Past exam of the physics A course of the University of Cambridge / 2010 / Paper 1 (Physics A (1), Part IB)

Let the common tension be T, the linear mass densities be ${\mu }_1<{\mu }_2$ and the speeds be $c_j=\sqrt{T/{\mu }_j}$, so $c_2<c_1$. Displacement and transverse-force continuity give the displacement coefficients at a string junctionThe reflected triangular pulse is inverted, has smaller height $|r|H$, unchanged spatial width and unchanged speed, and travels left. Its left-to-right asymmetric profile is reversed as well as vertically inverted. The transmitted pulse remains upright, has smaller height tH, travels more slowly, and has width $W{c}_2/c_1$. The string-pulse width mapping at a junction leaves its temporal duration unchanged. The transmitted slopes are steeper by $2c_1/(c_1+c_2)$; the reflected slope magnitudes are smaller by $|r|$.

The density ratio is unspecified, so only these inequalities are determined. The original plot below illustrates ${\mu }_2/{\mu }_1=4$, giving $r=-1/3$, $t=2/3$ and half-width transmission; it is not a numerical reading of the examination figure.

For the adjacent reflected minima of an air-film-air layer in thin-film interference, reflection at the front surface acquires a phase reversal, while reflection at the rear does not. Near normal incidence the reflected minima therefore obey $2nt=m\lambda$ for integral m. A single intermediate maximum means that the two stated minima are consecutive orders:Thus $2nt=3000\mathrm{nm}$, with orders6 and5, andThe intermediate reflected maximum is at $\lambda =3000/5.5\simeq 545\mathrm{nm}$. This assumes negligible dispersion of the supplied refractive index and an air environment on both sides. ^[ref]

The angular separation is $\Delta \theta =10^{11}/(2\times 10^{17})=5.00\times 10^{-7}\mathrm{rad}$. Each star's angular diameter is below $5\times 10^{-9}\mathrm{rad}$, so treating it as a point source is justified at these diffraction scales.

At300nm the Airy disk first-zero radius is $1.22\lambda /D=1.525\times 10^{-7}\mathrm{rad}$. The separation is3.28 times this radius: two well-separated bright peaks occur. The stars are mutually incoherent, so their point spread functions add as intensities, not amplitudes:The value at u=0 is defined by continuity. Faint diffraction rings remain on either side of each principal peak. ^[ref]

At1000nm the first-zero radius is $5.083\times 10^{-7}\mathrm{rad}$, slightly larger than the $5.00\times 10^{-7}\mathrm{rad}$ separation. The images therefore overlap strongly and just fail the conventional Rayleigh criterion, rather than forming two clean isolated peaks.

The summed Airy profile can still have a shallow central dip and two local maxima: Rayleigh's criterion is a resolution convention, not the exact condition for a profile to become single-peaked. The two-panel original plot in part(a) shows the correct intensity sums at both wavelengths. ^[ref]

For hard spheres of radius r in a face-centered cubic lattice, contact occurs along a face diagonal: $4r=\sqrt{2}a$. Four spheres belong to a conventional cell, so the hard-sphere cubic packing fractions giveFor a body-centered cubic lattice, contact occurs along a body diagonal, $4r=\sqrt{3}a$, and there are two spheres per conventional cell:Their ratio $8/(3\sqrt{6})=1.0887>1$ proves that FCC packs more densely. Coordination alone would not establish the occupied volume fraction.

