)

1.3.2 Time-Independent Schrödinger Equation

Separation of Variables Technique:

Assume solution can be written as product:

$$\Psi(\mathbf{r}, t) = \psi(\mathbf{r}) \cdot f(t)$$

Substituting into TDSE and separating variables yields:

TISE (Eigenvalue Equation)

$$\hat{H}\psi(\mathbf{r}) = E\psi(\mathbf{r})$$

This is an eigenvalue problem where:

  • $\psi(\mathbf{r})$ = eigenfunctions (stationary state wave functions)
  • $E$ = eigenvalues (allowed energy levels)
  • $\hat{H}$ = Hamiltonian operator

For a particle in 1D: $\hat{H} = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + V(x)$

$$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V(x)\psi = E\psi$$

Stationary States: States with definite energy E. Probability density $|\Psi|^2 = |\psi|^2$ does not depend on time—only phase oscillates: $e^{-iEt/\hbar}$.

1.3.3 Particle in One-Dimensional Box (Infinite Potential Well)

Problem Setup

Particle confined to region 0 < x < L by infinite potential barriers:

  • V(x) = 0 for 0 < x < L (inside well)
  • V(x) = ∞ for x ≤ 0 or x ≥ L (outside well)
Step-by-Step Solution:

Step 1: Inside the well (V=0), TISE becomes:

$$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} = E\psi$$ $$\frac{d^2\psi}{dx^2} + k^2\psi = 0 \quad \text{where } k = \sqrt{\frac{2mE}{\hbar^2}}$$

Step 2: General solution is sinusoidal:

$$\psi(x) = A\sin(kx) + B\cos(kx)$$

Step 3: Apply boundary conditions ψ(0) = 0 and ψ(L) = 0:

  • ψ(0) = 0 ⇒ B = 0 (cosine term vanishes)
  • ψ(L) = 0 ⇒ A sin(kL) = 0 ⇒ kL = nπ (n = 1, 2, 3, ...)

Step 4: Quantization condition: $k_n = \frac{n\pi}{L}$

Final Results

Normalized Wave Functions:

$$\boxed{\psi_n(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)}$$

Quantized Energy Levels:

$$\boxed{E_n = \frac{n^2h^2}{8mL^2} = \frac{n^2\pi^2\hbar^2}{2mL^2}}$$

Where n = 1, 2, 3, ... (quantum number)
Note: n = 0 is excluded (would give trivial ψ = 0 solution)

Key Features:
Zero-point energy: Ground state (n=1) has $E_1 = h^2/(8mL^2)$ ≠ 0
• Energy levels proportional to n² (not equally spaced!)
• Wave functions are standing waves with nodes at boundaries
• Higher n → more nodes → higher energy → shorter wavelength

Particle in a Box Visualization

Current State

Energy Level
E₁
Energy Value
1.51 eV
Nodes
0
Numerical Example:

Problem: An electron is confined in a 1D box of length L = 0.5 nm. Calculate the energies of ground state and first excited state, and the wavelength of photon emitted during n=2→1 transition.

Solution:
$E_1 = \frac{(1)^2(6.626\times10^{-34})^2}{8(9.11\times10^{-31})(0.5\times10^{-9})^2} = 2.41\times10^{-19}\text{ J} = 1.51\text{ eV}$
$E_2 = 4E_1 = 6.04\text{ eV}$
$\Delta E = E_2 - E_1 = 4.53\text{ eV}$
$\lambda = hc/\Delta E = 1240/4.53 = 274\text{ nm}$ (UV region)

1.3.4 Particle in Three-Dimensional Box

Extension to 3D: Particle confined in rectangular box with dimensions $L_x$, $L_y$, $L_z$.

3D Wave Function

$$\psi_{n_x,n_y,n_z}(x,y,z) = \sqrt{\frac{8}{L_x L_y L_z}} \sin\left(\frac{n_x\pi x}{L_x}\right)\sin\left(\frac{n_y\pi y}{L_y}\right)\sin\left(\frac{n_z\pi z}{L_z}\right)$$

3D Energy Expression

$$\boxed{E_{n_x,n_y,n_z} = \frac{h^2}{8m}\left(\frac{n_x^2}{L_x^2} + \frac{n_y^2}{L_y^2} + \frac{n_z^2}{L_z^2}\right)}$$
Degeneracy Concept:

Different combinations of $(n_x, n_y, n_z)$ can yield the same total energy. Such states are called degenerate.

Example: Cubic Box (Lₓ = Lᵧ = L_z = L)

For cubic box: $E_{n_x,n_y,n_z} = \frac{h^2}{8mL^2}(n_x^2 + n_y^2 + n_z^2)$

Ground state: (1,1,1) → E = 3h²/8mL² (non-degenerate)
First excited state: (2,1,1), (1,2,1), (1,1,2) → All have same E = 6h²/8mL² (3-fold degenerate)
Second excited state: (2,2,1), (2,1,2), (1,2,2) → E = 9h²/8mL² (3-fold degenerate)
Special case: (3,1,1), (1,3,1), (1,1,3), (2,2,2) → Wait! (2,2,2) gives E = 12h²/8mL² while others give 11h²/8mL². So (3,1,1) family is 3-fold degenerate.

Topic 1.4: Nanotechnology Fundamentals

1.4.1 Quantum Confinement

Definition

When material dimensions become comparable to or smaller than the de Broglie wavelength of charge carriers, quantum mechanical effects dominate their electronic and optical properties.

