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Write

x

for a comoving coordinate,

u

for the peculiar velocity,

H = \overset{a}{˙} / a

for the Hubble parameter, and

δ = δ ρ / \overset{ρ}{ˉ}

for the density contrast. For a homogeneous self-gravitating fluid, the linearized continuity equation, Euler equations for an inviscid fluid, and cosmological Poisson equation are

\dot{δ} = - \frac{1}{a} \nabla \cdot u, \dot{u} + H u = - \frac{c _{s}^{2}}{a} \nabla δ - \frac{1}{a} \nabla ϕ, \nabla^{2} ϕ = 4 π G a^{2} \overset{ρ}{ˉ} δ .

Here

c_{s}

is the physical adiabatic sound speed, gradients refer to comoving coordinates, and the homogeneous background force has already been subtracted. Differentiating the first equation and eliminating

u

and

ϕ

gives

\ddot{δ} + 2 H \dot{δ} - \frac{c _{s}^{2}}{a ^{2}} \nabla^{2} δ - 4 π G \overset{ρ}{ˉ} δ = 0.

A Fourier mode with comoving wavenumber

k

therefore obeys

\ddot{δ}_{k} + 2 H \dot{δ}_{k} + (\frac{c _{s}^{2} k ^{2}}{a ^{2}} - 4 π G \overset{ρ}{ˉ}) δ_{k} = 0.

The Jeans wavenumber occurs where the coefficient in parentheses vanishes. Since a comoving wavelength is

2 π / k

, the result is

k_{J} = \frac{a 4 π G ρ ˉ}{c _{s}}, λ_{J}^{com} = \frac{c _{s}}{a} \frac{π}{G ρ ˉ} .

Below this comoving Jeans length, pressure restores a displaced fluid element: there are acoustic waves rather than growing Jeans instability. In a static background they oscillate; expansion changes their amplitude and frequency, and dissipation can damp them. Pressure support alone does not erase them. In the pre-recombination photon-baryon fluid, Silk damping supplies a separate erasure mechanism.

For collisionless matter, replace the sound speed by a characteristic velocity dispersion. For an isotropic Maxwell-Boltzmann velocity distribution, defining

σ^{2} = ⟨ u_{x}^{2} ⟩ = ⟨ ∣ u ∣^{2} ⟩ /3

, the static collisionless marginal-stability calculation gives

λ_{J, collisionless}^{com} = \frac{σ}{a} \frac{π}{G ρ ˉ} = \frac{v _{rms}}{a} \frac{π}{3 G ρ ˉ} .

(5)

The coefficient depends on the distribution and the convention for velocity dispersion. Small-scale suppression is now collisionless free streaming and phase mixing, rather than collisional acoustic waves. An instantaneous collisionless Jeans length must be distinguished from the accumulated comoving distance

\int v d t / a

traveled since decoupling.

For the requested thermal histories, an Einstein-de Sitter universe describes the matter-era limit: it cannot literally also have a radiation-dominated epoch. Interpret the question as a spatially flat universe with negligible cosmological constant, passing from radiation domination,

a \propto t^{1/2}

, to matter domination,

a \propto t^{2/3}

. For the usual schematic dynamical Jeans scale use

\overset{ρ}{ˉ}_{tot}

in the estimate, so

λ_{dyn}^{com} \propto {c_{s} a, c_{s} a^{1/2}, t ≪ t_{eq}, \overset{ρ}{ˉ}_{tot} \propto a^{- 4}, t ≫ t_{eq}, \overset{ρ}{ˉ}_{tot} \propto a^{- 3} .

(6)

Near the particle horizon a relativistic perturbation calculation replaces the Newtonian derivation; the scaling estimate still identifies the relevant sound-crossing or streaming scale.

Before cosmological recombination, adiabatic initial conditions keep the entropy per baryon fixed in the tightly coupled photon-baryon fluid. With

R_{b} = 3 \overset{ρ}{ˉ}_{b} / (4 \overset{ρ}{ˉ}_{γ}) \propto a

, its photon-baryon sound speed is

c_{s}^{2} = \frac{c ^{2}}{3 ( 1 + R _{b} )} .