For a classical ideal monatomic gas, $C_{V,\mathrm{He}}=3N{k}_B/2$ and $C_{p,\mathrm{He}}=5N{k}_B/2$. Treat copper as having one conduction electron per atom. Its degenerate-electron contribution isConsequently the electron-only estimates at5K areThe constant-pressure gas versus electronic metal heat capacity convention makes the second numerator appropriate to a constant-pressure gas measurement. For the solid, the low-temperature difference between $C_p$ and $C_V$ is usually small, but its phonon contribution is not fixed by $E_F$. In a Debye model,The supplied $E_F$ alone therefore does not determine the ratio to copper's total heat capacity; the electronic-only result is conditional.^[ref] An estimate using additional material data is possible: the specific-heat fit in Material Properties: OFHC Copper by National Institute of Standards and Technology gives $c_{p,\mathrm{Cu}}(5\mathrm{K})\simeq 0.155\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. Using copper's rounded molar mass $0.0635\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$ gives $C_{p,\mathrm{Cu}}\simeq 9.85\times 10^{-3}\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, and henceThis supplementary estimate includes copper's lattice contribution and retains the ideal-gas approximation for helium. The NIST fit reports a 10% deviation from its source data below 15 K; neither that fit nor the stated pressure alone supplies helium nonideality corrections.^[ref]

Measure both vertical displacements upward from static equilibrium, so gravity cancels. The spring force changes by $k(x_1-x)$, while the damper is attached to a fixed surface and supplies $-b\dot{x}$. Newton's second law givesFor sinusoidal steady motion, substitution of $x=\mathrm{Re}(RA{e}^{i\omega t})$ gives the spring-base displacement response with a grounded damperA decaying homogeneous transient must be added unless initial conditions already select steady state. At low frequency, $R\to 1$ and the mass follows the slowly moving support; the first correction is $-ib\omega /k$. At high frequency, $R\sim -k/(m{\omega }^2)$: inertia suppresses motion and reverses its phase relative to the support. There is no $ib\omega$ in the numerator because the damper's other end does not move.

Minimize the squared denominator $D=(k-m{\omega }^2{)}^2+b^2{\omega }^2$. Its derivative is $2\omega [2m^2{\omega }^2-2mk+b^2]$, givingWrite ${\omega }_0=\sqrt{k/m}$ and $\zeta =b/(2m{\omega }_0)$. Then $|R|=[(1-r^2{)}^2+4{\zeta }^2{r}^2{]}^{-1/2}$ with $r=\omega /{\omega }_0$, and the finite-frequency peak is $1/[2\zeta \sqrt{1-{\zeta }^2}]$. Increasing light damping lowers and broadens the peak and shifts it left; the low-frequency limit remains1 and the high-frequency decay remains proportional to ${\omega }^{-2}$.

The resonance versus oscillatory damping thresholds differ. For $b^2\ge 2mk$, the maximum is at zero frequency and the curve decreases monotonically. This threshold is $\zeta =1/\sqrt{2}$, distinct from the free-oscillation critical damping threshold $\zeta =1$.

A dispersive wave has phase speed depending on frequency. Its phase velocity is $v_p=\omega /k$, whereas the group velocity $v_g=d\omega /dk$ describes a narrow wave packet in a lossless single branch. Examples are deep-water gravity waves, ${\omega }^2=gk$, and electromagnetic waves in a cold plasma, ${\omega }^2={\omega }_p^2+c^2{k}^2$.

The fixed-edge membrane waveguide has transverse profiles $\sin (n\pi y/b)$ withThere is no nonzero n=0 displacement satisfying both fixed-edge conditions. Even though the unconfined membrane is nondispersive, transverse confinement makes propagation along x dispersive. The membrane-guide phase and energy speeds are, for a propagating mode,For positive $k_x$, the phase speed exceeds v while group speed is below v. At cutoff $k_x=0$ the formal phase speed diverges, but group speed vanishes; no finite longitudinal energy flux is implied.

With the supplied width and speed, $f_{c,n}=10n\mathrm{kHz}$. At15kHz only n=1 propagates:The negative value represents travel in the opposite direction; higher transverse modes are evanescent.

At5kHz every transverse mode is below cutoff, so no propagating guided mode exists. A driven boundary can produce an evanescent near field. For n=1 the decay coefficient is $\sqrt{(20\pi {)}^2-(10\pi {)}^2}=10\pi \sqrt{3}=54.4{\mathrm{m}}^{-1}$, an e-folding distance of18.4mm. This is decay along the guide, not attenuation of an otherwise propagating lossless wave.