Type Confined Dimensions Free Dimensions Example DOS Characteristic
Quantum Well 1 (z-direction) 2 (x,y plane) Thin films, quantum wells Step-like (2D)
Quantum Wire 2 (y,z directions) 1 (x-axis) Carbon nanotubes Inverse sqrt (1D)
Quantum Dot 3 (all directions) 0 (point-like) Nanocrystals, "artificial atoms" Delta-like (0D)

Size-Dependent Optical Properties:

As particle size decreases below the exciton Bohr radius:

  • Band gap increases (blue shift in absorption/emission)
  • Discrete energy levels emerge (like particle in a box)
  • Optical properties become tunable by size control

Band Gap in Quantum Dots (Approximate)

$$E_g(dot) \approx E_g(bulk) + \frac{h^2}{8\mu R^2}$$

Where μ = reduced mass of exciton, R = dot radius

1.4.2 Carbon Nanomaterials

Fullerenes (C₆₀ Buckyball)
  • Structure: Truncated icosahedron (soccer ball shape)
  • Composition: 60 carbon atoms arranged in 20 hexagons + 12 pentagons
  • Discovery: 1985 by Kroto, Curl, Smalley (Nobel Prize 1996)
  • Applications: Drug delivery, antioxidants, solar cells, superconductivity (when doped)
Carbon Nanotubes (CNTs)
  • Structure: Cylindrical graphene sheets (rolled up)
  • Types: Single-walled (SWCNT) vs Multi-walled (MWCNT)
  • Diameter: 0.4-3 nm (SWCNT), 2-100 nm (MWCNT)
  • Electronic properties depend on chirality: Can be metallic or semiconducting
  • Mechanical strength: Tensile strength ~100× stronger than steel
  • Applications: Nanoelectronics, composites, drug delivery, field emission displays
Graphene
  • Structure: Single atomic layer of carbon in honeycomb lattice
  • Discovery: 2004 by Geim and Novoselov (Nobel Prize 2010) — isolated using Scotch tape!
  • Remarkable Properties:
Property Graphene Value Comparison
Tensile Strength 130 GPa 200× Steel
Electrical Conductivity 10⁸ S/m Better than copper
Thermal Conductivity 5000 W/mK 10× Copper
Optical Transparency 97.7% Nearly transparent
Surface Area 2630 m²/g Highest known

Applications: Flexible electronics, touchscreens, high-frequency transistors, composite materials, energy storage (supercapacitors), biosensors, water desalination membranes.

Quick Check
6. Quantum dots are also called "artificial atoms" because:
Correct! Quantum confinement creates discrete energy levels similar to atomic orbitals.

Unit 1 Summary

Essential Formulas

de Broglie Wavelength:$\lambda = h/p = h/mv$
Uncertainty Principle:$\Delta x \cdot \Delta p \geq \hbar/2$, $\Delta E \cdot \Delta t \geq \hbar/2$
Time-Dependent SE:$i\hbar \frac{\partial\Psi}{\partial t} = \hat{H}\Psi$
Time-Independent SE:$\hat{H}\psi = E\psi$
Particle in 1D Box:$\psi_n = \sqrt{2/L}\sin(n\pi x/L)$, $E_n = n^2h^2/8mL^2$
Particle in 3D Box:$E = \frac{h^2}{8m}(\frac{n_x^2}{L_x^2}+\frac{n_y^2}{L_y^2}+\frac{n_z^2}{L_z^2})$
Common Mistakes to Avoid:
• Don't confuse h (Planck constant) with ℏ (reduced Planck constant)! ℏ = h/2π
• Remember normalization factor √(2/L) for particle in box—not just sin(nπx/L)
• Zero-point energy exists: E₁ ≠ 0 (ground state has non-zero energy)
• n starts from 1, not 0 (n=0 gives trivial solution ψ=0)
UNIT 2

Solid State Physics

Crystal structures, free electron theory, band theory of solids, and superconductivity with technological applications.

15 Contact Hours 35% Weightage 4 Topics
Unit 2 Syllabus Overview

Recent Breakthroughs

Graphene Superconductivity 2018

Discovery of superconductivity in twisted bilayer graphene at "magic angle" (~1.1°). Potential pathway to room-temperature superconductors and novel quantum devices.

High-Temperature Superconductors 2015-present

Cuprate superconductors, iron-based materials. Hydride compounds under extreme pressure (H₃S, LaH₁₀) reaching near-room temperature superconductivity (>250 K).

Topic 2.1: Classification of Solids

2.1.1 Crystalline Solids

Crystalline Solids

Solids with long-range periodic order of atoms, ions, or molecules. Atoms arranged in regular repeating pattern extending throughout the material.

Key Concepts:

  • Lattice: Infinite array of points with identical surroundings
  • Basis: Atom or group of atoms associated with each lattice point
  • Crystal Structure: = Lattice + Basis
  • Unit Cell: Smallest repeating unit that generates entire crystal by translation
  • Primitive Cell: Smallest possible unit cell containing exactly one lattice point

Bravais Lattices (14 types in 3D):

There are only 14 unique ways to arrange points in 3D space with translational symmetry. These belong to 7 crystal systems:

Crystal System Axis Relations Angle Relations Bravais Types
Cubica=b=cα=β=γ=90°P, I, F (3)
Tetragonala=b≠cα=β=γ=90°P, I (2)
Orthorhombica≠b≠cα=β=γ=90°P, I, F, C (4)
Rhombohedrala=b=cα=β=γ≠90°P (1)
Monoclinica≠b≠cα=γ=90°≠βP, C (2)
Triclinica≠b≠cα≠β≠γ≠90°P (1)
Hexagonala=b≠cα=β=90°, γ=120°P (1)

2.1.2 Amorphous Solids

Amorphous Solids

Solids lacking long-range order. Only short-range order exists (local coordination similar to crystals). Also called "glassy" or "non-crystalline" solids.

Property Crystalline Amorphous
Atomic Arrangement Long-range periodic order Short-range order only
Melting Behavior Sharp melting point Softening range (glass transition)
X-ray Diffraction Sharp peaks Broad halos
Anisotropy Anisotropic (direction-dependent) Isotropic
Examples Metals, diamond, NaCl, quartz Glass, rubber, plastics, gels

2.1.3 Crystal Structures

Four fundamental crystal structures form the basis for most metallic elements:

Property Simple Cubic (SC) Body-Centered Cubic (BCC) Face-Centered Cubic (FCC) Hexagonal Close-Packed (HCP)
Atoms per Unit Cell 1 2 4 6
Coordination Number 6 8 12 12
Atomic Packing Factor 52% 68% 74% 74%
a-r Relationship a = 2r √3a = 4r √2a = 4r c/a = 1.633
Examples Polonium (only!) Fe, Cr, W, Na, K Cu, Al, Ag, Au, Ni Mg, Zn, Ti, Co

APF Calculation Method

$$APF = \frac{\text{Volume of atoms in unit cell}}{\text{Volume of unit cell}} = \frac{n \cdot \frac{4}{3}\pi r^3}{a^3}$$

For FCC: APF = [4 × (4/3)πr³] / [(4r/√2)³] = π√2/6 ≈ 0.74 (74%)
Maximum packing efficiency for equal spheres!