(7)

For example, this follows from

δ ρ_{γ} / \overset{ρ}{ˉ}_{γ} = (4/3) δ ρ_{b} / \overset{ρ}{ˉ}_{b}

and

δ P = c^{2} δ ρ_{γ} /3

. In the radiation-dominated limit

R_{b} ≪ 1

c_{s} ≃ c / 3

; in the strongly baryon-loaded matter-era limit

R_{b} ≫ 1

c_{s} \propto a^{- 1/2}

. Thus the baryon Jeans length across recombination has the asymptotic history

λ_{J, b}^{com} \propto ⎩ ⎨ ⎧ a \propto t^{1/2}, approximately constant, a^{- 1/2} \propto t^{- 1/3}, t ≪ t_{eq}, t_{eq} ≪ t < t_{rec}, t > t_{rec} with adiabatic gas cooling .

(8)

The transition to the plateau is smooth: retaining

1 + R_{b}

gives

λ_{J, b}^{com} \propto a^{1/2} / 1 + R_{b}

in matter domination. If baryon loading is still small, this intermediate segment instead rises as

a^{1/2}

until loading becomes important.

At photon decoupling, radiation ceases to provide pressure support to the baryons, so the sound speed drops from the coupled-fluid value to that of a nonrelativistic ideal gas,

c_{s, b}^{2} = γ k_{B} T_{b} / (μ m_{p})

, with

γ = 5/3

. The drop in comoving Jeans length at fixed density is the same ratio of sound speeds, typically several orders of magnitude. Subsequently an isentropic process gives

T_{b} \propto \overset{ρ}{ˉ}_{b}^{γ - 1} \propto a^{- 2}

, so

c_{s, b} \propto a^{- 1}

and

λ_{J, b}^{com} \propto a^{- 1/2}

. Residual energy exchange through Compton scattering can initially keep

T_{b} \propto a^{- 1}

, producing a short approximately flat gas segment before the decreasing segment. The idealized sketch assumes immediate adiabatic cooling:

log(comoving baryon Jeans scale)
  ^
  |                   __________________
  |                 /                   |
  |               /                     |  loss of photon pressure
  |             /                       |
  |           /                         +---\
  |         /                                \
  +-------------------|-----------------|----------> log(time)
                    t_eq              t_rec
        slope 1/2          slope ~0        slope -1/3

For the cold-dark-matter kinetic Jeans scale, take the stated toy history to mean that collisions maintain a particle temperature equal to the photon temperature until kinetic decoupling. This is an assumption about thermal coupling; a negligible collision rate after decoupling makes the subsequent treatment collisionless. Assume the ordering

t_{NR} < t_{dec} < t_{eq} < t_{rec}

. While relativistic, the characteristic particle speed is constant; while nonrelativistic but thermally coupled,

σ \propto T \propto a^{- 1/2}

; after kinetic decoupling, particle momentum redshifts as

a^{- 1}

and

σ \propto a^{- 1}

. Combining these with the background dynamical density gives

λ_{dyn, CDM}^{com} \propto ⎩ ⎨ ⎧ a \propto t^{1/2}, a^{1/2} \propto t^{1/4}, constant, a^{- 1/2} \propto t^{- 1/3}, t < t_{NR}, t_{NR} < t < t_{dec}, t_{dec} < t < t_{eq}, t > t_{eq} .

(9)

There is no sudden change at cosmological recombination: cold dark matter has already decoupled. Its small initial velocity dispersion makes its suppression scale much smaller than the pre-recombination baryon scale.

log(comoving CDM dynamical Jeans scale)
  ^
  |                        _______________
  |                    ___/               \
  |                ___/                    \       no jump
  |              /                          \      at t_rec
  |            /                             \
  +-----------|------------|--------------|----|------> log(time)
            t_NR         t_dec          t_eq t_rec
   slope 1/2    slope 1/4       slope 0       slope -1/3

The density convention matters. If one instead defines a species' self-gravitating collisionless Jeans length using only its own nonrelativistic density

\overset{ρ}{ˉ}_{m} \propto a^{- 3}

, its thermally coupled segment is constant and its decoupled segment decreases as

a^{- 1/2}

even during radiation domination. The constant radiation-era segment in the second sketch is the usual instantaneous background dynamical streaming estimate, not an exact single-species self-gravity threshold. A complete multicomponent calculation uses the separate perturbation equations and their combined cosmological Poisson equation.

