In a crystal, atoms are arranged in a regular, repeating 3-D pattern. This topic explains lattice, basis and unit cell, the seven crystal systems and fourteen Bravais lattices, the simple cubic, BCC and FCC structures, Miller indices and the spacing between lattice planes.
Key points
Crystalline solids have long-range order and sharp melting points; amorphous solids (glass) do not.
Lattice: a regular array of points in space, each with identical surroundings.
Basis: the atom or group of atoms attached to each lattice point. Crystal structure = lattice + basis.
Unit cell: the smallest repeating block; described by edges a, b, c and angles α, β, γ.
Coordination number: number of nearest neighbours of an atom.
Packing fraction: fraction of unit-cell volume filled by atoms.
Miller indices (h k l) describe a plane: take intercepts in units of a, b, c, take reciprocals, clear fractions to smallest integers.
Planes parallel to an axis have intercept ∞ on it, so index 0.
Examples: Polonium is SC; iron (α), sodium are BCC; copper, aluminium, gold are FCC; NaCl is FCC lattice with a two-ion basis.
Important terms
Lattice parameters – a, b, c and α, β, γ.
Primitive cell – unit cell with lattice points only at corners (one lattice point per cell).
Symmetry elements – centre, planes and axes of symmetry.
Crystal (स्फटिक) – a solid with regular repeating arrangement of atoms.
Formulas
Property
SC
BCC
FCC
Atoms per cell
1
2
4
Coordination number
6
8
12
Relation a and r
a = 2r
a = 4r/√3
a = 2√2 r
Packing fraction
π/6 ≈ 0.52
√3π/8 ≈ 0.68
π/(3√2) ≈ 0.74
Interplanar spacing (cubic): d = a/√(h² + k² + l²)
Density: ρ = nM/(N_A a³) (n = atoms per cell, M = molar mass in kg/mol)
5 important questions
Q1. What is a Bravais lattice? How many are there?
A Bravais lattice is an infinite array of points in which every point has exactly the same surroundings. In three dimensions there are 14 Bravais lattices grouped in 7 crystal systems.
Q2. Find the Miller indices of a plane with intercepts 2a, 3b and ∞ (parallel to the z-axis).
Intercepts in units: 2, 3, ∞. Reciprocals: 1/2, 1/3, 0. Multiply by LCM 6: 3, 2, 0. Miller indices = (3 2 0).
Q3. Calculate the packing fraction of a BCC lattice.
Atoms per cell = 8 × (1/8) + 1 = 2.
Atoms touch along the body diagonal: √3 a = 4r, so r = √3a/4.
Q5. Explain Miller indices and find the spacing of (1 1 0) planes in a cubic crystal with a = 4 Å.
Steps: find intercepts, express in units of a, b, c; take reciprocals; reduce to smallest integers; enclose in ( ).
Equally spaced parallel planes share the same indices.
For cubic crystals, d = a/√(h² + k² + l²).
d₁₁₀ = 4/√(1 + 1 + 0) = 4/√2 ≈ 2.83 Å.
Note: d₁₀₀ = 4 Å, d₁₁₁ = 4/√3 ≈ 2.31 Å — higher indices give closer planes.
Exam tip
Learn the SC–BCC–FCC table thoroughly; most short questions and numericals come from it. Draw the three cubic unit cells and the (100), (110), (111) planes neatly.
Atoms in a solid are held together by bonds. The type of bond decides whether the solid is hard or soft, a conductor or an insulator, and what its melting point is. This topic covers interatomic forces and the potential-energy curve, cohesive energy, and the five types of bonding: ionic, covalent, metallic, Van der Waals and hydrogen bonds, with the lattice energy of ionic crystals.
Key points
Between two atoms there is a long-range attraction and a short-range repulsion (due to overlap of electron clouds and the Pauli principle).
Potential energy U(r) = −A/rᵐ + B/rⁿ, with n greater than m. At equilibrium separation r₀, dU/dr = 0 and U is minimum; the depth of this minimum is the bond energy.
Cohesive energy: energy needed to separate a solid into free neutral atoms.
Ionic bond: transfer of electrons (Na⁺Cl⁻). Hard, brittle, high melting point, insulators as solids, conduct when molten; soluble in water.
Covalent bond: sharing of electron pairs (diamond, Si, Ge). Very hard, very high melting point, directional bonds, poor conductors.
Metallic bond: positive ions in a "sea" of free electrons (Cu, Na). Good electrical and thermal conductors, malleable, ductile, shiny.
Van der Waals (molecular) bond: weak attraction between induced dipoles (solid argon, iodine). Soft, low melting point.
Hydrogen bond: H atom between two strongly electronegative atoms (ice, DNA). Strength between Van der Waals and covalent.
Bond strength order (roughly): covalent ≈ ionic greater than metallic greater than hydrogen greater than Van der Waals.
Madelung constant α accounts for the attraction and repulsion from all ions in the crystal; for NaCl α = 1.748.
Important terms
Equilibrium spacing (r₀) – separation at which the net force between atoms is zero.
Lattice energy – energy released when free ions come together to form one mole of an ionic crystal.
Ionisation energy – energy needed to remove an electron from a neutral atom.
Electron affinity – energy released when a neutral atom gains an electron.
Born exponent (n) – power in the repulsive term, usually 6–12.
Formulas
U(r) = −A/rᵐ + B/rⁿ
Force F = −dU/dr; at r₀, F = 0
Ionic crystal energy per ion pair: U(r) = −αe²/(4πε₀r) + B/rⁿ
Equilibrium energy: U₀ = −(αe²/4πε₀r₀)(1 − 1/n)
Energy to form Na⁺Cl⁻ pair from atoms: ΔE = (ionisation energy of Na) − (electron affinity of Cl) + Coulomb energy
5 important questions
Q1. Compare ionic and covalent bonds.
Ionic
Covalent
Electron transfer
Electron sharing
Non-directional
Directional
Between metal and non-metal
Between non-metals
Conduct when molten
Do not conduct (except graphite)
Example: NaCl, KCl
Example: diamond, Si
Q2. Why are metals good conductors of electricity?
In a metal, valence electrons are not tied to any one atom. They form a free electron gas that moves through the lattice of positive ions. A small electric field makes these electrons drift, so current flows easily. The same free electrons also carry heat, so metals are good thermal conductors.
Q3. Derive the expression for equilibrium spacing and bond energy from U(r) = −A/rᵐ + B/rⁿ.
At equilibrium dU/dr = 0: mA/r₀^(m+1) − nB/r₀^(n+1) = 0.
So r₀^(n−m) = nB/(mA), i.e. r₀ = (nB/mA)^(1/(n−m)).
Put B = (mA/n) r₀^(n−m) back: U₀ = −A/r₀ᵐ + (m/n)A/r₀ᵐ = −(A/r₀ᵐ)(1 − m/n).
For a stable bond, d²U/dr² must be positive at r₀, which needs n greater than m (repulsion is shorter-ranged).
Q4. Calculate the energy needed to form Na⁺ and Cl⁻ ions from neutral atoms, and the energy released when they come to 0.236 nm apart. (Ionisation energy of Na = 5.14 eV, electron affinity of Cl = 3.61 eV.)
Forming ions: 5.14 − 3.61 = 1.53 eV must be supplied.
Coulomb energy at r = 0.236 nm: e²/(4πε₀r) = 1.44 eV nm ÷ 0.236 nm ≈ 6.10 eV released.
Net energy released ≈ 6.10 − 1.53 = 4.57 eV, so the NaCl molecule is stable (ignoring repulsion).
Q5. Explain Van der Waals bonding and hydrogen bonding with examples.
Van der Waals: in inert-gas atoms the electron cloud fluctuates, creating a momentary dipole that induces a dipole in the neighbour; the two attract. The energy varies as 1/r⁶ and is very small (about 0.01–0.1 eV), so solid argon melts at 84 K. Hydrogen bond: a hydrogen atom covalently bonded to an electronegative atom (O, N, F) is attracted to another such atom nearby. Energy about 0.1–0.5 eV. It explains the open structure of ice (less dense than water), the high boiling point of water and the double helix of DNA.
Exam tip
Use the value e²/(4πε₀) = 1.44 eV nm to do ionic-energy numericals quickly. Always sketch the U(r) curve and mark r₀ and the bond energy on it.
X-rays have wavelengths (about 0.1–10 Å) similar to the spacing between atoms in crystals, so crystals act as natural 3-D gratings for them. Bragg's law explains X-ray diffraction and lets us find crystal structure. This topic also covers production of X-rays, continuous and characteristic spectra, and Moseley's law.
Key points
X-rays were discovered by W. C. Röntgen in 1895. They are produced when fast electrons strike a heavy-metal target (Coolidge tube).
Continuous spectrum (Bremsstrahlung, braking radiation): electrons slow down near nuclei; it has a sharp short-wavelength limit λ_min that depends only on the accelerating voltage (Duane–Hunt law).
Characteristic spectrum: sharp lines (Kα, Kβ, Lα…) when an inner-shell vacancy is filled by an outer electron; they depend on the target element.
Moseley's law (1913): √ν of a characteristic line is proportional to (Z − b). It made atomic number, not atomic mass, the basis of the periodic table.
Laue (1912) first showed X-ray diffraction by a crystal.
Bragg's law (W. H. and W. L. Bragg, 1913): reflections from successive parallel planes add up when the path difference 2d sin θ equals a whole number of wavelengths.
θ is the glancing angle (angle with the plane, not with the normal).
Bragg's spectrometer measures θ for successive orders to find d or λ.
Uses: crystal structure, medical radiography, detecting flaws in metals.