Worked Example: Verify FCC APF

Given: 4 atoms per cell, relation √2a = 4r → a = 4r/√2 = 2√2 r
Volume of atoms = 4 × (4/3)πr³ = (16/3)πr³
Volume of cell = a³ = (2√2 r)³ = 16√2 r³
APF = [(16/3)πr³] / [16√2 r³] = π/(3√2) = π√2/6 ≈ 0.7405 = 74.05% ✓

2.1.4 Miller Indices

Miller Indices (hkl)

Notation system for describing crystallographic planes and directions. Denoted as (hkl) for planes, [hkl] for directions.

Procedure to Determine Miller Indices:

  1. Find intercepts of plane with crystallographic axes (in units of lattice constants a, b, c)
  2. Take reciprocals of these intercepts
  3. Convert to smallest set of integers having same ratio
  4. Enclose in parentheses (hkl)

Interplanar Spacing (Cubic Systems)

$$d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}}$$

Used in X-ray diffraction analysis via Bragg's law: 2d sinθ = nλ

Example: Miller Indices of (110) Plane

Plane intersects x-axis at a, y-axis at a, parallel to z-axis (∞)
Intercepts: (a, a, ∞) → (1, 1, ∞) in units of lattice constants
Reciprocals: (1, 1, 0)
Miller indices: (110) ✓

d-spacing calculation: For Cu (a = 3.61 Å), d₁₁₀ = 3.61/√(1+1+0) = 3.61/√2 = 2.55 Å
Test Yourself
7. The coordination number of FCC structure is:
Correct! Each atom in FCC touches 12 nearest neighbors (4 in same layer, 4 above, 4 below).
8. Which element crystallizes in simple cubic structure?
Correct! Polonium is the only element that naturally crystallizes in simple cubic structure. SC has very low packing efficiency (52%), making it rare.

Topic 2.2: Free Electron Theory of Metals

2.2.1 Classical Drude Model (1900)

Drude Model Assumptions

  1. Metal consists of positive ion cores with free valence electrons
  2. Electrons move freely between collisions (no forces except at collisions)
  3. Collisions are instantaneous and randomize velocity direction
  4. Electron distribution follows Maxwell-Boltzmann statistics
  5. Thermal equilibrium achieved via collisions with ions

Key Results of Drude Model

Drift Velocity:

$$v_d = -\frac{eE\tau}{m} = \mu E$$

Electrical Conductivity:

$$\sigma = \frac{ne^2\tau}{m} = ne\mu$$

Ohm's Law Derived: $J = \sigma E$

Thermal Conductivity:

$$\kappa = \frac{1}{3}nv\ell C_V = \frac{1}{3}v_F^2\tau C_V$$

Wiedemann-Franz Law:

$$\frac{\kappa}{\sigma} = LT \quad \text{where } L = \frac{\pi^2}{3}\left(\frac{k_B}{e}\right)^2 \approx 2.44 \times 10^{-8} \text{ WΩ/K}^2$$
Limitations of Classical Model:
• Predicts wrong temperature dependence: σ ∝ T⁻¹/² (observed σ ∝ T⁻¹)
• Wrong heat capacity: predicts C_V = (3/2)R per mole electrons (observed ~0.01R)
• Cannot explain Hall effect sign anomalies in some metals
• Assumes classical statistics (should use quantum Fermi-Dirac statistics!)

2.2.2 Quantum Sommerfeld Model (1928)

Key Improvement

Replace classical Maxwell-Boltzmann statistics with quantum Fermi-Dirac statistics. Electrons are indistinguishable fermions obeying Pauli exclusion principle.

Fermi-Dirac Distribution

$$f(E) = \frac{1}{\exp\left(\frac{E-E_F}{k_BT}\right) + 1}$$

At T=0: f(E)=1 for EE_F (all empty)
E_F = Fermi energy (chemical potential at T=0)

Fermi Energy for Free Electron Gas (3D)

$$E_F = \frac{\hbar^2}{2m}(3\pi^2 n)^{2/3}$$

Where n = electron density (number per unit volume)

Density of States

$$g(E) = \frac{V}{2\pi^2}\left(\frac{2m}{\hbar^2}\right)^{3/2}\sqrt{E} \propto \sqrt{E}$$

Number of available states per unit energy interval. Increases with √E.

Electron Heat Capacity (Major Success!)

$$C_e = \gamma T = \frac{\pi^2}{2}\frac{k_BT}{E_F}Nk_B$$

Linear in T (unlike classical 3/2 Nk_B constant)!
Explains why electron contribution to heat capacity is tiny at room temperature.
Only electrons near E_F (~k_BT range) can be thermally excited.

Example: Sodium (Na)

Given: n = 2.65 × 10²⁸ m⁻³ (one valence electron per atom)
$E_F = \frac{(1.055\times10^{-34})^2}{2(9.11\times10^{-31})}(3\pi^2 \times 2.65\times10^{28})^{2/3}$
$E_F = 3.15$ eV
$T_F = E_F/k_B = 36,500$ K (very high!)
$v_F = \sqrt{2E_F/m} = 1.07 \times 10^6$ m/s (~0.003c)
γ = π²k_B²N/(2E_F) = 1.38 × 10⁻³ J/(mol·K) (matches experiment!)
Aspect Classical (Drude) Quantum (Sommerfeld)
Statistics Maxwell-Boltzmann Fermi-Dirac
Heat Capacity (3/2)R per mole (wrong!) γT (linear, correct!)
σ(T) Dependence T⁻¹/² (wrong) T⁻¹ (correct, due to τ ∝ T⁻¹)
Wiedemann-Franz L = constant (works) L = constant (still works!)
Mean Free Path ~0.1 nm (too small) ~10-100 nm (matches experiment)

Topic 2.3: Band Theory of Solids

2.3.1 Origin of Energy Bands

How Bands Form

As atoms come together to form a solid, their atomic orbitals overlap and split into closely spaced energy levels, forming quasi-continuous bands.