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 39B / iii / Solution Created 2026-09-24 Updated 2026-09-29

The stationary contribution travels with group velocity

V = ω^{'} (k_{0}) = \frac{3 S}{2} k_{0},

k_{0} = 4 V^{2} / (9 S^{2})

. The condition

\frac{ϵ V ^{2}}{S ^{2}} ≪ 1

is therefore equivalent, up to a constant factor, to

k_{0} ϵ ≪ 1

: the dominant wavelength is much longer than the impacted interval. Its Fourier transform can then be replaced by

- \frac{2 W s i n ( k ϵ )}{k} \sim - 2 W ϵ,

which is the total initial vertical velocity integrated over space. The far field cannot resolve

W

and

ϵ

separately and depends only on their product

W ϵ

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 4 / 38B / i / Solution Created 2026-09-24 Updated 2026-09-29

Let the disturbed sheet be

y = h + η

and take a normal mode proportional to

e^{i (k x - ω t)}

, with

K = ∣ k ∣

. Irrotational flow and the rigid-wall condition at

y = 0

give perturbation velocity potentials

ϕ_{-}^{'} = A cosh (Ky) e^{i (k x - ω t)}, ϕ_{+}^{'} = B e^{- K (y - h)} e^{i (k x - ω t)} .

The two linearized kinematic boundary conditions at

y = h

are

- i (ω - k U) η = A K sinh (K h), - iω η = - K B .

The Unsteady Bernoulli equation gives

p_{-}^{'} = i ρ (ω - k U) A cosh (K h)

and

p_{+}^{'} = i ρ ω B

. Continuity of pressure therefore yields the finite-depth vortex-sheet dispersion relation

(ω - k U)^{2} coth (K h) + ω^{2} = 0,

or, with

T = tanh (K h)

(ω - k U)^{2} + T ω^{2} = 0.

For

k > 0

its two roots are

ω = \frac{k U}{1 + T} (1 \pm i T) .

(5)

One root has positive temporal growth for every

k \neq = 0

, so every finite wavelength is subject to the Kelvin-Helmholtz instability.

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 345 / 1 / d / Solution 2026-09-28

Set

r = tan θ = \frac{ω}{N ^{2} - ω ^{2}}, s = \frac{H _{0}}{L},

where

r

is the magnitude of the internal-wave ray slope and

s

is the bottom slope. A rightward ray remains in the triangular basin precisely when its bottom reflection is subcritical internal-wave reflection, namely

r > s

. Therefore

\frac{N H _{0}}{L ^{2} + H _{0}^{2}} < ω < N .

Conservation of frequency and of the component of the wavevector tangent to the slope gives the focusing power of internal-wave reflection

γ = \frac{k _{r}}{k _{i}} = \frac{sin ( θ + α )}{sin ( θ - α )} = \frac{r + s}{r - s} = \frac{L ω + H _{0} N ^{2} - ω ^{2}}{L ω - H _{0} N ^{2} - ω ^{2}},

where

tan α = s

. In this range

γ > 1

, so the reflected wavelength is shorter by the factor

γ

. The reflected normal group velocity is smaller by

γ^{2}

; conservation of normal energy flux therefore increases the wavelength-averaged energy density by

\overline{E}_{r} = γ^{2} \overline{E}_{i} .

Past exam of the optics and electrodynamics course of the University of Cambridge / 2017 / 2 / a / Solution 2026-10-18

Molecules in clean air are much smaller than visible wavelengths. Their induced electric dipole moments undergo Rayleigh scattering. Away from molecular resonances, the induced dipole amplitude is approximately

p = α E

, with only weak wavelength variation in the polarizability

α

. The radiation-zone electric field is proportional to

\overset{p}{¨}

, so the scattered intensity satisfies

I_{sc} \propto ω^{4} ∣ α ∣^{2} I_{in} ≃ λ^{- 4} I_{in} .

This is the Rayleigh scattering wavelength dependence. For equal incident intensities,

450 nm

light scatters about

(650/450)^{4} = 4.35

times as strongly as

650 nm

light. Thus light scattered out of the direct solar beam and into our line of sight is enriched in blue wavelengths.