Important terms
Glancing angle (θ) – angle between incident ray and crystal plane.
Order of reflection (n) – 1, 2, 3, … for successive maxima.
Hard and soft X-rays – short wavelength (high energy) and long wavelength (low energy).
Diffraction (विवर्तन) – bending and spreading of waves by obstacles or openings.
Formulas
Bragg's law: 2d sin θ = nλ
Duane–Hunt: λ_min = hc/(eV); λ_min (Å) ≈ 12400/V (V in volts)
Moseley: √ν = a(Z − b); for Kα, b ≈ 1 and ν = (3/4)Rc(Z − 1)²
Cubic spacing: d = a/√(h² + k² + l²)
5 important questions
Q1. State Bragg's law and define the glancing angle.
When X-rays of wavelength λ fall on crystal planes of spacing d, strong reflection occurs when 2d sin θ = nλ, where n is an integer. θ, the glancing angle, is the angle between the incident X-ray and the crystal plane.
Q2. Why can an ordinary optical grating not diffract X-rays?
Diffraction is noticeable only when the spacing is comparable to the wavelength. X-ray wavelengths are about 1 Å, but ruled grating spacings are about 10⁴ Å. Crystal planes (spacing ~1 Å) are suitable.
Q3. Derive Bragg's law.
Parallel X-rays fall at glancing angle θ on parallel planes separated by d.
Ray 2 reflected from the lower plane travels extra distance compared with ray 1.
Drop perpendiculars from the point on the upper plane; extra path = 2d sin θ (d sin θ before and after reflection).
Constructive interference when path difference = nλ.
So 2d sin θ = nλ, n = 1, 2, 3 …
Since sin θ ≤ 1, nλ ≤ 2d — this limits the possible orders.
Q4. X-rays of wavelength 1.54 Å are reflected by planes with d = 2.0 Å. Find the glancing angle for first-order reflection and the highest possible order.
sin θ = nλ/(2d) = 1 × 1.54/(2 × 2.0) = 0.385.
θ = sin⁻¹(0.385) ≈ 22.6°.
Highest order: n ≤ 2d/λ = 4.0/1.54 = 2.6, so n_max = 2.
Q5. An X-ray tube works at 30 kV. Find the shortest wavelength emitted (h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m/s, e = 1.6 × 10⁻¹⁹ C).
Remember θ in Bragg's law is measured from the plane, not the normal. Draw the two-plane ray diagram clearly with the path difference marked — it carries most of the marks in the derivation.
Atoms in a crystal are not still; they vibrate about their lattice positions. These vibrations carry heat and decide the specific heat of a solid. This topic covers the classical Dulong–Petit law and its failure at low temperature, Einstein's quantum theory, Debye's theory with the T³ law, and the idea of phonons and the dispersion relation for a one-dimensional chain.
Key points
Dulong–Petit law (classical): molar specific heat of a solid Cv = 3R ≈ 24.9 J mol⁻¹ K⁻¹ (about 6 cal mol⁻¹ K⁻¹), the same for all solids and independent of temperature.
Experiments show Cv falls towards zero as T → 0 K. Classical physics cannot explain this.
Einstein (1907): each atom is a 3-D quantum oscillator with one frequency ν_E; energy levels are quantised. Cv agrees with 3R at high T and falls at low T, but it falls exponentially, faster than experiment.
Debye (1912): the solid is a continuous elastic medium with a range of frequencies up to a maximum ν_D (Debye frequency). The total number of modes is 3N.
Debye's result: at high T, Cv → 3R; at low T, Cv ∝ T³ (Debye T³ law), which agrees well with experiment.
Debye temperature θ_D = hν_D/k_B. Above θ_D the solid behaves classically.
A quantum of lattice vibration energy is called a phonon, with energy hν = ħω.
For a 1-D monatomic chain, ω = 2√(C/m) |sin(ka/2)|: frequency has an upper limit (cut-off).
A diatomic chain has two branches: acoustic (atoms move together) and optical (atoms move opposite; can interact with light).
Important terms
Specific heat at constant volume (Cv) – heat to raise 1 mole by 1 K at constant volume.
Einstein temperature (θ_E) – hν_E/k_B.
Debye frequency – highest vibration frequency allowed in the solid.
Dispersion relation – relation between ω and wave vector k.
Brillouin zone – range of k from −π/a to +π/a in a 1-D chain.
Every solid element has a molar specific heat of about 3R (≈ 25 J mol⁻¹ K⁻¹). It is based on classical equipartition: each atom has 3 vibrational degrees of freedom, each with average energy k_BT, so U = 3RT and Cv = 3R. It fails at low temperatures (and for light, hard solids like diamond even at room temperature) because vibrational energy is quantised; low-frequency modes alone can be excited when kT is small.
Q2. Compare Einstein's and Debye's theories.
Einstein
Debye
All atoms vibrate with one frequency
A continuous spectrum of frequencies up to ν_D
Atoms independent
Atoms coupled; whole crystal vibrates
Cv falls exponentially at low T
Cv ∝ T³ at low T
Poor fit at very low T
Good fit with experiment
Q3. Derive Einstein's expression for specific heat (outline).
Mean energy of one oscillator, using Planck's result: ε̄ = hν/(e^(hν/kT) − 1).
For N atoms × 3 directions: U = 3N hν/(e^(hν/kT) − 1).
Put x = θ_E/T. High T (x small): e^x ≈ 1 + x, so Cv → 3R (Dulong–Petit). Low T (x large): Cv ≈ 3R x² e^(−x) → 0.
Q4. The Debye temperature of copper is 343 K. Find its molar specific heat at 10 K.
Cv = 1944 (T/θ_D)³ = 1944 × (10/343)³ = 1944 × 2.478 × 10⁻⁵ ≈ 0.048 J mol⁻¹ K⁻¹. This is about 500 times smaller than 3R, showing the strong fall at low temperature.
Q5. Obtain the dispersion relation for a one-dimensional monatomic lattice.
Atoms of mass m, spacing a, joined by springs of constant C. Force on atom s: m d²u_s/dt² = C(u_(s+1) + u_(s−1) − 2u_s).
So ω = 2√(C/m) |sin(ka/2)|. ω is maximum at k = π/a (edge of the Brillouin zone), so the lattice acts as a low-pass filter. For small k, ω ≈ a√(C/m) k: the speed of sound is constant.
Exam tip
For low-temperature numericals use 1944 (T/θ_D)³ directly. In the theory answer, always state the high-T and low-T limits — that is where the marks are.
Free electron theory treats the valence electrons of a metal as a gas of particles moving freely inside the metal. The classical (Drude–Lorentz) model explains Ohm's law and the Wiedemann–Franz law but fails for specific heat. The quantum (Sommerfeld) model uses Fermi–Dirac statistics and fixes these failures. This topic covers drift velocity, relaxation time, conductivity, mobility, the Fermi energy and the density of states.
Key points
Drude–Lorentz model: free electrons move randomly like gas molecules and collide with the positive ions. In an electric field they gain a small drift velocity opposite to the field.
Conductivity σ = ne²τ/m, where n is electron density and τ is relaxation time (mean time between collisions).
Mobility μ = v_d/E = eτ/m; σ = neμ.
Wiedemann–Franz law: the ratio of thermal to electrical conductivity is proportional to temperature, K/σ = LT (L = Lorenz number).
Failures of the classical model: (1) predicts electronic specific heat (3/2)R per mole, but measured is about 100 times smaller; (2) wrong temperature dependence of resistivity (predicts ρ ∝ √T, measured ρ ∝ T); (3) cannot explain why some solids are insulators.
Sommerfeld (quantum) model: electrons obey the Pauli exclusion principle and Fermi–Dirac statistics. At 0 K they fill energy levels up to the Fermi energy E_F.
Only electrons within about k_BT of E_F can absorb heat, so the electronic specific heat is small and ∝ T.
E_F for metals is a few eV (Cu ≈ 7.0 eV), much larger than k_BT at room temperature (0.025 eV).
Important terms
Relaxation time (τ) – average time between collisions, about 10⁻¹⁴ s.
Mean free path (λ) – average distance travelled between collisions.
Fermi energy (E_F) – highest filled energy level at 0 K.
Fermi–Dirac distribution – probability f(E) = 1/(e^((E − E_F)/kT) + 1) that a level E is occupied.
Density of states g(E) – number of states per unit energy per unit volume; g(E) ∝ √E.
Formulas
Current density J = nev_d = σE
σ = ne²τ/m; resistivity ρ = 1/σ = m/(ne²τ)
μ = eτ/m; σ = neμ
Wiedemann–Franz: K/σT = L; quantum value L = (π²/3)(k_B/e)² = 2.44 × 10⁻⁸ W Ω K⁻²
Fermi energy (0 K): E_F = (ħ²/2m)(3π²n)^(2/3)
Fermi temperature T_F = E_F/k_B; Fermi velocity v_F = √(2E_F/m)
Electronic specific heat (quantum): C_e = (π²/2) R (T/T_F) per mole of electrons
5 important questions
Q1. Define drift velocity and mobility.
Drift velocity is the small average velocity that free electrons acquire opposite to an applied electric field, on top of their random motion. Mobility is the drift velocity per unit electric field, μ = v_d/E, in m² V⁻¹ s⁻¹.
Q2. State the Fermi–Dirac distribution. What is its value at E = E_F?
f(E) = 1/[e^((E − E_F)/k_BT) + 1]. At E = E_F, f = 1/(1 + 1) = ½ at any temperature above 0 K. At 0 K, f = 1 for E below E_F and f = 0 above it.