Molecular Orbital Approach:

  1. Isolated atom: discrete, sharp energy levels
  2. Two atoms: each level splits into two (bonding/antibonding)
  3. N atoms: each level splits into N closely spaced levels → forms an energy band
  4. Band width increases with orbital overlap (smaller interatomic distance)

Kronig-Penney Model (Qualitative):

Periodic potential V(x+a) = V(x) leads to:

  • Allowed bands: Energy ranges where propagating solutions exist
  • Forbidden gaps (band gaps): Energy ranges with no allowed states
  • Brillouin zones: Regions in k-space separated by zone boundaries at k = ±nπ/a

Bloch Theorem

$$\psi_k(x) = u_k(x)e^{ikx}$$

Where u_k(x) has periodicity of lattice: u_k(x+a) = u_k(x)
Wave function = plane wave modulated by periodic function

2.3.2 Classification by Band Structure

Material Type Band Structure Typical E_g Examples σ Range (S/m)
Conductors (Metals) Partially filled OR overlapping bands 0 (or negative) Cu, Al, Ag, Au, Fe 10⁶ - 10⁸
Semiconductors Small gap (filled VB, empty CB) 0.1 - 3 eV Si (1.1), Ge (0.67), GaAs (1.43) 10⁻⁶ - 10⁴
Insulators Large gap (filled VB, empty CB) > 3 eV Diamond (5.5), SiO₂ (9), Al₂O₃ (8) < 10⁻¹⁰
Direct vs Indirect Band Gap:
Direct: CB minimum and VB maximum at same k-value → efficient light emission (LEDs, laser diodes)
Indirect: Minima at different k-values → requires phonon assistance for transitions (less efficient optically)
• GaAs = direct (good for optoelectronics), Si = indirect (poor light emitter)

2.3.3 Effective Mass

Definition

The effective mass describes how an electron in a crystal responds to external forces, accounting for the influence of the periodic lattice potential.

Effective Mass Formula

$$m^* = \frac{\hbar^2}{d^2E/dk^2}$$

Depends on curvature of E-k diagram:
• High curvature (large d²E/dk²) → small |m*| ("light" carrier)
• Low curvature → large |m*| ("heavy" carrier)
• Near bottom of conduction band: m* > 0
• Near top of valence band: m* < 0

Holes: Absence of electron in nearly-filled valence band behaves as positive charge carrier with positive effective mass: m*_h = -m*_e (where m*_e is negative near VB top).

Importance: Determines mobility μ = eτ/m*, transport properties, optical response, and device performance.

Topic 2.4: Superconductivity

2.4.1 Discovery and Basic Properties

Discovery (1911)

Heike Kamerlingh Onnes (Leiden University) discovered that mercury's electrical resistance suddenly drops to zero below 4.2 K. Nobel Prize in Physics 1913.

Fundamental Properties:

Property Description Significance
Zero Resistance R = 0 for T < T_c Persistent currents flow without decay (observed for years)
Meissner Effect (1933) Perfect diamagnetism: B = 0 inside Distinguishes SC from perfect conductor
Critical Temperature (T_c) Below this, SC occurs Material-dependent (0.001 K to >150 K)
Critical Field H_c Above this, SC destroyed H_c(T) = H_c(0)[1-(T/T_c)²]

Critical Magnetic Field vs Temperature

$$H_c(T) = H_c(0)\left[1 - \left(\frac{T}{T_c}\right)^2\right]$$

Parabolic decrease from H_c(0) at T=0 to zero at T=T_c

Type I vs Type II Superconductors:

Feature Type I Type II
Meissner Effect Complete flux expulsion Partial (mixed/vortex state)
Critical Fields Single H_c Two: H_c1 (lower), H_c2 (upper)
Transition Abrupt (first-order) Gradual (second-order)
T_c Range Low (<10 K) Can be high (>77 K)
Examples Pb (7.2K), Hg (4.2K), Al (1.2K) Nb (9.2K), NbTi (10K), YBCO (93K)
Applications Limited Magnets, cables, electronics

2.4.2 BCS Theory (Qualitative, 1957)

Nobel Prize 1972

Bardeen, Cooper, Schrieffer developed microscopic theory explaining conventional superconductivity.

Key Concepts:

Cooper Pairs:
  • Two electrons with opposite momenta and spins: (k↑, -k↓)
  • Bound together via phonon exchange (lattice vibration mediated attraction)
  • Pair binding energy ~meV scale (much smaller than Fermi energy ~eV)
  • All Cooper pairs condense into same quantum ground state (macroscopic coherence)
Electron-Phonon Interaction Mechanism:
  1. Electron 1 passes through lattice, attracting positive ions
  2. Region of enhanced positive density forms
  3. This attracts Electron 2
  4. Net effect: effective attraction overcoming Coulomb repulsion

BCS Energy Gap Prediction

$$2\Delta(0) = 3.528\,k_B T_c$$

Universal ratio! Energy gap at T=0 is proportional to T_c.
No electronic states exist within ±Δ of Fermi level → explains zero resistance (insufficient energy to break pairs).

Isotope Effect: $T_c \propto M^{-1/2}$ confirms role of lattice vibrations (phonons) in pairing mechanism.

Limitations of BCS Theory:
• Only explains conventional (low-T_c) superconductors
• Cannot explain high-T_c cuprates (T_c > 30 K)
• Fails for iron-based superconductors
• Room-temperature superconductivity remains elusive

2.4.3 Applications of Superconductors

Application SC Property Exploited Technology Details Status
MRI Machines Zero resistance + High current NbTi coils, 1.5-7 Tesla fields, liquid He cooling Widespread medical use
Maglev Trains Meissner effect (levitation) Shanghai (431 km/h), Japan MLX01 (581 km/h test) Commercial operation
Particle Accelerators High magnetic fields LHC: 1232 dipole magnets, 8.33 T, 1.9 K Operational (CERN)
SQUID Magnetometers Flux quantization Most sensitive detector: 10⁻¹⁵ T resolution Research & medical
Power Transmission Zero resistance (no loss) Prototype cables demonstrated Pilot projects
Fusion Reactors Plasma confinement magnets ITER project, Nb₃Sn coils Under construction
Unit 2 Review Questions
9. The Meissner effect refers to:
Correct! Perfect diamagnetism (B=0 inside) distinguishes superconductor from perfect conductor.
10. In BCS theory, superconductivity arises due to formation of:
Correct! Electron-lattice-electron interaction creates attractive force forming Cooper pairs.