The dipole mechanism also determines the Rayleigh scattering angular pattern. For unpolarized light, its angular factor is

1 + cos^{2} ϑ

; near

9 0^{\circ}

scattering, the scattered light has strong linear polarization. The sky's perceived colour depends additionally on the solar spectrum, atmospheric transmission and the eye's spectral response, so the

λ^{- 4}

law alone does not predict a violet sky. Over a long near-horizontal path the direct beam loses more short-wavelength light, explaining red sunsets. Large aerosol particles instead produce Mie scattering, which is less strongly wavelength-selective and can make haze look whitish.

Past exam of the optics and electrodynamics course of the University of Cambridge / 2018 / Section B / B2 / Spectrometer and resolution 2026-10-18

A non-polarizing beam splitter divides the incident spectrum into a fixed arm and an arm with a translating mirror. Recombining their returned fields gives a detector signal as a function of mirror position

x

. The round-trip optical path difference is

D = 2 x

relative to the equal-arm position, so the delay is

τ = 2 x / c

in air approximated as vacuum. Record both detector signal and calibrated position, subtract the constant background, and take a Fourier transform of the interferogram.

Figure 1.
Scanning Michelson spectrometer: mirror position sets delay, and a Fourier transform of the recorded interferogram recovers the spectrum
. **Details**.

For matched arm polarizations and balanced throughput, write the ideal signal as

I (τ) = I_{dc} + C \int_{0}^{\infty} S (ω) cos (ω τ) d ω,

where

S (ω)

is the incident spectral power density and

C

is a constant set by the beam splitter and detector. This is the cosine-transform form of Fourier transform spectroscopy, or equivalently the Wiener-Khinchin theorem applied to delayed self-interference.

The delay-range resolution in Fourier transform spectroscopy follows because recording a finite delay range multiplies the correlation by a window. Its transform broadens each spectral line by an instrumental width of order the reciprocal delay range. A mirror travel range

L

creates a delay range of order

2 L / c

, so

δ ν \sim \frac{c}{2 L}, δ λ ≃ \frac{λ ^{2}}{c} δ ν, δ λ \propto \frac{λ ^{2}}{L} .

At fixed wavelength, absolute wavelength resolution therefore improves inversely with travel range. The numerical coefficient depends on whether

L

is a one-sided or total travel, on the window or apodization, and on whether resolution means first zero, half width or another criterion. The proportionality is the requested convention-independent result. Position sampling must also be fine enough to sample the fastest fringe variation without aliasing.

Past exam of the optics and electrodynamics course of the University of Cambridge / 2018 / Section B / B3 / Gain and effective collecting area 2026-10-18

Antenna gain in direction

(θ, ϕ)

compares the far-field power per solid angle with an isotropic radiator using the same accepted input power:

G (θ, ϕ) = \frac{4 π}{P _{in}} \frac{d P}{d Ω} .

It includes radiation efficiency; for the lossless dipole

P_{in} = P_{rad}

. The antenna effective area is defined for incidence from a specified direction with matched polarization and conjugate impedance matching:

P_{available} = A_{eff} (θ, ϕ) S_{inc} .

Its units are square metres and it need not equal the physical metal area.

For ordinary reciprocal antennas, electromagnetic reciprocity makes the coupling between two matched ports the same when transmitter and receiver are exchanged. Free-space transfer in one direction is proportional to

G_{1} A_{2}

, while the reverse transfer is proportional to

G_{2} A_{1}

. Thus

A / G

must be a universal factor at that wavelength, independent of antenna or direction. The supplied relation fixes it as

A_{eff} (θ, ϕ) = \frac{λ ^{2}}{4 π} G (θ, ϕ) .

Equivalently, thermal detailed balance prevents an angular receiving pattern from collecting more radiation than its reciprocal emitting pattern would return in equilibrium. Reciprocity and the specified matching are the assumptions behind this statement; it is not a general theorem for active nonreciprocal devices.

Past exam of the optics and electrodynamics course of the University of Cambridge / 2018 / Section B / B3 / Returned pulse power 2026-10-18

The receiving dipole remains matched. To model its actual reradiation, let its feed impedance be

R_{rad} + i X

with no ohmic loss, and its matched load be

R_{rad} - i X

. The induced current dissipates equal powers in the load and in radiation resistance. Thus matched-load scattering of a Hertzian dipole has

P_{sca} = P_{2}

, even though its port is matched.