Q3. Derive the expression for electrical conductivity in the Drude model.
Force on an electron: F = −eE, so acceleration a = −eE/m.
After a collision the electron starts with random velocity (average zero). In average time τ it gains v_d = −eEτ/m.
Current density J = −nev_d = (ne²τ/m) E.
Comparing with J = σE (Ohm's law): σ = ne²τ/m.
Q4. Copper has n = 8.5 × 10²⁸ m⁻³ and resistivity 1.7 × 10⁻⁸ Ω m. Find the relaxation time and the drift velocity for a current of 1 A in a wire of area 1 mm².
Write the failures of the classical model as numbered points — a common 4-mark question. In numericals keep powers of ten in a separate line to avoid slips.
In a solid, the energy levels of many atoms spread into bands. The way electrons fill these bands, and the size of the gap between them, decides whether a solid is a conductor, semiconductor or insulator. This topic covers formation of bands, the Kronig–Penney idea, Fermi level, intrinsic and extrinsic semiconductors and the Hall effect.
Key points
In an isolated atom, electrons have sharp energy levels. When N atoms come close, each level splits into N very close levels, forming an energy band.
Valence band: the highest band filled (fully or partly) with electrons at 0 K. Conduction band: the next higher band.
Forbidden gap (E_g): energy range between them with no allowed states.
Conductors: bands overlap or the band is half-filled (metals) — many free electrons.
Insulators: large gap (diamond, about 5–6 eV); electrons cannot jump at room temperature.
Semiconductors: small gap (Si ≈ 1.1 eV, Ge ≈ 0.7 eV); some electrons cross at room temperature. Conductivity rises with temperature.
The free-electron model could not explain why some solids are insulators. Kronig–Penney model (periodic potential) shows that allowed energy bands are separated by forbidden gaps.
Fermi level: in intrinsic, near the middle of the gap; moves towards the conduction band in n-type and towards the valence band in p-type.
Hall effect: a current-carrying conductor in a perpendicular magnetic field develops a transverse voltage. The sign of the Hall voltage tells the carrier type; its size gives carrier density.
Important terms
Hole – a vacancy left in the valence band, acting as a positive charge carrier.
Fermi energy – highest occupied level at 0 K; at T > 0 the level with ½ probability of occupation.
Mobility (μ) – drift velocity per unit electric field, m² V⁻¹ s⁻¹.
Hall voltage: V_H = BI/(net); Hall coefficient R_H = 1/(ne) (negative for electrons)
Mobility from Hall: μ = σR_H
5 important questions
Q1. Distinguish between conductors, semiconductors and insulators on the basis of band theory.
Conductors: valence and conduction bands overlap (or partly filled band), no gap. Semiconductors: small gap (about 1 eV), few electrons cross at room temperature. Insulators: large gap (over about 3 eV), almost no electrons cross.
Q2. What is the Fermi level? Where does it lie in an intrinsic semiconductor?
The Fermi level is the energy at which the probability of finding an electron is ½ (at T > 0). In an intrinsic semiconductor it lies near the middle of the forbidden gap.
Q3. Explain the Hall effect and derive the expression for the Hall coefficient.
A slab carries current I along x; magnetic field B along z.
Carriers (charge q, drift velocity v) feel Lorentz force qvB along y and pile up on one face.
This builds a Hall field E_H; equilibrium when qE_H = qvB.
Current density J = nqv, so E_H = JB/(nq).
Hall coefficient R_H = E_H/(JB) = 1/(nq); for electrons R_H = −1/(ne).
With width w and thickness t: V_H = E_H w = BI/(nqt).
Q4. For germanium, nᵢ = 2.5 × 10¹⁹ m⁻³, μₑ = 0.39 and μₕ = 0.19 m² V⁻¹ s⁻¹. Find its intrinsic conductivity.
σ = nᵢe(μₑ + μₕ).
= 2.5 × 10¹⁹ × 1.6 × 10⁻¹⁹ × (0.39 + 0.19).
= 4.0 × 0.58 = 2.32 S/m.
Resistivity ρ = 1/σ ≈ 0.43 Ω m.
Q5. The Hall coefficient of a sample is 3.66 × 10⁻⁴ m³/C. Find the carrier concentration.
If R_H is positive, the carriers are holes (p-type).
Exam tip
Draw the three band diagrams (overlap, small gap, large gap) side by side. In Hall effect answers, always mention that the sign of V_H shows whether carriers are electrons or holes.
Some materials lose all electrical resistance when cooled below a certain critical temperature. This state is called superconductivity, discovered by Kamerlingh Onnes in 1911 in mercury at 4.2 K. This topic covers zero resistance, the Meissner effect, critical field, Type I and Type II superconductors, the isotope effect, an outline of BCS theory, high-Tc superconductors and applications.
Key points
Below the critical temperature Tc, resistance drops sharply to zero. A current set up in a superconducting ring can flow for years without loss.
Meissner effect (1933): a superconductor expels magnetic flux from its interior when cooled below Tc in a field. It behaves as a perfect diamagnet (B = 0 inside, susceptibility χ = −1).
A strong enough magnetic field destroys superconductivity. The critical field Hc depends on temperature: Hc(T) = H₀[1 − (T/Tc)²].
A large current also destroys it, because its own field exceeds Hc (Silsbee's rule).
Type I (soft): one critical field; sudden, complete transition (Pb, Hg, Sn). Low Hc.
Type II (hard): two critical fields Hc1 and Hc2; between them a mixed (vortex) state where flux enters as quantised tubes (Nb₃Sn, NbTi). High Hc2, so they are used in magnets.
Isotope effect: Tc ∝ M^(−½), showing that lattice vibrations (phonons) are involved.
BCS theory (Bardeen, Cooper, Schrieffer, 1957): two electrons with opposite momentum and spin attract each other through the lattice and form a Cooper pair. Pairs move without scattering. An energy gap of about 3.5 k_BTc opens at the Fermi level.
High-Tc superconductors (1986 onwards): copper-oxide ceramics such as YBa₂Cu₃O₇ with Tc ≈ 92 K, above the boiling point of liquid nitrogen (77 K).
Other properties: specific heat jumps at Tc; entropy falls (more ordered state); thermal conductivity changes.
Important terms
Critical temperature (Tc) – temperature below which resistance is zero.
Persistent current – current that keeps flowing in a superconducting loop without a source.
Cooper pair – bound pair of electrons responsible for superconductivity.
Flux quantum – smallest unit of trapped flux, Φ₀ = h/2e = 2.07 × 10⁻¹⁵ Wb.
Josephson effect – tunnelling of Cooper pairs through a thin insulator between two superconductors.
Formulas
Critical field: Hc(T) = H₀[1 − (T/Tc)²]
Critical current of a wire of radius r: Ic = 2πr Hc
Isotope effect: Tc M^α = constant, α ≈ ½
BCS energy gap at 0 K: Eg = 2Δ ≈ 3.52 k_BTc
Flux quantum Φ₀ = h/(2e)
5 important questions
Q1. What is the Meissner effect? Show that a superconductor is a perfect diamagnet.
When a superconductor is cooled below Tc in a magnetic field, the field lines are pushed out of it. Inside, B = μ₀(H + M) = 0, so M = −H and the susceptibility χ = M/H = −1. A susceptibility of −1 means perfect diamagnetism. This shows that superconductivity is more than zero resistance: a perfect conductor would only trap the existing flux, not expel it.
Q2. Distinguish between Type I and Type II superconductors.
Type I
Type II
One critical field Hc
Two fields Hc1 and Hc2
Complete Meissner effect up to Hc
Full Meissner effect only below Hc1; mixed state between Hc1 and Hc2
Low Hc (below about 0.1 T)
High Hc2 (up to tens of tesla)
Pure metals: Pb, Hg
Alloys and compounds: NbTi, Nb₃Sn, YBCO
Not used for magnets
Used in strong magnets
Q3. The critical field of lead is 6.5 × 10⁴ A m⁻¹ at 0 K and its Tc = 7.2 K. Find the critical field at 5 K.
Q4. Explain the BCS theory of superconductivity (outline).
An electron passing through the lattice attracts nearby positive ions, creating a region of extra positive charge.
A second electron is attracted to this region. So, through a phonon, the two electrons attract each other.
Below Tc this attraction beats their Coulomb repulsion, and electrons near the Fermi surface with opposite momenta and spins bind into Cooper pairs.
All pairs share one quantum state. Breaking a pair needs the gap energy 2Δ, so ordinary small scatterings cannot slow the current: resistance is zero.
The theory explains the isotope effect (phonons depend on ion mass) and the energy gap.
Q5. Give the applications of superconductors.
Superconducting magnets for MRI scanners, particle accelerators (LHC) and fusion research.
Maglev trains that float using strong magnetic repulsion.
SQUIDs (based on the Josephson effect) that measure extremely weak magnetic fields, such as those from the brain.
Loss-free power cables and fault-current limiters.
Fast switching elements and qubits in quantum computers.
Exam tip
Always say that zero resistance and the Meissner effect are the two defining properties. Draw the Hc–T parabola and the M–H curves for Type I and Type II; diagrams fetch easy marks.
Real optical instruments use thick lenses or combinations of lenses. Their behaviour can be described simply by six cardinal points. Real lenses also form imperfect images; these defects are called aberrations. This topic covers the cardinal points of a coaxial system, the equivalent focal length of two thin lenses, and the main types of aberration with methods to reduce them, including the achromatic doublet.
Key points
A coaxial optical system is fully described by six cardinal points: two focal points (F₁, F₂), two principal points (P₁, P₂) and two nodal points (N₁, N₂).