Unit 2 Summary

Essential Formulas

APF (FCC):π√2/6 ≈ 0.74
Miller Indices d-spacing:d_{hkl} = a/√(h²+k²+l²)
Drude Conductivity:σ = ne²τ/m
Fermi Energy:E_F = (ℏ²/2m)(3π²n)^(2/3)
Effective Mass:m* = ℏ²/(d²E/dk²)
Critical Field:H_c(T) = H_c(0)[1-(T/T_c)²]
BCS Gap Ratio:2Δ(0) = 3.528 k_B T_c
Exam-Critical Points:
• SC: CN=6, BCC: CN=8, FCC/HCP: CN=12 (memorize!)
• Atoms/cell: SC=1, BCC=2, FCC=4 (count carefully!)
• FCC a-r relation: √2a = 4r (most commonly tested)
• Type I: abrupt transition, Type II: mixed state between H_c1-H_c2
• BCS theory applies ONLY to conventional (low-T_c) superconductors
UNIT 3

Lasers and Fiber Optics

Principles of laser operation, types of lasers, optical fiber technology, and fiber optic communication systems.

15 Contact Hours 30% Weightage 4 Topics
Unit 3 Syllabus Overview

Cutting-Edge Developments

Advanced Laser Technology 2010-present

Ultrashort pulse lasers (femtosecond, attosecond), LIDAR for autonomous vehicles, laser-driven inertial confinement fusion (National Ignition Facility), quantum cascade lasers for sensing and communication.

Fiber Optic Innovations 2015-present

Photonic crystal fibers with unprecedented properties, space-division multiplexing for massive capacity increase, submarine cables connecting continents, 400G/800G transmission systems deployed globally.

Topic 3.1: Basics of Light-Matter Interaction

3.1.1 Spontaneous and Stimulated Emission

Three Fundamental Processes

Interaction of radiation with matter involves three distinct processes, each with different characteristics.

Process Description Rate Characteristics
Absorption Atom in ground state absorbs photon → excited $B_{12}N_1\rho(\nu)$ Requires incoming photon
Spontaneous Emission Excited atom decays randomly emitting photon $A_{21}N_2$ Random direction, phase, polarization (incoherent)
Stimulated Emission ★ Incident photon triggers coherent emission $B_{21}N_2\rho(\nu)$ Same direction, phase, frequency, polarization (coherent)

Einstein Coefficients Relations

At thermal equilibrium, detailed balance gives:

$$B_{12} = B_{21} \quad \text{(absorption = stimulated emission coefficient)}$$ $$\frac{A_{21}}{B_{21}} = \frac{8\pi h\nu^3}{c^3}$$

Key insight: Spontaneous emission rate A₂₁ increases as ν³ (cubic frequency dependence). This makes X-ray and gamma-ray lasers extremely difficult to build!

Why Stimulated Emission is Special:
The emitted photon is an exact clone of the stimulating photon:
• Same frequency (ν)
• Same direction of propagation
• Same phase (coherent)
• Same polarization
→ This is what enables amplification and laser action!

3.1.2 Population Inversion

Critical Condition for Lasing

Population inversion: N₂ > N₁ (more atoms in upper state than lower state). Required for net stimulated emission (optical gain).

Normal Boltzmann Distribution: At thermal equilibrium, N₂ < N₁ always. Net absorption dominates. Cannot achieve lasing!

Why Two-Level System Fails: Even with intense pumping, best case is N₁ ≈ N₂ (saturation), never N₂ > N₁.

Three-Level System (e.g., Ruby Laser):

  • Pump: Ground (E₁) → E₃ (excited state)
  • Fast decay: E₃ → E₂ (metastable, long lifetime ~ms)
  • Accumulation in E₂ → Inversion between E₂ and E₁
  • Disadvantage: Must pump >50% of atoms to E₂ (high threshold)

Four-Level System (e.g., He-Ne Laser):

  • Pump: E₀ → E₃ → Fast decay to E₂ (metastable upper laser level)
  • Laser transition: E₂ → E₁
  • Fast decay: E₁ → E₀ (ground state)
  • Advantage: E₁ normally empty → easy inversion (low threshold!)
Aspect Two-Level Three-Level Four-Level
Inversion Possible? ✗ No ✓ Yes (hard) ✓ Yes (easy)
Pump Requirement N/A >50% atoms excited Much lower threshold
Laser Type Example None Ruby laser He-Ne, semiconductor lasers
Operation Mode - Pulsed typically CW possible
Quick Check
11. Stimulated emission produces light that is:
Correct! Stimulated emission produces photons identical to stimulating photon in frequency, direction, phase, and polarization.

Topic 3.2: Laser Principles

3.2.1 Components of a Laser

Component Function Examples/Types
Active Medium (Gain Medium) Provides energy levels for lasing; determines output wavelength Ruby, Nd:YAG, He-Ne gas, CO₂, GaAs semiconductor, dye solutions
Pumping Mechanism Populates upper laser level; creates population inversion Optical (flash lamp), Electrical (discharge/injection), Chemical, Particle beam
Optical Resonator (Cavity) Provides feedback; stores energy; determines output characteristics Plane-parallel (Fabry-Perot), Confocal, Hemispherical, Unstable
Output Coupler Partially reflecting mirror; transmits useful output while maintaining feedback 1-99% reflectivity depending on gain medium

3.2.2 Lasing Action & Threshold Condition

Threshold Condition

For lasing to occur, round-trip gain must equal or exceed losses:

$$R_1 R_2 \exp[2(\gamma_{th} - \alpha)L] = 1$$

Where: R₁, R₂ = mirror reflectivities
γ_th = threshold gain coefficient
α = distributed loss coefficient
L = cavity length

Mode Selection by Cavity:

  • Longitudinal modes: Standing waves along cavity axis. Resonance condition: $2L = m\lambda$ (m = integer)
  • Frequency spacing: $\Delta\nu = c/(2L)$ (free spectral range)
  • Transverse modes (TEM_mn): Intensity patterns perpendicular to axis. TEM₀₀ (Gaussian) is most desirable.