The scattered wave has the same dipole gain toward the original antenna. Applying the reciprocal link factor a second time gives the reciprocal Hertzian-dipole echo:

η = G_{t} G_{r} (\frac{λ}{4 π r})^{2} ≃ 1.28 \times 1 0^{- 5}, P_{echo} = P_{0} η^{2} ≃ 1.64 \times 1 0^{- 10} W = 0.164 nW, P_{echo} \propto r^{- 4} .

Each leg supplies an inverse-square spreading factor. This is the carrier-averaged power during a returned pulse with the stated transmitted power during its pulse. The pulse-train average additionally depends on duty cycle, which is not given. Treating the pulses as narrowband keeps the same gain and wavelength approximation.

A matched feed does not make a physical dipole a perfect absorber with zero external scattering. A model retaining only an antenna load-mismatch echo would set its reflected travelling-port wave to zero and would not determine this reradiation. Conversely, an arbitrary antenna structure can add scattering not fixed by gain and effective area alone. The numerical answer uses the ideal lossless Hertzian-dipole scattering model just stated.

Past exam of the optics and electrodynamics course of the University of Cambridge / 2018 / Section B / B3 / Solution 2026-10-18

Use the ideal electrically short Hertzian dipole,

d ≪ λ

, with negligible ohmic loss. Far-field statements additionally require an observation distance large compared with the wavelength and source dimensions. These assumptions distinguish its radiation from near-field coupling and from scattering by a general extended antenna structure.

Past exam of the optics and electrodynamics course of the University of Cambridge / 2019 / Section A / A2 / Retardation mechanism 2026-10-18

Birefringence gives two orthogonal linear eigenpolarizations with different refractive indices,

n_{f}

and

n_{s}

. They accumulate different wave phases through the same physical thickness

d

. Their relative phase retardation is

δ = \frac{ω d}{c} (n_{s} - n_{f}) = \frac{2 π d}{λ _{0}} (n_{s} - n_{f}), δ = \pm \frac{π}{2} (mod 2 π) .

Here

λ_{0}

is the vacuum wavelength. For the thinnest positive-retardance plate,

d = λ_{0} / [4 (n_{s} - n_{f})]

. The plate uses a difference in phase velocity, rather than differential wave absorption. Equal wave amplitudes along the two axes, obtained from input linear polarization at

4 5^{\circ}

to them, acquire phase quadrature and become circular polarization. Reversing that conversion gives linear polarization from the corresponding circular input. A different input angle generally gives elliptical polarization on one pass.

Past exam of the optics and electrodynamics course of the University of Cambridge / 2019 / Section A / A2 / Solution 2026-10-18

Use a transparent reciprocal wave plate at its design wavelength. For the return calculation, assume a mirror at normal incidence with the same reflection coefficient for both transverse components; this is the usual ideal double-pass quarter-wave plate arrangement.

Past exam of the optics and electrodynamics course of the University of Cambridge / 2019 / Section B / B5 / Thomson limit 2026-10-18

Thomson scattering describes the low-recoil scattering of electromagnetic radiation by a free charge. For charge

q

and mass

m

, the weak incident field gives peak acceleration

∣ q ∣ E_{0} / m

. The mean Larmor formula power is therefore

⟨ P ⟩ = \frac{q ^{4} ∣ E _{0} ∣ ^{2}}{12 π ϵ _{0} m ^{2} c ^{3}} .

There is no binding force to introduce a characteristic frequency. Dividing by the same incident light intensity gives the Thomson cross-section:

σ_{T} = \frac{q ^{4}}{6 π ϵ _{0}^{2} m ^{2} c ^{4}} = \frac{8 π}{3} (\frac{q ^{2}}{4 π ϵ _{0} m c ^{2}})^{2} .

For an Electron, the length inside parentheses is the classical electron radius. The cross section has units

m^{2}

and is independent of wavelength in this classical limit. The driven velocity must remain small compared with

c

, and

ℏ ω ≪ m c^{2}

must make Compton recoil kinematics negligible.