Principal planes: planes of unit positive lateral magnification; a ray entering at a height h on the first principal plane leaves from the same height on the second.
Nodal points: a ray aimed at N₁ leaves from N₂ parallel to its original direction (unit angular magnification).
If the same medium is on both sides of the system, the nodal points coincide with the principal points.
Two thin lenses of focal lengths f₁ and f₂ separated by d act like one lens of focal length F, with 1/F = 1/f₁ + 1/f₂ − d/(f₁f₂).
Aberrations are of two kinds: chromatic (because μ depends on wavelength) and monochromatic (because of the lens shape; Seidel aberrations).
Five monochromatic aberrations: spherical aberration, coma, astigmatism, curvature of field, distortion.
Spherical aberration: marginal rays focus nearer the lens than paraxial rays. Reduced by stops, by using two lenses separated by d = f₁ − f₂, or by a plano-convex lens with its curved side facing the parallel beam.
Chromatic aberration: violet light focuses nearer the lens than red. Removed by an achromatic doublet (crown convex + flint concave lens in contact) satisfying ω₁/f₁ + ω₂/f₂ = 0.
Two lenses of the same glass separated by d = (f₁ + f₂)/2 are achromatic for focal length (used in the Huygens eyepiece).
Important terms
Equivalent lens – single thin lens giving the same magnification and image as the system.
Dispersive power (ω) – ω = (μ_v − μ_r)/(μ_y − 1).
Longitudinal chromatic aberration – distance between the violet and red focal points, = ωf.
Coma – off-axis point appears as a comet-shaped blur.
Astigmatism – off-axis point gives two line images at different distances.
Q1. Define the six cardinal points of an optical system.
Focal points: F₁, where rays parallel on the image side meet the axis on the object side, and F₂, where an incident parallel beam converges. Principal points: axial points of the two principal planes, which are conjugate planes of unit lateral magnification. Nodal points: axial conjugate points with unit angular magnification; a ray towards N₁ emerges from N₂ parallel to itself.
Q2. Two thin convex lenses of focal lengths 20 cm and 10 cm are 5 cm apart. Find the equivalent focal length and the positions of the principal points.
1/F = 1/20 + 1/10 − 5/(20 × 10) = 0.05 + 0.10 − 0.025 = 0.125, so F = 8 cm. α = Fd/f₂ = 8 × 5/10 = 4 cm (first principal point 4 cm to the right of the first lens). β = −Fd/f₁ = −8 × 5/20 = −2 cm (second principal point 2 cm to the left of the second lens).
Q3. Derive the condition for achromatism of two thin lenses in contact.
Combined power: 1/F = 1/f₁ + 1/f₂.
For each lens, 1/f = (μ − 1)K, so d(1/f) = K dμ = (dμ/(μ − 1))(1/f) = ω/f.
For F to be the same for all colours, d(1/F) = 0: ω₁/f₁ + ω₂/f₂ = 0.
Since ω₁, ω₂ are positive, f₁ and f₂ must have opposite signs: a convex lens of crown glass (low ω) and a concave lens of flint glass (high ω). The combination is still converging because the convex lens is stronger.
Q4. Design an achromatic doublet of focal length 50 cm using crown (ω₁ = 0.015) and flint (ω₂ = 0.030) glass.
From ω₁/f₁ + ω₂/f₂ = 0: f₂ = −(ω₂/ω₁)f₁ = −2f₁. From 1/f₁ + 1/f₂ = 1/50: 1/f₁ − 1/(2f₁) = 1/(2f₁) = 1/50, so f₁ = 25 cm (crown convex) and f₂ = −50 cm (flint concave). Check: 0.015/25 + 0.030/(−50) = 0.0006 − 0.0006 = 0 ✔.
Q5. What is spherical aberration? How can it be reduced?
Rays from a point on the axis meeting the lens at different heights focus at different points: marginal rays come to focus closer than paraxial rays, so the image is a blurred disc. Methods: (1) use a stop to block marginal rays (dims the image); (2) use a plano-convex lens with the curved face towards the incident parallel light, so the deviation is shared equally at both surfaces; (3) use two lenses separated by d = f₁ − f₂; (4) use a crossed lens or an aspheric lens; (5) combine a convex and a suitable concave lens (aplanatic combination).
Exam tip
In cardinal-point numericals, keep the sign convention (distances measured left to right are positive). Always check an achromatic design by putting the values back into ω₁/f₁ + ω₂/f₂ = 0.
Interference is the redistribution of light energy when two coherent waves overlap, giving bright and dark fringes. This topic covers coherent sources, Young's double-slit experiment, interference in thin films and wedge-shaped films, Newton's rings and the Michelson interferometer.
Key points
Principle of superposition: the resultant displacement is the sum of the individual displacements.
Coherent sources have the same frequency and a constant phase difference. Two independent lamps are not coherent.
Coherent sources are made by division of wavefront (Young's slits, Fresnel's biprism) or division of amplitude (thin films, Newton's rings, Michelson interferometer).
Energy is not destroyed; it is only redistributed.
Thin film (soap bubble, oil on water): rays reflected from top and bottom surfaces interfere. Reflection from a denser medium adds an extra path of λ/2 (phase change π).
Wedge-shaped film: straight, equally spaced fringes parallel to the edge; used to test flatness and to find thin wire diameter.
Newton's rings: a plano-convex lens on a glass plate forms an air film; circular fringes with a dark centre in reflected light. Ring diameters give λ or R or the refractive index of a liquid.
Michelson interferometer: splits a beam into two perpendicular arms; moving one mirror by λ/2 shifts one fringe. Used to measure λ, the difference of close wavelengths and thickness of thin sheets.
Important terms
Fringe width (β) – distance between two successive bright (or dark) fringes.
Optical path – μ × geometric path.
Coherence (सुसंगतता) – constant phase relation between waves.
Visibility – contrast of fringes, (I_max − I_min)/(I_max + I_min).
Q1. What are coherent sources? Why can two separate bulbs not produce interference?
Coherent sources emit waves of the same frequency with a constant phase difference. Two bulbs emit light from many atoms with random, rapidly changing phases, so the fringe pattern shifts too fast to be seen.
Q2. Why is the centre of Newton's rings dark in reflected light?
At the point of contact film thickness t = 0. The ray reflected from the lower glass surface (denser medium) gets an extra λ/2 path. So the path difference is λ/2 and destructive interference makes the centre dark.
Q3. Derive the expression for the diameter of dark rings in Newton's rings.
Lens of radius R on a plate; at distance r from contact, air film thickness t.
Geometry: r² = (2R − t)t ≈ 2Rt, so t = r²/(2R).
For normal incidence in air (μ = 1, cos r = 1), dark condition: 2t = nλ.
2 × r²/(2R) = nλ → rₙ² = nλR.
With D = 2r: Dₙ² = 4nλR, so Dₙ ∝ √n — rings get closer as n increases.
Q4. In a Newton's rings experiment, the diameters of the 5th and 15th dark rings are 0.336 cm and 0.590 cm. If R = 100 cm, find λ.
Q5. When the movable mirror of a Michelson interferometer is moved by 0.0589 mm, 200 fringes cross the field. Find λ.
2d = Nλ → λ = 2d/N.
d = 0.0589 mm = 5.89 × 10⁻⁵ m.
λ = 2 × 5.89 × 10⁻⁵/200 = 5.89 × 10⁻⁷ m.
λ = 5890 Å (sodium light).
Exam tip
Remember the extra λ/2 for reflection at a denser medium — it decides whether the centre is dark or bright. In Newton's rings numericals, using the difference of squares (Dₙ₊ₚ² − Dₙ²) cancels errors at the centre.
Diffraction is the bending of light around the edges of obstacles or openings and its spreading into the geometrical shadow. It shows that light is a wave. This topic covers Fresnel and Fraunhofer diffraction, the single slit, the plane transmission grating, and resolving power with Rayleigh's criterion.
Key points
Diffraction is noticeable when the size of the opening or obstacle is comparable to the wavelength.
Fresnel diffraction: source and/or screen at finite distance; wavefronts are spherical or cylindrical; no lenses needed.
Fraunhofer diffraction: source and screen effectively at infinity; plane wavefronts; achieved using lenses. Simpler to analyse.
Fresnel's half-period zones explain the rectilinear propagation of light and the zone plate.
Single slit (Fraunhofer): a bright central maximum with weaker secondary maxima on both sides. Minima at a sin θ = nλ.
The central maximum is twice as wide as each secondary maximum; intensities of secondary maxima fall quickly.
Plane transmission grating: many equally spaced parallel slits. Gives very sharp, bright principal maxima; used to measure wavelength.
Grating element (a + b) = width of slit + width of opaque part = 1/N (N = lines per unit length).
Rayleigh's criterion: two close images are just resolved when the central maximum of one falls on the first minimum of the other.
Resolving power of a grating increases with the order and the total number of lines.
Important terms
Diffraction (विवर्तन) – bending of waves around edges.
Resolving power – ability of an instrument to show two close objects or wavelengths separately.
Dispersive power of grating – dθ/dλ = n/((a + b) cos θ).
Zone plate – plate with alternate zones blocked; acts like a lens.
Formulas
Single slit minima: a sin θ = nλ (n = 1, 2, …); intensity I = I₀(sin α/α)², α = πa sin θ/λ
Width of central maximum on screen: 2λD/a (small angles)
Grating: (a + b) sin θ = nλ; maximum order n_max < (a + b)/λ
Resolving power of grating: λ/dλ = nN (N = total lines illuminated)
Telescope: limit of resolution θ = 1.22λ/D (D = aperture)
5 important questions
Q1. Distinguish between Fresnel and Fraunhofer diffraction.