Quality Factor (Q-factor)

$$Q = \frac{\nu}{\Delta\nu} = \frac{\text{Energy stored}}{\text{Energy lost per cycle}}$$

High Q → narrow linewidth, low threshold, high coherence

3.2.3 Laser Characteristics

Characteristic Description Quantitative Measure Comparison to Ordinary Light
Monochromaticity Very narrow spectral width $\Delta\lambda \sim 10^{-10}$ m (gas lasers) ~nm range for ordinary sources
Coherence Fixed phase relationship Coherence length: meters to km ~μm (incoherent)
Directionality Highly collimated beam Divergence: ~mrad Isotropic (4π sr)
Brightness High power per area/solid angle >10¹⁵ W (pulsed) Many orders lower
Coherence Length Formula:
$$l_c = \frac{c}{\Delta\nu} = \frac{\lambda^2}{\Delta\lambda}$$
For He-Ne laser (λ=632.8 nm, Δλ≈10⁻¹⁵ m): l_c ≈ 400 m!
For ordinary light (Δλ~10 nm): l_c ≈ 40 μm
Test Yourself
12. The function of optical resonator in a laser is:
Correct! The resonator provides optical feedback (photons make multiple passes) and determines beam properties.

Topic 3.3: Types of Lasers

3.3.1 Ruby Laser (Three-Level, 1960 - First Laser!)

The World's First Laser

Built by Theodore Maiman in 1960 at Hughes Research Laboratories. Used synthetic ruby crystal (Al₂O₃ doped with Cr³⁺ ions).

Specifications:

  • Active medium: Ruby rod (Al₂O₃ + 0.05% Cr³⁺) - pink color from Cr³⁺ absorption
  • Pumping: Optical - xenon flash lamp (green/blue light absorbed)
  • Output wavelength: 694.3 nm (deep red)
  • Operation mode: Pulsed only (ms pulse duration)
  • Efficiency: ~1% (quite low)

Energy Level Diagram Summary

Ground state: ⁴A₂ → Pump bands: ⁴F₁, ⁴F₂ (wide, short-lived)

Fast non-radiative decayMetastable ²E level (lifetime ~3 ms)

Laser transition: ²E → ⁴A₂ emitting 694.3 nm photon

Applications: Holography, welding, range finding, research (historical significance). Largely replaced by more efficient lasers today.

3.3.2 He-Ne Laser (Four-Level, 1961)

Most Common Educational Laser

Helium-Neon gas mixture laser. Continuous wave operation with excellent beam quality.

Specifications:

  • Active medium: He-Ne gas mixture (~10:1 ratio) at ~1 torr pressure
  • Pumping: Electrical discharge (RF or DC glow discharge)
  • Primary output: 632.8 nm (bright red) - most common
  • Other wavelengths: 543.5 nm (green), 594.1 nm (yellow), 1523 nm (IR)
  • Operation mode: Continuous wave (CW)
  • Output power: 0.5 - 50 mW typical
  • Efficiency: <0.1% (very low but acceptable)

How It Works (Energy Transfer):

  1. Electrical discharge excites He atoms to metastable states 2¹S and 2³S
  2. Resonant energy transfer to Ne atoms (nearly exact energy match!)
  3. Neon populates 5s and 4s states (upper laser levels)
  4. Laser transition occurs (e.g., 5s → 3p for 632.8 nm)
  5. Fast decay from lower laser level maintains inversion

Advantages of Four-Level System: Low threshold, CW operation, excellent beam quality (TEM₀₀), visible output, relatively inexpensive.

Applications: Barcode scanners, alignment tools, interferometry, holography, demonstration labs, laser printers (older models), surveying.

3.3.3 Semiconductor Laser (Diode Laser, 1962)

The Workhorse of Modern Photonics

p-n junction laser using direct bandgap semiconductor. Compact, efficient, and mass-producible.

Structure & Operation:

  • Active region: p-n junction in direct bandgap material (GaAs, InGaAsP, GaN)
  • Design: Double heterostructure (confinement layer between wider bandgap cladding)
  • Pumping: Forward bias injection (electrons from n-side, holes from p-side)
  • Population inversion: At junction under forward bias
  • Emission wavelength: $\lambda \approx hc/E_g$ (determined by bandgap)
Material Wavelength Application
GaAs~850 nmCD players, optical mice
InGaAsP1300-1550 nmFiber optic communications
GaN~405 nmBlu-ray drives
AlGaInP~650 nmLaser pointers, DVD

Key Advantages: Extremely compact (mm scale), high efficiency (30-50%), low power consumption, direct current modulation (GHz speeds), inexpensive (mass-produced), robust.

Limitations: Highly divergent beam (needs collimation), temperature-sensitive wavelength drift, lower coherence than gas lasers.

Applications: Fiber optic transmitters, CD/DVD/Blu-ray drives, laser pointers, barcode scanners, laser printers, free-space communication, pumping solid-state lasers, lidar, medical devices.

Complete Laser Comparison:
Property Ruby Laser He-Ne Laser Semiconductor Laser
Type 3-level 4-level 4-level
Active Medium Ruby (Cr³⁺:Al₂O₃) He-Ne gas mix GaAs/InGaN junction
Pumping Optical (flash lamp) Electrical discharge Current injection
Wavelength 694.3 nm (red) 632.8 nm (red, primary) Varies (UV-IR)
Output Mode Pulsed Continuous (CW) CW or pulsed
Efficiency ~1% <0.1% 30-50%
Size Large (cm scale) Medium (10-100 cm) Tiny (<1 mm chip)
Cost High Moderate Very low (mass-produced)
Beam Quality Good Excellent (TEM₀₀) Fair (needs optics)
Main Applications Holography, welding Scanners, alignment, labs Comms, storage, pointers
Review Questions
13. Ruby laser operates on how many energy levels?
Correct! Ruby is a three-level laser: ground state → pump band → metastable → ground state (laser transition).
14. Which laser has the highest efficiency among these?
Correct! Semiconductor lasers achieve 30-50% efficiency due to direct electrical-to-optical conversion without intermediate steps.

Topic 3.4: Optical Fibers

3.4.1 Total Internal Reflection (TIR)

Foundation of Fiber Optics

Total internal reflection traps light within the fiber core, enabling long-distance signal transmission with minimal loss.

Critical Angle Condition

From Snell's Law: $n_1\sin\theta_1 = n_2\sin\theta_2$

Critical Angle:

$$\theta_c = \sin^{-1}\left(\frac{n_2}{n_1}\right) \quad \text{where } n_1 > n_2$$

Condition for TIR:

  1. Light travels from denser to rarer medium ($n_1 > n_2$)
  2. Angle of incidence $\theta_i > \theta_c$

Result: 100% reflection, no transmission into rarer medium. Evanescent wave penetrates slightly but carries no net energy.