Past exam of the optics and electrodynamics course of the University of Cambridge / 2021 / Section A / A2 / a / Solution 2026-10-18

The transmitting antenna is matched to its input transmission line, so its real terminal electrical resistance is

Z_{i}

. The identical receiving antenna has the same electrical impedance at the operating frequency. Its available received power for a matched load is the incident energy flux times its antenna effective area:

P_{av} = \frac{P _{i} G _{a}}{4 π d ^{2}} A_{e} = P_{i} (\frac{G _{a} λ}{4 π d})^{2} \equiv P_{i} F .

The real receiving load is

Z_{t}

, so its voltage-wave reflection coefficient relative to

Z_{i}

Γ_{L} = (Z_{t} - Z_{i}) / (Z_{t} + Z_{i})

. The fraction of available power delivered to the output line is

1 - ∣ Γ_{L} ∣^{2} = \frac{4 Z _{i} Z _{t}}{( Z _{i} + Z _{t} ) ^{2}} .

Consequently, neglecting path losses, mismatch of polarization of an electromagnetic wave and line losses,

P_{t} = P_{i} (\frac{G _{a} λ}{4 π d})^{2} \frac{4 Z _{i} Z _{t}}{( Z _{i} + Z _{t} ) ^{2}} .

This is the Friis transmission equation with a receiving impedance-mismatch factor. For fixed

P_{i}, G_{a}, d, Z_{i}, Z_{t}

, it scales as

λ^{2}

. That scaling is conditional on holding antenna gain fixed: a fixed physical aperture can instead have

G_{a} \propto λ^{- 2}

, and any wavelength dependence of the impedances also enters the expression. The matched-output check

Z_{t} = Z_{i}

gives

P_{t} = P_{i} F

Past exam of the optics and electrodynamics course of the University of Cambridge / 2022 / Section A / 3 / Solution 2026-10-18

Each mutually incoherent patch of an extended star produces a displaced double-slit interference pattern. The detector adds their light intensities. For two apertures separated by a projected interferometric baseline

D

, a source direction offset by a small angle

ϑ_{x}

changes their phase difference by

2 π D ϑ_{x} / λ

. The normalized sum over the source's angular brightness distribution is its first-order optical coherence, whose magnitude is the interferometric visibility for equally bright channels.

For a uniform circular stellar disk of full angular diameter

α

, the normalized angular Fourier transform gives the uniform-disk stellar visibility

V (D) = \frac{2 J _{1} ( x )}{x}, x = \frac{π D α}{λ} .

Here

J_{1}

is the order-one Bessel function of the first kind. This follows by integrating

e^{i (2 π D / λ) ϑ_{x}}

over the disk and dividing by its area. In particular, its power series gives

2 J_{1} (x) / x = 1 - x^{2} /8 + O (x^{4})

For an ordinary slit separation

d = 1 mm

, using the paper's

α = 22.6 \times 1 0^{- 8} rad

and observing wavelength

λ = 570 nm

\frac{d α}{λ} = 3.965 \times 1 0^{- 4}, x = 1.2456 \times 1 0^{- 3}, 1 - V ≃ 1.94 \times 1 0^{- 7} .

Even

d = 10 mm

gives a visibility loss of only

1.94 \times 1 0^{- 5}

. Thus millimetre separation leaves the interference fringes almost indistinguishable from those of a point source; it gives very little sensitivity to the diameter.

The movable mirrors in the Michelson stellar interferometer collect light at a large adjustable separation

D

and redirect it into closely spaced output beams. They increase the sampling interferometric baseline independently of the fixed output slit spacing

d

D

sets sensitivity to source structure, whereas the output optics set the screen's interference fringe spacing. The first Bessel function zero is

x_{1} = 3.83170597

, hence

D_{null} = \frac{x _{1}}{π} \frac{λ}{α} = 1.21967 \frac{570 \times 1 0 ^{- 9} m}{22.6 \times 1 0 ^{- 8}} = 3.076 m ≃ 3.08 m .

The interference fringes disappear at this baseline even though each beam separately remains bright.

Figure 1.
Original calculated interferometric visibility for the given stellar diameter and wavelength
. **Details**.