In Fresnel diffraction the source and screen are at finite distances, wavefronts are spherical and no lens is used. In Fraunhofer diffraction they are effectively at infinity, wavefronts are plane and lenses are used; the analysis is simpler.
Q2. State Rayleigh's criterion of resolution.
Two point sources or two spectral lines are just resolved when the principal maximum of one coincides with the first minimum of the other.
Q3. Light of 6000 Å falls normally on a slit 0.2 mm wide. Find the angular position of the first minimum and the width of the central maximum on a screen 1 m away.
Distance of first minimum from centre: y = Dθ = 1 × 3 × 10⁻³ = 3 mm.
Width of central maximum = 2y = 6 mm.
Q4. A grating has 5000 lines per cm. Find the angle of first-order diffraction for λ = 5893 Å and the highest order visible.
(a + b) = 1/5000 cm = 2 × 10⁻⁴ cm = 2 × 10⁻⁶ m.
sin θ = nλ/(a + b) = 5.893 × 10⁻⁷/(2 × 10⁻⁶) = 0.2947.
θ = sin⁻¹(0.2947) ≈ 17.1°.
n_max < (a + b)/λ = 2 × 10⁻⁶/5.893 × 10⁻⁷ = 3.39, so the highest order is 3.
Q5. Explain diffraction at a plane transmission grating and find the lines needed to resolve the sodium D lines (5890 Å and 5896 Å) in the first order.
Parallel light falls normally; each slit sends secondary waves.
A lens focuses waves going at angle θ; path difference between corresponding points of adjacent slits = (a + b) sin θ.
Principal maxima when (a + b) sin θ = nλ; between them, very weak secondary maxima, so lines are sharp.
Resolving power needed: λ/dλ = 5893/6 ≈ 982.
nN ≥ 982; for n = 1, N ≥ 982 lines must be illuminated.
Exam tip
Diffraction minima in a single slit use a sin θ = nλ, while grating maxima use (a + b) sin θ = nλ — do not mix them. Always convert lines per cm into grating element in metres.
Polarisation shows that light is a transverse wave. In polarised light, the electric vector vibrates in one particular direction. This topic covers methods of producing polarised light (reflection, refraction, double refraction, selective absorption), Malus's law, Brewster's law, Nicol prism, wave plates and optical activity.
Key points
In unpolarised light the electric vector vibrates in all directions perpendicular to propagation. In plane-polarised light it vibrates in one plane only.
Only transverse waves can be polarised; sound (longitudinal) cannot.
Malus's law: intensity through an analyser varies as cos²θ.
Brewster's law: at the polarising angle θₚ, the reflected light is completely plane-polarised, and tan θₚ = μ. At this angle the reflected and refracted rays are perpendicular.
Double refraction: in calcite and quartz a ray splits into the ordinary (O) ray, which obeys Snell's law, and the extraordinary (E) ray, which does not. Both are plane-polarised at right angles to each other.
Optic axis: direction in the crystal along which there is no double refraction.
Calcite is a negative crystal (μₒ > μₑ); quartz is a positive crystal (μₑ > μₒ).
Nicol prism: a calcite crystal cut and joined with Canada balsam; the O-ray is removed by total internal reflection and only the E-ray passes. Used as polariser and analyser.
Quarter-wave plate introduces a path difference λ/4 (turns plane-polarised light at 45° into circularly polarised light). Half-wave plate introduces λ/2.
Optical activity: some substances (quartz, sugar solution) rotate the plane of polarisation. Measured with a polarimeter.
Dextro-rotatory / laevo-rotatory – rotates the plane clockwise / anticlockwise (looking towards the source).
Specific rotation – rotation by a 1 dm column of solution of unit concentration (1 g/cc).
Formulas
Malus: I = I₀ cos²θ
Brewster: μ = tan θₚ; θₚ + r = 90°
Unpolarised light through a polaroid: I = I₀/2
Quarter-wave plate thickness: t = λ/(4(μₒ − μₑ)) (use the magnitude of the difference); half-wave: t = λ/(2(μₒ − μₑ))
Specific rotation: [α] = θ/(l × c) (l in dm, c in g/cc)
5 important questions
Q1. State Brewster's law.
When unpolarised light falls on a transparent surface at the polarising angle θₚ, the reflected light is completely plane-polarised and the refractive index μ = tan θₚ. Then the reflected and refracted rays are at 90°.
Q2. Find the polarising angle for glass of μ = 1.5.
θₚ = tan⁻¹(1.5) ≈ 56.3°. The angle of refraction is 90° − 56.3° = 33.7°.
Q3. Prove that at the polarising angle the reflected and refracted rays are perpendicular.
Brewster: μ = tan θₚ = sin θₚ/cos θₚ.
Snell: μ = sin θₚ/sin r.
Equate: cos θₚ = sin r = cos(90° − r).
So θₚ = 90° − r, i.e. θₚ + r = 90°.
The angle between reflected and refracted rays = 180° − (θₚ + r) = 90°.
Q4. Calculate the thickness of a quartz quarter-wave plate for λ = 5893 Å (μₒ = 1.544, μₑ = 1.553).
t = λ/(4(μₑ − μₒ)) (quartz is positive, so μₑ > μₒ).
Q5. A 20 cm tube of sugar solution of concentration 0.1 g/cc rotates the plane of polarisation by 13°. Find the specific rotation. Also find the intensity through two polaroids at 60° if the incident light is unpolarised with intensity I₀.
l = 20 cm = 2 dm, c = 0.1 g/cc.
[α] = θ/(lc) = 13/(2 × 0.1) = 65° per dm per (g/cc).
First polaroid: I₁ = I₀/2.
Second (Malus): I₂ = I₁ cos²60° = (I₀/2)(1/4) = I₀/8.
Exam tip
In polarimeter numericals, change the tube length to decimetres. Draw the Nicol prism with the Canada balsam layer, O-ray and E-ray labelled — it is a common diagram question.
LASER stands for Light Amplification by Stimulated Emission of Radiation. A laser gives an intense, highly directional, monochromatic and coherent beam. This topic covers absorption, spontaneous and stimulated emission, Einstein's coefficients, population inversion, pumping, the ruby and He–Ne lasers, and applications.
Key points
Absorption: an atom in lower level E₁ absorbs a photon of energy hν = E₂ − E₁ and goes to E₂.
Spontaneous emission: an excited atom returns to E₁ on its own, emitting a photon in a random direction and phase (ordinary light).
Stimulated emission: an incoming photon of energy hν makes an excited atom emit an identical photon — same frequency, phase, direction and polarisation. This is the basis of the laser (idea given by Einstein in 1917).
Normally lower levels have more atoms (Boltzmann distribution). Population inversion (more atoms in the upper level) is needed for amplification.
A metastable state has a long lifetime (about 10⁻³ s compared with about 10⁻⁸ s for ordinary excited states); it helps build population inversion.
Pumping: supplying energy to achieve inversion — optical (flash lamp), electrical discharge, chemical.
Optical resonator: two parallel mirrors (one fully, one partly reflecting) bounce light back and forth, building up the beam.
Ruby laser (Maiman, 1960): first laser; Al₂O₃ doped with Cr³⁺ ions; three-level system; optical pumping by flash lamp; pulsed output at 694.3 nm (red).
He–Ne laser: gas laser; electrical discharge excites He atoms which transfer energy to Ne by collision; four-level system; continuous output at 632.8 nm (red).
Four-level lasers achieve inversion more easily than three-level ones.
Applications: optical fibre communication, surgery (eye), cutting and welding, barcode readers, holography, distance measurement, CD/DVD.
Important terms
Coherence – constant phase relation; laser has high temporal and spatial coherence.
Monochromaticity – very narrow range of wavelengths.
Active medium (सक्रिय माध्यम) – material in which inversion is created.
Ratio of spontaneous to stimulated rate: R = e^(hν/kT) − 1
Number of photons per second: n = P/(hν) = Pλ/(hc)
5 important questions
Q1. What is population inversion? Why is it necessary?
Population inversion is the state where the number of atoms in a higher energy level exceeds that in a lower level. It is needed because only then do stimulated emissions outnumber absorptions, so light is amplified.
Q2. Distinguish between spontaneous and stimulated emission.
Spontaneous emission happens on its own, giving photons with random phase and direction (incoherent). Stimulated emission is triggered by an incident photon, and the emitted photon is identical to it in frequency, phase and direction (coherent).
Q3. Describe the construction and working of the He–Ne laser.
Quartz tube filled with He and Ne (ratio about 10:1 He:Ne); electrodes for discharge; mirrors at both ends (one partly transparent).
Discharge excites He atoms to metastable levels.
Excited He transfers energy to Ne atoms by collision (levels nearly equal), creating inversion in Ne.
Ne atoms drop to a lower level by stimulated emission giving 632.8 nm light; then return to ground state quickly.
Mirrors provide feedback; continuous coherent beam comes out.
Four-level system, so inversion is easy and output is continuous.
Q4. Find the energy of a photon from a He–Ne laser (λ = 632.8 nm) and the number of photons emitted per second by a 1 mW laser.
n = P/E = 10⁻³/3.14 × 10⁻¹⁹ ≈ 3.2 × 10¹⁵ photons per second.
Q5. Explain the ruby laser with an energy level diagram.
Ruby rod (Al₂O₃ with about 0.05% Cr³⁺); ends polished and silvered (one partially); helical xenon flash lamp around it.
Flash lamp pumps Cr³⁺ ions from ground level E₁ to a broad band E₃.