Example Calculations:

Glass-Air interface: n₁=1.5, n₂=1.0
θ_c = sin⁻¹(1.0/1.5) = sin⁻¹(0.667) = 41.8°

Glass-Water interface: n₁=1.5, n₂=1.33
θ_c = sin⁻¹(1.33/1.5) = sin⁻¹(0.887) = 62.5°

Note: Higher n₂ → larger θ_c → TIR harder to achieve!

3.4.2 Fiber Structure and Types

Basic Structure (from center outward):

  1. Core: Central cylinder where light propagates. Diameter: 8-62.5 µm. Refractive index n₁ (higher).
  2. Cladding: Surrounding layer. Standard diameter: 125 µm. Index n₂ < n₁. Provides TIR condition.
  3. Jacket/Coating: Protective polymer layer. Diameter: 250-900 µm. Mechanical protection.

Numerical Aperture (NA)

$$NA = \sqrt{n_1^2 - n_2^2} = n_0\sin\theta_a$$

NA measures light-gathering ability. Larger NA accepts more light but increases modal dispersion.
θ_a = acceptance angle (maximum entrance angle from fiber axis)

Acceptance Cone

All rays entering within cone of half-angle θ_a will be guided by TIR:

$$\theta_a = \sin^{-1}(NA) = \sin^{-1}\left(\sqrt{n_1^2 - n_2^2}\right)$$
Feature Single-Mode Fiber (SMF) Multi-Mode Fiber (MMF)
Core diameter 8-10 µm 50 µm or 62.5 µm
Number of modes 1 (fundamental HE₁₁ only) Many (hundreds to thousands)
Modal dispersion None (major advantage!) Significant (limits bandwidth)
Bandwidth Very high (THz·km potential) Moderate (MHz·km to GHz·km)
Light source Laser diode required LED or laser diode
Distance capability 100+ km (telecom) < 2 km (LANs, data centers)
Cost Higher (precision components) Lower
Applications Telecom, CATV, long-haul LANs, data centers, short links

Index Profiles:

  • Step-index: Abrupt change n₁→n₂ at core-cladding boundary. Simple but higher dispersion in MMF.
  • Graded-index: n decreases parabolically from center. Reduces modal dispersion by equalizing mode velocities.

3.4.3 Attenuation and Dispersion

Attenuation (Signal Power Loss):

Exponential Decay Law

$$P(z) = P_0 \exp(-\alpha z)$$

In dB/km: $\alpha_{dB} = \frac{10}{L}\log_{10}(P_{in}/P_{out})$ dB/km

Loss Mechanisms:

Type Cause Characteristics
Absorption Intrinsic (UV/IR tails), Impurities (OH⁻ ions, metals) OH⁻ peak at 1383 nm problematic historically
Rayleigh Scattering Density fluctuations (frozen-in) ∝ λ⁻⁴ dominant loss at telecom wavelengths
Bending Losses Macrobending (sharp bends), Microbending Increases sharply below critical bend radius

Telecom Windows:

  • 850 nm: ~2 dB/km, LED sources, MMF, short reach
  • 1310 nm: ~0.4 dB/km, zero material dispersion wavelength, SMF
  • 1550 nm: ~0.2 dB/km (minimum attenuation!), SMF with EDFA amplifiers
Dispersion (Pulse Broadening):
Dispersion Type Cause Affecting Factor Mitigation
Material Dispersion n varies with λ Source spectral width Use narrow-linewidth laser
Modal Dispersion Different path lengths per mode MMF only Use SMF or graded-index MMF
Waveguide Dispersion k-dependent propagation constant SMF near cutoff Dispersion-shifted fibers
PMD Birefringence (polarization modes) Very high-speed systems Polarization-maintaining fiber
Numerical Example:

Problem: Input power = 1 mW into 20 km fiber with α = 0.25 dB/km. Find output power.

Solution:
Total loss = α × L = 0.25 × 20 = 5 dB
P_out = P_in × 10^(-loss/10) = 1 × 10^(-0.5) = 0.316 mW
Or: P_out/P_in = 10^(-5/10) = 0.316 → P_out = 0.316 mW ✓

Verification: About 68% power lost over 20 km (reasonable for older fiber)

3.4.4 Fiber Optic Communication System

Complete System Overview

End-to-end conversion of electrical signals to optical transmission and back to electrical reception.

Block Diagram:

Information Input → Transmitter → Optical Signal → Fiber CableReceiver → Output
Component Function Types/Options
Light Source (Tx) Converts electrical → optical signals LED (cheap, broad spectrum), LD (high power, coherent)
Optical Fiber Transmission medium with minimal loss SMF (long haul), MMF (short reach)
Photodetector (Rx) Converts optical → electrical signals PIN diode (moderate sensitivity), APD (high sensitivity, gain)
Optical Amplifier Boosts signal without O-E-O conversion EDFA (Erbium-Doped Fiber Amplifier) for 1550 nm window
WDM System Multiple channels on single fiber DWDM (dense), CWDM (coarse)

Wavelength Division Multiplexing (WDM):

  • Multiple signals at different wavelengths transmitted simultaneously
  • DWDM: 40-96 channels, 0.4-0.8 nm spacing (50-100 GHz)
  • CWDM: 18 channels, 20 nm spacing (lower cost)
  • Enables massive capacity increase (Tb/s on single fiber!)

Advantages over Copper:

  • 📡 Enormous bandwidth (THz potential vs MHz for copper)
  • 📉 Ultra-low attenuation (0.2 dB/km vs 10+ dB/km for high-frequency copper)
  • 🛡️ Immunity to EMI/RFI interference
  • ⚡ No ground loops or spark hazards
  • 💪 Lightweight and small diameter
  • 🔒 Security (very difficult to tap without detection)
  • 💰 Long-term cost savings despite higher initial cost
Unit 3 Final Questions
15. Numerical aperture of an optical fiber depends on:
Correct! NA = √(n₁² - n₂²) depends only on refractive indices of core (n₁) and cladding (n₂).
16. The attenuation minimum for silica optical fiber occurs at approximately:
Correct! 1550 nm is the third telecom window with minimum attenuation (~0.2 dB/km), used for long-haul communication with EDFAs.