The value

3.08 m

assumes a uniform disk. The original question supplies a diameter without a surface brightness law; real limb darkening changes the precise visibility zero. A one-dimensional uniformly bright strip of the same full angular width instead has a sinc function visibility and first zero

D = λ / α = 2.522 m

. That is a different source model, not the circular-disk result. The robust conclusion without a surface brightness model is that a baseline of a few metres, rather than millimetres, is required.

Past exam of the optics and electrodynamics course of the University of Cambridge / 2022 / Section B / 4 / a / Solution 2026-10-18

Birefringence means that two independent polarization eigenmodes have different refractive indices at the same frequency and propagation direction. The rotatory-power formula relevant here requires circular birefringence: the eigenmodes are opposite circular polarizations. General linear birefringence instead introduces phase retardation between linearly polarized components and need not rotate every incident linear polarization through one common angle.

Use the complex amplitude convention

Re [E e^{- iω t}]

and transverse unit vectors

e_{1}, e_{2}

. Define

u_{\pm} = \frac{e _{1} \mp i e _{2}}{2}, e_{1} = \frac{u _{+} + u _{-}}{2} .

Propagation through thickness

d

adds phases

ϕ_{\pm} = (2 π / λ_{0}) n_{\pm} d

, where

λ_{0}

is the vacuum wavelength. Recombining the two components gives

\frac{e ^{i ϕ_{+}} u _{+} + e ^{i ϕ_{-}} u _{-}}{2} = e^{i \overset{ˉ}{ϕ}} (e_{1} cos θ + e_{2} sin θ), θ = \frac{ϕ _{+} - ϕ _{-}}{2} .

Thus the rotation and the specific rotatory power per unit length are

θ = \frac{π d}{λ _{0}} (n_{+} - n_{-}), ρ \equiv \frac{θ}{d} = \frac{π}{λ _{0}} (n_{+} - n_{-}) = \frac{ω}{2 c} (n_{+} - n_{-}) .

The units of

ρ

are

rad m^{- 1}

. The sign follows the specified circular basis and viewing convention. Equal attenuation of the two eigenmodes has been assumed; unequal attenuation also produces elliptical polarization.

Past exam of the physics A course of the University of Cambridge / 2017 / Paper 1 / Section C / C11 / a / Solution 2026-10-18

The Bohr model combines classical circular motion in the Coulomb potential with explicitly quantum postulates. A one-electron atom of nuclear charge

+ Z e

has electron–nucleus reduced mass

μ

. With

C = Z e^{2} / (4 π ε_{0})

, the relative circular orbit obeys

\frac{μ v ^{2}}{r} = \frac{C}{r ^{2}}, L = μv r .

Bohr postulates permitted nonradiating stationary orbits with

L = n ℏ

, for

n = 1, 2, \dots

. These assumptions give

r_{n} = \frac{n ^{2} ℏ ^{2}}{μ C} = \frac{4 π ε _{0} ℏ ^{2}}{μ Z e ^{2}} n^{2} .

For hydrogen with the nucleus taken infinitely heavy, the

n = 1

radius is the Bohr radius,

a_{0} ≃ 5.29 \times 1 0^{- 11} m

. The finite reduced mass gives the small nuclear-motion correction, and increasing

Z

contracts the one-electron orbit.

The circular-motion equation gives

T = C / (2 r)

while the potential energy is

- C / r

, so the total energy is

- C / (2 r)

. Substituting the quantized radius yields

E_{n} = - \frac{μ C ^{2}}{2 ℏ ^{2} n ^{2}} = - \frac{μ Z ^{2} e ^{4}}{2 ( 4 π ε _{0} ) ^{2} ℏ ^{2} n ^{2}} .

For the heavy-nucleus hydrogen approximation,

E_{n} ≃ - 13.6 eV / n^{2}

. The ionization threshold is zero energy, so removing the ground-state electron requires about

13.6 eV

The model further postulates that radiation is emitted or absorbed only in transitions between stationary orbits, with photon energy

h ν = ∣ E_{i} - E_{f} ∣

. Emission from

n_{i} > n_{f}

gives

\frac{1}{λ} = R_{μ} Z^{2} (\frac{1}{n _{f}^{2}} - \frac{1}{n _{i}^{2}}), R_{μ} = \frac{μ e ^{4}}{8 ε _{0}^{2} h ^{3} c} .