Ions drop quickly (non-radiative) to metastable level E₂.
Population inversion between E₂ and E₁ builds up.
A spontaneous photon triggers stimulated emission at 694.3 nm; it is amplified by reflections.
Output is pulsed, because more than half of the ions must be lifted to E₂ (three-level system).
Exam tip
Always draw the energy-level diagram with the pumping, non-radiative and laser transitions labelled. Remember the wavelengths: ruby 694.3 nm, He–Ne 632.8 nm.
Holography is a method of recording and reconstructing the full three-dimensional image of an object using laser light. Unlike a photograph, which records only intensity, a hologram records both the amplitude and the phase of light waves. This topic covers the principle of holography, recording and reconstruction, the properties of holograms, and the main applications of lasers in industry, medicine, communication and science.
Key points
Holography was invented by Dennis Gabor (1947; Nobel Prize 1971). It became practical only after the laser (1960) gave highly coherent light.
A hologram is an interference pattern between the light scattered by the object (object beam) and a direct beam (reference beam) from the same laser.
Recording: a laser beam is split by a beam splitter; one part lights the object, the other goes straight to the photographic plate. The two beams interfere on the plate.
Reconstruction: the developed plate is lit by the reference beam alone. The fringes act like a diffraction grating and recreate the original object wave, so a 3-D virtual image is seen.
A hologram looks like a random pattern to the eye; it does not resemble the object.
Every small piece of a hologram contains information about the whole object (with lower resolution).
Coherence is essential: the path difference between the two beams must be less than the coherence length of the laser.
Applications of lasers: cutting, welding and drilling; eye surgery (LASIK), removal of tumours, bloodless surgery; optical-fibre communication; barcode scanners, CD/DVD; LIDAR and range finding; holographic security marks on cards and notes; measuring very small distances (interferometry).
Important terms
Object beam – laser light reflected or scattered from the object.
Reference beam – unscattered laser light falling directly on the plate.
Coherence length – maximum path difference over which light stays coherent.
Holographic interferometry – comparing two holograms of an object to detect tiny deformations.
LIDAR – light detection and ranging, radar with laser light.
Formulas
Intensity on the plate: I = |O + R|² = |O|² + |R|² + O R* + O* R (the cross terms store the phase)
Fringe spacing for two plane beams at angle θ: d = λ/[2 sin(θ/2)]
Laser range finding: distance D = ct/2 (t = round-trip time)
Coherence length L = c × Δt = c/Δν
5 important questions
Q1. How does a hologram differ from an ordinary photograph?
Photograph
Hologram
Records intensity only
Records intensity and phase
2-D image
3-D image with parallax
Ordinary light
Coherent laser light
A piece shows only part of the scene
Each piece shows the whole scene
Image looks like the object
Plate looks like random fringes
Q2. A laser pulse returns from the Moon after 2.56 s. Find the Earth–Moon distance.
D = ct/2 = 3 × 10⁸ × 2.56/2 = 3.84 × 10⁸ m (about 3.84 lakh km).
Q3. Explain the recording and reconstruction of a hologram with the principle.
Principle: the interference of two coherent waves records the phase of one relative to the other as fringe positions.
Recording: laser → beam splitter. Beam 1 (object beam) is reflected from the object to the plate; beam 2 (reference beam) reaches the plate via a mirror. The plate records the fringe pattern I = |O + R|².
Reconstruction: the developed plate is lit by a beam identical to R. Its transmittance is proportional to I, so the transmitted light contains a term proportional to |R|²O — a copy of the original object wave.
An observer looking through the plate sees a virtual 3-D image at the object's position; a real image (conjugate) also forms on the other side.
Q4. Two plane laser beams (λ = 633 nm) meet at 30° on a plate. Find the fringe spacing recorded.
d = λ/[2 sin(θ/2)] = 633 × 10⁻⁹/(2 × sin 15°) = 633 × 10⁻⁹/(2 × 0.2588) = 633 × 10⁻⁹/0.5176 ≈ 1.22 × 10⁻⁶ m. The plate must resolve about 800 lines per mm, so special fine-grain plates are used.
Q5. Describe the applications of lasers in medicine and industry.
Medicine: CO₂ laser as a bloodless scalpel (light seals blood vessels while cutting); excimer laser for LASIK eye correction; argon laser to repair detached retina; lasers for kidney-stone breaking, tattoo and tumour removal; photodynamic therapy for cancer. Industry: Nd:YAG and CO₂ lasers for cutting, welding and drilling hard metals and ceramics with a narrow heat-affected zone; laser marking; alignment in construction; measuring length and flatness by interferometry; non-destructive testing by holographic interferometry.
Exam tip
Remember the one-line idea: a hologram stores the phase as fringe positions. Draw the recording set-up with beam splitter, mirror, object and plate — the diagram alone earns marks.
An optical fibre is a thin, flexible strand of glass or plastic that guides light by total internal reflection. Fibres carry huge amounts of information over long distances with low loss. This topic covers structure, acceptance angle and numerical aperture, types of fibres, attenuation and dispersion, and applications.
Key points
Structure: a central core of higher refractive index n₁, surrounded by cladding of lower index n₂, covered by a protective buffer/jacket.
Light travels by total internal reflection (TIR) at the core–cladding boundary. Conditions: light goes from denser to rarer medium, and the angle of incidence exceeds the critical angle.
Acceptance angle: the maximum angle with the fibre axis at which light can enter and still be guided. Rays inside this cone are guided.
Numerical aperture (NA) = sine of acceptance angle (in air); measures light-gathering ability.
Step-index fibre: core index is uniform; sudden change at the cladding; rays follow zig-zag paths.
Graded-index fibre: core index decreases gradually from axis to edge; rays follow curved paths; less dispersion.
Single-mode fibre: very thin core (a few μm); one mode; no intermodal dispersion; used for long-distance communication.
Multimode fibre: larger core; many modes; easier to couple light; used for short links.
Attenuation (loss): due to absorption, scattering (Rayleigh) and bending losses; measured in dB/km.
Dispersion: spreading of light pulses — intermodal (different paths) and material/chromatic (different wavelengths). It limits the data rate.
Advantages over copper: huge bandwidth, low loss, no electromagnetic interference, light weight, secure.
Applications: telecommunication and internet, endoscopy in medicine, sensors, decorative lighting.
Important terms
Critical angle (क्रांतिक कोन) – angle of incidence in the denser medium for which the refracted ray grazes the surface.
Mode – an allowed path (pattern) of light in a fibre.
V-number – fibre parameter that decides the number of modes.
Fractional index change (Δ) – (n₁ − n₂)/n₁.
Formulas
Critical angle: sin θc = n₂/n₁
NA = √(n₁² − n₂²) ≈ n₁√(2Δ); sin θₐ = NA/n₀ (n₀ = 1 for air)
Δ = (n₁ − n₂)/n₁
V = (2πa/λ) × NA (a = core radius); single-mode if V < 2.405
Number of modes (step index, large V) ≈ V²/2
Attenuation: α = (10/L) log₁₀(P_in/P_out) dB/km
5 important questions
Q1. What is total internal reflection? State its conditions.
When light in a denser medium meets a rarer medium at an angle greater than the critical angle, all of it is reflected back. Conditions: light must go from denser to rarer medium, and the angle of incidence must exceed the critical angle.
Q2. Distinguish between step-index and graded-index fibres.
In a step-index fibre the core has a uniform index that drops suddenly at the cladding, and rays move in zig-zag paths. In a graded-index fibre the core index decreases gradually from the axis, rays bend smoothly, and dispersion is less.
Q3. Derive the expression for numerical aperture.
Ray enters from air (n₀) at angle θₐ, refracts at angle θᵣ in core: n₀ sin θₐ = n₁ sin θᵣ.
It meets the core–cladding surface at angle (90° − θᵣ), which must be at least θc for TIR.
At the limit, sin(90° − θᵣ) = cos θᵣ = n₂/n₁.
sin θᵣ = √(1 − n₂²/n₁²) = √(n₁² − n₂²)/n₁.
n₀ sin θₐ = n₁ sin θᵣ = √(n₁² − n₂²).
NA = n₀ sin θₐ = √(n₁² − n₂²).
Q4. A fibre has n₁ = 1.50 and n₂ = 1.47. Find the critical angle, NA and acceptance angle in air.
Q5. The power input to a fibre is 10 mW and output after 5 km is 5 mW. Find the attenuation.
α = (10/L) log₁₀(P_in/P_out).
P_in/P_out = 10/5 = 2; log₁₀ 2 = 0.301.
α = (10/5) × 0.301 = 0.602 dB/km.
Total loss over 5 km = 3.01 dB.
Exam tip
Draw a fibre showing core, cladding, the acceptance cone and a zig-zag ray — it supports both the NA derivation and short answers. Learn NA = √(n₁² − n₂²) and the attenuation formula; numericals use them directly.
Vector calculus describes how fields change in space. Three operations built from the del operator ∇ — gradient, divergence and curl — appear throughout electricity, magnetism, fluids and heat flow. This topic covers scalar and vector fields, the meaning and calculation of grad, div and curl, the Laplacian, and the important identities, with worked examples.
Key points
A scalar field gives a number at every point (temperature T(x, y, z), potential V). A vector field gives a vector at every point (electric field E, velocity of a fluid).
Del operator: ∇ = î ∂/∂x + ĵ ∂/∂y + k̂ ∂/∂z. It is a vector differential operator.
Gradient ∇φ of a scalar field is a vector. It points in the direction of the fastest increase of φ, and its magnitude is that maximum rate. It is perpendicular to the level surfaces φ = constant.