Unit 3 Summary

Essential Formulas

Einstein Relations:B₁₂=B₂₁, A₂₁/B₂₁=8πhν³/c³
Critical Angle:θ_c = sin⁻¹(n₂/n₁)
Numerical Aperture:NA = √(n₁²-n₂²) = sinθ_a
Longitudinal Mode Spacing:Δν = c/(2L)
Attenuation (dB):α_dB = (10/L)log(P_in/P_out)
V-parameter:V = (2πa/λ)NA (determines mode count)
Exam-Critical Points:
• Remember: 3-level lasers need >50% excitation (hard!), 4-level much easier
• Ruby = 694.3 nm red, He-Ne = 632.8 nm red (most common)
• Semiconductor lasers: highest efficiency, smallest size, cheapest
• TIR requires: n_core > n_clad AND θ_i > θ_c
• SMF: no modal dispersion, uses laser only; MMF: can use LED
• Three telecom windows: 850, 1310, 1550 nm (know their purposes!)

Flashcard Study System

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de Broglie Wavelength

Formula: λ = h/p = h/mv

Relates particle momentum to its wave-like wavelength. Observable only at atomic scales.

Comprehensive Formula Sheet

Print-ready reference sheet containing all essential formulas organized by unit. Use this for quick revision before exams.

Unit 1: Quantum Physics & Nanotechnology

Wave-Particle Duality

de Broglie wavelength$\lambda = \dfrac{h}{p} = \dfrac{h}{mv}$
Electron wavelength shortcut$\lambda = \dfrac{1.226}{\sqrt{V}}$ nm (V in volts)

Uncertainty Principle

Position-Momentum$\Delta x \cdot \Delta p_x \geq \dfrac{\hbar}{2}$
Energy-Time$\Delta E \cdot \Delta t \geq \dfrac{\hbar}{2}$
Confined particle min KE$E_{min} \geq \dfrac{\hbar^2}{8mL^2}$

Schrödinger Equation

Time-Dependent$i\hbar\dfrac{\partial\Psi}{\partial t} = \hat{H}\Psi$
Time-Independent$\hat{H}\psi = E\psi$
Normalization$\int|\Psi|^2 dV = 1$

Particle in 1D Box

Wave function$\psi_n(x) = \sqrt{\dfrac{2}{L}}\sin\left(\dfrac{n\pi x}{L}\right)$
Energy eigenvalues$E_n = \dfrac{n^2h^2}{8mL^2} = \dfrac{n^2\pi^2\hbar^2}{2mL^2}$
Zero-point energy$E_1 = \dfrac{h^2}{8mL^2} \neq 0$

Particle in 3D Box

Energy$E_{n_x,n_y,n_z} = \dfrac{h^2}{8m}\left(\dfrac{n_x^2}{L_x^2}+\dfrac{n_y^2}{L_y^2}+\dfrac{n_z^2}{L_z^2}\right)$
Cubic box special case$E = \dfrac{h^2}{8mL^2}(n_x^2+n_y^2+n_z^2)$

Unit 2: Solid State Physics

Crystal Structures

SC: atoms/cell=1, CN=6, APF=52%, a=2r
BCC: atoms/cell=2, CN=8, APF=68%, √3a=4r
FCC: atoms/cell=4, CN=12, APF=74%, √2a=4r
HCP: atoms/cell=6, CN=12, APF=74%, c/a=1.633

Miller Indices

Interplanar spacing (cubic)$d_{hkl} = \dfrac{a}{\sqrt{h^2+k^2+l^2}}$
Bragg's law$n\lambda = 2d\sin\theta$

Free Electron Theory

Drude conductivity$\sigma = \dfrac{ne^2\tau}{m}$
Wiedemann-Franz$\dfrac{\kappa}{\sigma} = LT$, $L = 2.44\times10^{-8}$ WΩ/K²
Fermi energy (3D)$E_F = \dfrac{\hbar^2}{2m}(3\pi^2 n)^{2/3}$
Density of states$g(E) = \dfrac{V}{2\pi^2}\left(\dfrac{2m}{\hbar^2}\right)^{3/2}\sqrt{E}$
Electron heat capacity$C_e = \gamma T = \dfrac{\pi^2}{2}\dfrac{k_BT}{E_F}Nk_B$

Band Theory & Superconductivity

Effective mass$m^* = \dfrac{\hbar^2}{d^2E/dk^2}$
Critical field$H_c(T) = H_c(0)\left[1-\left(\dfrac{T}{T_c}\right)^2\right]$
BCS gap ratio$2\Delta(0) = 3.528\,k_B T_c$

Unit 3: Lasers & Fiber Optics

Einstein Coefficients & Laser Action

Coefficient relation$B_{12}=B_{21}, \quad \dfrac{A_{21}}{B_{21}}=\dfrac{8\pi h\nu^3}{c^3}$
Threshold condition$R_1R_2\exp[2(\gamma_{th}-\alpha)L]=1$
Longitudinal mode spacing$\Delta\nu = \dfrac{c}{2L}$
Coherence length$l_c = \dfrac{\lambda^2}{\Delta\lambda}$

Optical Fibers

Critical angle$\theta_c = \sin^{-1}\left(\dfrac{n_2}{n_1}\right)$
Numerical aperture$NA = \sqrt{n_1^2-n_2^2} = \sin\theta_a$
Acceptance angle$\theta_a = \sin^{-1}(NA)$
Attenuation (dB/km)$\alpha_{dB} = \dfrac{10}{L}\log_{10}\left(\dfrac{P_{in}}{P_{out}}\right)$
Power decay$P(z) = P_0 e^{-\alpha z}$
V-parameter$V = \dfrac{2\pi a}{\lambda}NA$
Number of modes (MMF)$M \approx V^2/2$ (step-index)

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UNIT 1 | 5 MARKS | DERIVATION

Derive the expression for energy eigenvalues of a particle confined in a one-dimensional infinite potential well.

UNIT 2 | 5 MARKS | THEORY

Explain the differences between Type I and Type II superconductors with examples.

UNIT 3 | 5 MARKS | THEORY

Describe the working principle of He-Ne laser with energy level diagram. Why is it called a four-level laser?

UNIT 3 | 5 MARKS | NUMERICAL

An optical fiber has core refractive index 1.50 and cladding index 1.46. Calculate: (a) Numerical aperture, (b) Acceptance angle, (c) Critical angle.

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