Thus it reproduces the principal hydrogen spectral series, their limiting wavelengths, and the gross hydrogenic energy levels. For example,

n = 3 \to 2

in hydrogen has

Δ E ≃ 1.89 eV

and

λ ≃ 656 nm

, a visible line. Absorption reverses the transition when the photon matches the energy gap.

The achievement is that quantized energies explain discrete spectra, but the model's stationary orbits are postulates rather than consequences of classical electrodynamics: an accelerating classical charge would radiate. It does not provide a general calculation of line intensities or transition selection rules and fails as a quantitative theory of many-electron atoms. Spin, fine structure and the full Zeeman effect require further physics. Modern quantum mechanics replaces definite electron orbits by wavefunctions and quantum probabilities. In particular, the true hydrogen ground state has

ℓ = 0

, so its orbital angular momentum vanishes; the model's rule

L = n ℏ

should not be identified with the modern orbital eigenvalue rule

L^{2} = ℓ (ℓ + 1) ℏ^{2}

. Bohr's energy quantization remains a successful hydrogenic approximation, while its literal orbital picture does not.

Past exam of the physics A course of the University of Cambridge / 2018 / Paper 1 / Section A / A2 / Solution 2026-10-18

The infinite square well has energy levels

E_{n} = n^{2} h^{2} / (8 m_{e} L^{2})

, with

n = 1, 2, \dots

. The longest absorption wavelength corresponds to the smallest available energy difference from the ground state, namely

n = 1 \to 2

. Its position matrix element is nonzero because the two states have opposite parity about the center of the well, so it is an allowed electric dipole transition.

Equating the photon energy to this gap gives

\frac{h c}{λ} = E_{2} - E_{1} = \frac{3 h ^{2}}{8 m _{e} L ^{2}}, L = \frac{3 hλ}{8 m _{e} c} = 8.53 \times 1 0^{- 10} m = 0.853 nm .

Here the photon gap is approximately

1.55 eV

, not the ground-state energy itself.

Past exam of the physics A course of the University of Cambridge / 2018 / Paper 2 / Section A / A4 / Solution 2026-10-18

Take nearest-neighbour separation

d

, a common spring constant

K

, and one scalar displacement per atom. For a monatomic chain, the harmonic approximation near a potential minimum, Newton's second law and a normal mode give

m \overset{u}{¨}_{j} = K (u_{j + 1} + u_{j - 1} - 2 u_{j}), u_{j} = u_{0} e^{i (kj d - ω t)}, ω = 2 K / m ∣ sin (k d /2) ∣ .

The monatomic-chain dispersion relation has one acoustic phonon branch, with

ω ≃ d K / m ∣ k ∣

near the origin. The first Brillouin zone ends at

k = \pm π / d

, where

ω = 2 K / m

and the group velocity vanishes.

For alternating masses

m, M

, choose a two-atom cell of length

2 d

. The coupled normal modes give the diatomic-chain dispersion relation with equal springs,

ω_{\pm}^{2} = K (\frac{1}{m} + \frac{1}{M}) \pm K (\frac{1}{m} + \frac{1}{M})^{2} - \frac{4}{m M} sin^{2} (k d) .

The first Brillouin zone now ends at

k = \pm π / (2 d)

. The minus branch is the acoustic phonon branch; its two atoms move together at long wavelength. The plus branch is the optical phonon branch; at

k = 0

the atoms move oppositely with a stationary centre of mass and

ω_{+}^{2} = 2 K (1/ m + 1/ M)

. For

M > m

the edge frequencies are

2 K / M

and

2 K / m

, with a gap between them.

M = m

, the two branches reduce to

2 K / m ∣ sin (k d /2) ∣

and

2 K / m ∣ cos (k d /2) ∣

, touching at

2 K / m

. This is zone folding in an equal-mass diatomic chain: identical sites actually have a primitive unit cell of length

d

. A doubled cell folds the monatomic branch but opens no gap. Equal masses alone do not specify all possible spring models; the sketches assume equal nearest-neighbour spring constants.

Figure 1.
Monatomic and two-atom-chain dispersion relations, with explicitly chosen cell lengths and zone edges
. **Details**.