Directional derivative of φ along unit vector â = ∇φ · â.
Divergence ∇·A of a vector field is a scalar: the net outward flux per unit volume. Positive → source, negative → sink.
A field with ∇·A = 0 is solenoidal (e.g. magnetic field B).
Curl ∇×A is a vector: it measures the rotation (circulation per unit area) of the field.
A field with ∇×A = 0 is irrotational or conservative, and can be written as A = −∇V (e.g. electrostatic field).
For r = xî + yĵ + zk̂: ∇·r = 3, ∇×r = 0, ∇r = r/r, ∇(1/r) = −r/r³
5 important questions
Q1. Give the physical meaning of divergence and curl.
Divergence measures how much a field spreads out from a point: the outward flux per unit volume around it. In a fluid, positive divergence means fluid is being created (source). Curl measures how much the field circulates around a point; for fluid flow it equals twice the angular velocity of a small paddle wheel placed there. A uniform flow has zero curl; a whirlpool has non-zero curl.
Q2. Find the gradient of φ = x²y + yz³ at (1, 2, 1) and the directional derivative along (î + 2ĵ + 2k̂)/3.
Q4. Prove that curl grad φ = 0 and div curl A = 0.
∇×(∇φ): the x-component is ∂/∂y(∂φ/∂z) − ∂/∂z(∂φ/∂y) = ∂²φ/∂y∂z − ∂²φ/∂z∂y = 0, since mixed partial derivatives are equal. The y and z components vanish the same way. So ∇×∇φ = 0.
∇·(∇×A) = ∂/∂x(∂Az/∂y − ∂Ay/∂z) + ∂/∂y(∂Ax/∂z − ∂Az/∂x) + ∂/∂z(∂Ay/∂x − ∂Ax/∂y). The six terms cancel in pairs, so ∇·(∇×A) = 0.
Physical meaning: a gradient (conservative) field has no rotation; a curl field (like B) has no sources.
Q5. Show that the field E = (2xy + z) î + x² ĵ + x k̂ is conservative and find its potential.
∇×E: î(∂x/∂y − ∂x²/∂z) = 0; ĵ(∂(2xy + z)/∂z − ∂x/∂x) = ĵ(1 − 1) = 0; k̂(∂x²/∂x − ∂(2xy + z)/∂y) = k̂(2x − 2x) = 0. So E is irrotational, hence conservative. Find φ with ∇φ = E: ∂φ/∂x = 2xy + z → φ = x²y + xz + f(y, z). Then ∂φ/∂y = x² + ∂f/∂y = x², so f depends only on z; ∂φ/∂z = x + f′(z) = x, so f is constant. φ = x²y + xz + C (and V = −φ if E = −∇V).
Exam tip
Write the curl as a 3×3 determinant with ∂/∂x, ∂/∂y, ∂/∂z in the middle row; it prevents sign errors. Remember: grad acts on a scalar and gives a vector; div gives a scalar; curl gives a vector.
Vector integral theorems connect integrals over a region with integrals over its boundary. Gauss's divergence theorem links a volume integral to a surface integral, Stokes' theorem links a surface integral to a line integral, and Green's theorem is the plane form of Stokes' theorem. This topic covers line, surface and volume integrals, the three theorems with proofs in outline, and worked examples, along with their use in electromagnetism.
Key points
Line integral ∫A·dl along a curve; for a force field it gives the work done. Around a closed loop, ∮A·dl is called the circulation.
For a conservative field (∇×A = 0) the line integral depends only on the end points, and ∮A·dl = 0.
Surface integral ∫A·dS gives the flux of A through a surface; dS = n̂ dS points along the outward normal.
Volume integral ∫φ dV adds a scalar over a volume (for example total charge = ∫ρ dV).
Gauss's divergence theorem: ∮_S A·dS = ∫_V (∇·A) dV. Total outward flux = total "source strength" inside.
Stokes' theorem: ∮_C A·dl = ∫_S (∇×A)·dS. Circulation around a loop = flux of the curl through any surface bounded by it.
Green's theorem in the plane: ∮(M dx + N dy) = ∬(∂N/∂x − ∂M/∂y) dx dy.
Physics uses: Gauss's theorem turns Gauss's law into ∇·E = ρ/ε₀; Stokes' theorem turns Ampere's law into ∇×B = μ₀J; the equation of continuity ∇·J + ∂ρ/∂t = 0 follows from charge conservation.
Important terms
Closed surface – a surface that encloses a volume (sphere, cube).
Circulation – line integral of a field around a closed path.
Simply connected region – a region in which every closed curve can be shrunk to a point without leaving it.
Outward normal – unit vector perpendicular to a surface, pointing out of the enclosed volume.
Formulas
Work W = ∫ F·dl
Gauss: ∮_S A·dS = ∫_V ∇·A dV
Stokes: ∮_C A·dl = ∫_S (∇×A)·dS
Green (plane): ∮_C (M dx + N dy) = ∬_R (∂N/∂x − ∂M/∂y) dx dy
Area of a plane region: A = ½ ∮ (x dy − y dx)
Differential form of Gauss's law: ∇·E = ρ/ε₀
5 important questions
Q1. State Gauss's divergence theorem and Stokes' theorem.
Gauss: the surface integral of the normal component of a vector field over a closed surface equals the volume integral of its divergence over the enclosed volume, ∮A·dS = ∫∇·A dV. Stokes: the line integral of the tangential component of a vector field around a closed curve equals the surface integral of the normal component of its curl over any surface bounded by that curve, ∮A·dl = ∫(∇×A)·dS.
Q2. Find the work done by F = 3x² î + (2xz − y) ĵ + z k̂ along the straight line from (0, 0, 0) to (2, 1, 3).
Path: x = 2t, y = t, z = 3t, t from 0 to 1; dx = 2dt, dy = dt, dz = 3dt. F·dl = 3(4t²)(2dt) + (2·2t·3t − t)dt + 3t(3dt) = (24t² + 12t² − t + 9t)dt = (36t² + 8t)dt. W = ∫₀¹ (36t² + 8t) dt = 12 + 4 = 16 units.
Q3. Verify Gauss's divergence theorem for A = x î + y ĵ + z k̂ over a cube of side a with one corner at the origin.
Volume side: ∇·A = 3, so ∫∇·A dV = 3a³. Surface side: on the face x = a, A·n̂ = x = a, flux = a × a² = a³; on x = 0, flux = 0. The same holds for y and z faces. Total flux = 3a³. Both sides equal 3a³, so the theorem is verified.
Q4. Prove Gauss's divergence theorem (outline).
Divide the volume V into tiny cubes of volume dx dy dz.
For one cube, the net outflow in the x-direction is [Ax(x + dx) − Ax(x)] dy dz = (∂Ax/∂x) dx dy dz. Adding y and z: net outflow = (∇·A) dV.
Add over all cubes. Across every inner face the flux leaving one cube enters its neighbour, so these cancel.
Only the outer faces remain, giving the flux through the boundary surface S.
Hence ∮_S A·dS = ∫_V (∇·A) dV.
Q5. Using Green's theorem, find the area of an ellipse x²/a² + y²/b² = 1.
Area = ½ ∮ (x dy − y dx). Parametrise: x = a cos θ, y = b sin θ, θ from 0 to 2π. Then dx = −a sin θ dθ, dy = b cos θ dθ. x dy − y dx = ab cos²θ dθ + ab sin²θ dθ = ab dθ. Area = ½ × ab × 2π = πab.
Exam tip
For a "verify" question, always compute both sides separately and show they are equal. Mention the electromagnetic use (Gauss's law, Ampere's law) — it shows the examiner you know why the theorem matters.
Most laws of physics are written as differential equations: radioactive decay, cooling, motion of a spring, current in a circuit and waves. This topic covers the order and degree of a differential equation, first-order equations (variable separable and linear), second-order linear equations with constant coefficients (complementary function and particular integral), and an introduction to partial differential equations and separation of variables, all with physics examples.
Key points
Order = order of the highest derivative; degree = power of the highest derivative after removing fractions and radicals.
A linear equation has the dependent variable and its derivatives only to the first power and not multiplied together.
Variable separable: write f(y) dy = g(x) dx and integrate both sides. Example: radioactive decay dN/dt = −λN gives N = N₀e^(−λt).
Linear first-order equation dy/dx + P(x)y = Q(x): multiply by the integrating factor e^(∫P dx). Example: an LR circuit.
Real, different roots m₁, m₂: y = C₁e^(m₁x) + C₂e^(m₂x)
Equal roots m: y = (C₁ + C₂x)e^(mx)
Complex roots α ± iβ: y = e^(αx)(C₁ cos βx + C₂ sin βx)
Second-order homogeneous a y″ + b y′ + c y = 0: try y = e^(mx) to get the auxiliary equation am² + bm + c = 0.
Non-homogeneous: general solution = complementary function (CF) + particular integral (PI).
Partial differential equations contain partial derivatives: wave equation, heat (diffusion) equation, Laplace's equation. They are often solved by separation of variables, u(x, t) = X(x)T(t).
Important terms
General solution – solution with as many arbitrary constants as the order.
Particular solution – solution after fixing constants using initial or boundary conditions.
Integrating factor (IF) – factor that makes the left side an exact derivative.
Auxiliary equation – algebraic equation obtained by putting y = e^(mx).
Boundary conditions – values of the solution fixed at the edges of the region.
Formulas
Linear first order: y × IF = ∫Q × IF dx + C, IF = e^(∫P dx)
String fixed at x = 0 and x = L: y = Σ A_n sin(nπx/L) cos(nπvt/L)
5 important questions
Q1. Find the order and degree of (d²y/dx²)³ + (dy/dx)⁴ + y = 0.
The highest derivative is d²y/dx², so the order is 2. Its power is 3, so the degree is 3.
Q2. Solve the LR circuit equation L dI/dt + RI = E with I = 0 at t = 0.
Write dI/dt + (R/L)I = E/L. IF = e^(Rt/L). I e^(Rt/L) = ∫(E/L)e^(Rt/L) dt = (E/R)e^(Rt/L) + C. At t = 0, I = 0 → C = −E/R. So I = (E/R)(1 − e^(−Rt/L)). The current grows to E/R with time constant L/R.
Q3. Solve y″ − 5y′ + 6y = 0 with y(0) = 2 and y′(0) = 5.
Q4. Solve the forced equation ẍ + 4x = 8 cos t (find the general solution).
CF: m² + 4 = 0 → m = ±2i → x_c = C₁ cos 2t + C₂ sin 2t. PI: try x_p = A cos t → −A cos t + 4A cos t = 3A cos t = 8 cos t → A = 8/3. General solution: x = C₁ cos 2t + C₂ sin 2t + (8/3) cos t. The first part is the natural oscillation, the second follows the driving force.
Q5. Solve the one-dimensional wave equation for a string fixed at both ends by separation of variables.
Put y(x, t) = X(x)T(t): X T″ = v² X″ T → X″/X = T″/(v²T) = −k² (a constant).
X″ + k²X = 0 → X = A sin kx + B cos kx. y(0, t) = 0 → B = 0. y(L, t) = 0 → sin kL = 0 → k = nπ/L, n = 1, 2, 3 …
T″ + k²v²T = 0 → T = C cos(nπvt/L) + D sin(nπvt/L).
y = Σ sin(nπx/L)[C_n cos(nπvt/L) + D_n sin(nπvt/L)]: the normal modes with frequencies f_n = nv/(2L).
The constants C_n and D_n come from the initial shape and velocity using Fourier series.
Exam tip
In second-order problems, write the auxiliary equation first and name the case (real, equal or complex roots). Always put the initial conditions back to check your constants.
Any periodic function that meets certain conditions can be written as a sum of sines and cosines — its Fourier series. In physics this lets us break complex waveforms (square wave, sawtooth, rectified AC) into a fundamental and its harmonics. This topic covers Dirichlet conditions, Fourier coefficients, even and odd functions and standard waveforms.
Key points
A function is periodic with period T if f(t + T) = f(t).
Fourier's theorem: a periodic function can be expressed as a constant plus a series of sine and cosine terms of frequencies that are whole multiples of the fundamental.
Dirichlet conditions: in one period, f must be single-valued, finite, have a finite number of discontinuities, and a finite number of maxima and minima.
At a jump discontinuity, the series converges to the average of left and right values.
Even function (f(−x) = f(x)): only cosine terms (bₙ = 0).
Odd function (f(−x) = −f(x)): only sine terms (a₀ = aₙ = 0).
Square wave and sawtooth wave are odd (if centred suitably), so they contain only sine terms.
A square wave contains only odd harmonics with amplitudes falling as 1/n.
Uses: analysing musical sounds (quality/timbre depends on harmonics), electrical signals and filters, heat conduction, vibrating strings.
Fourier analysis: finding coefficients from a waveform; synthesis: building the waveform from harmonics.
Important terms
Fundamental – the lowest frequency term (n = 1).
Harmonics (संनादी) – terms with frequencies 2, 3, 4 … times the fundamental.
Fourier coefficients – a₀, aₙ, bₙ.
Orthogonality – ∫ sin mx sin nx dx over a period = 0 for m ≠ n (basis for finding coefficients).
Period T in time (ω = 2π/T): f(t) = a₀/2 + Σ(aₙ cos nωt + bₙ sin nωt), aₙ = (2/T) ∫₀^T f(t) cos nωt dt
Full-wave rectified wave: |sin x| = 2/π − (4/π)[cos 2x/3 + cos 4x/15 + cos 6x/35 + …]
5 important questions
Q1. State Dirichlet's conditions for a Fourier series.
In one period the function must be single-valued and finite, have a finite number of discontinuities and a finite number of maxima and minima. Then its Fourier series converges to f(x), and to the mean value at a discontinuity.
Q2. Why does the Fourier series of an odd function contain only sine terms?
For odd f, f(x) cos nx is odd, so its integral over (−π, π) is zero; hence a₀ = aₙ = 0. Only bₙ survives, giving a pure sine series.
Q3. Find the Fourier series of the square wave f(x) = −1 for −π < x < 0 and +1 for 0 < x < π.
f is odd, so a₀ = aₙ = 0.
bₙ = (2/π) ∫₀^π sin nx dx = (2/π)[−cos nx/n]₀^π = (2/(nπ))(1 − cos nπ).
cos nπ = (−1)ⁿ: for even n, bₙ = 0; for odd n, bₙ = 4/(nπ).
A Fourier series breaks a periodic function into sine and cosine terms with discrete frequencies. The Fourier transform extends this idea to non-periodic functions, giving a continuous spread (spectrum) of frequencies. This topic covers the Fourier integral, the Fourier transform and its inverse, sine and cosine transforms, important properties, transforms of standard functions such as the rectangular pulse and Gaussian, and uses in optics, signals and quantum mechanics.
Key points
A non-periodic function can be treated as a periodic one whose period L → ∞. The sum over harmonics then becomes an integral (the Fourier integral).
Fourier transform: F(k) = (1/√2π) ∫ f(x) e^(−ikx) dx over all x. Inverse: f(x) = (1/√2π) ∫ F(k) e^(ikx) dk over all k.
f(x) and F(k) form a transform pair: one describes the function in space (or time), the other in wave number (or frequency).
For an even function the transform reduces to the cosine transform; for an odd function to the sine transform.
Properties: linearity; shifting f(x − a) multiplies F(k) by e^(−ika); scaling f(ax) gives (1/|a|)F(k/a); the transform of df/dx is ik F(k).
Convolution theorem: the transform of a convolution is the product of transforms (times √2π).
Parseval's theorem: ∫|f(x)|² dx = ∫|F(k)|² dk — total energy is the same in both pictures.
A narrow pulse has a wide spectrum and a wide pulse has a narrow spectrum: Δx·Δk ≈ constant. This is the mathematical root of the Heisenberg uncertainty principle.
Uses: Fraunhofer diffraction pattern is the Fourier transform of the aperture; spectrum analysers; image processing; solving differential equations; momentum-space wavefunctions.
Important terms
Spectrum – distribution of amplitude (or energy) over frequency.
Transform pair – f(x) and its Fourier transform F(k).
Rectangular pulse (top-hat) – f = 1 for |x| ≤ a, 0 otherwise.
Sinc function – sin(u)/u, the transform of a rectangular pulse.
Bandwidth – spread of frequencies present in a signal.
Q1. Distinguish between a Fourier series and a Fourier transform.
Fourier series
Fourier transform
For periodic functions
For non-periodic functions
Discrete frequencies (harmonics)
Continuous range of frequencies
Coefficients a_n, b_n
Function F(k)
Sum over n
Integral over k
Q2. Find the Fourier transform of a rectangular pulse f(x) = 1 for |x| ≤ a and 0 elsewhere.
F(k) = (1/√2π) ∫ from −a to a of e^(−ikx) dx = (1/√2π) [e^(−ikx)/(−ik)] from −a to a = (1/√2π)(e^(ika) − e^(−ika))/(ik) = (1/√2π)(2 sin ka)/k. So F(k) = √(2/π) sin(ka)/k. Its first zeros are at k = ±π/a. A wider pulse (bigger a) gives a narrower central peak: this is the single-slit diffraction pattern.
Q3. Find the Fourier transform of f(x) = e^(−a|x|), a positive.
F(k) = (1/√2π)[∫ from −∞ to 0 of e^(ax)e^(−ikx) dx + ∫ from 0 to ∞ of e^(−ax)e^(−ikx) dx] = (1/√2π)[1/(a − ik) + 1/(a + ik)] = (1/√2π) × 2a/(a² + k²). So F(k) = √(2/π) a/(a² + k²), a Lorentzian shape. Natural spectral lines have this shape because excited states decay exponentially.
Q4. Prove the shifting property and the derivative property.
Shifting: transform of f(x − a) = (1/√2π)∫f(x − a)e^(−ikx) dx. Put u = x − a: = e^(−ika)(1/√2π)∫f(u)e^(−iku) du = e^(−ika) F(k). A shift changes only the phase, not the amplitude spectrum.
Derivative: transform of f′(x) = (1/√2π)∫f′(x)e^(−ikx) dx. Integrate by parts; the boundary term vanishes when f → 0 at ±∞: = (1/√2π)(ik)∫f e^(−ikx) dx = ik F(k). This turns differential equations into algebraic ones.
Q5. Show how the Fourier transform explains the uncertainty principle.
Take a Gaussian wave packet of width σ in x: f(x) = e^(−x²/2σ²). Its transform is F(k) = σe^(−σ²k²/2), a Gaussian of width 1/σ in k. So Δx ≈ σ and Δk ≈ 1/σ, giving Δx·Δk ≈ 1. With p = ħk (de Broglie), Δx·Δp ≈ ħ. A particle confined to a small region must contain a wide range of momenta. (With standard deviations the exact limit is Δx·Δp ≥ ħ/2.)
Exam tip
Learn three standard pairs by heart — rectangular pulse → sinc, exponential → Lorentzian, Gaussian → Gaussian. Most numerical questions are one of these with different constants.