At25kHz, the n=1 and n=2 modes can propagate, withThey have different phase and group speeds, so a launch exciting both produces multimode propagation and possible beating. Actual mode amplitudes depend on the transverse launch profile; their existence does not guarantee both are excited. Modes n≥3 remain evanescent.

In Fraunhofer diffraction, the phase difference from aperture coordinate y to observation angle theta is $qy$, with $q=2\pi \sin \theta /\lambda$. Coherent superposition of the transmitted field elements therefore givesThe common propagation factor is omitted. This is the Fourier transform of the complex amplitude transmission, not of its intensity.

For a uniform slit of total width $W=0.2\mathrm{mm}$, integration gives ${\psi }_0=W\mathrm{sinc}(qW/2)$, where $\mathrm{sinc}s=\sin s/s$. Its normalized intensity is ${\mathrm{sinc}}^2(\pi W\sin \theta /\lambda )$, with a central maximum and first zeros at $\sin \theta =\pm \lambda /W=\pm 0.0025$.

For a half-covered slit choose $T=+1$ on the left half and $T=-1$ on the right. Put $h=W/2$ and $s=qh/2$. The two equal rectangles have centres separated by h, soThis half-slit phase-step diffraction factorization is the phase-reversed two-slit interference factor multiplied by the diffraction envelope of each half-slit. Here the half-slit width and centre separation both equal the earlier0.1mm separation. The central intensity is zero; the largest off-axis pair satisfies $\tan s=2s$, giving $|\sin \theta |\simeq 0.00186$. Other zeros are at $s=m\pi$. The plot includes the requested uniform-slit and half-covered patterns.

The slit transmissions have the same magnitude but opposite phase. An ideal narrow-slit representation isEquivalently draw two narrow rectangles at $y=\pm d/2$, one at+1 and the other at−1, and zero elsewhere. A finite rectangle width gives each slit its own diffraction envelope. The original aperture plot shows such rectangles with explicitly illustrative width; the negative value is a field phase reversal, not negative transmitted intensity.

For the narrow pair, the amplitude is $e^{-iqd/2}-e^{iqd/2}=-2i\sin (qd/2)$. Hence phase-reversed double-slit diffraction givesThere is a central dark fringe. Minima obey $d\sin \theta =m\lambda$, maxima $d\sin \theta =(m+1/2)\lambda$, and the near-axis angular fringe period is $\lambda /d=0.005\mathrm{rad}$. Finite slit widths multiply this interference profile by their single-slit envelope. The parent plot includes the narrow-pair profile.

Fresnel diffraction retains the quadratic transverse propagation phase; Fraunhofer diffraction neglects it and leaves an aperture Fourier transform. Expanding the propagation distance gives $z+y^2/(2z)+\cdots$. The aperture phase variation is of order $\pi {\rho }^2/(\lambda z)$, so the crossover scale is $z\sim {\rho }^2/\lambda$, with far-field propagation requiring a much smaller Fresnel number than1.

Here $\lambda \simeq 0.0100\mathrm{m}$ and the full gap width gives ${\rho }^2/\lambda =25\mathrm{m}$, comparable with the16m range. More decisively, measured from the grazing line of sight to the far gap edge, the quadratic phase reaches $\pi (0.5{)}^2/(0.01\times 16)=4.91\mathrm{rad}$. Fresnel diffraction is required; conventions using a half-width change the quoted Fresnel number but not this phase test.

Define $F(w)=C(w)+iS(w)$ with the normalization in the paper. The grazing-edge gap spans $w=0$ toThe Cornu chord power normalization uses a received field proportional to $F(w_g)-F(0)$, while no fence gives $F(\infty )-F(-\infty )=1+i$. Numerical Fresnel integration gives $C(w_g)=0.3245$, $S(w_g)=0.4818$, and thereforeThe two labelled child answers specify the corresponding vectors.

For the largest contiguous Fresnel aperture, a freely chosen single contiguous opening must maximize the chord between two points of the Cornu spiral. Its longest chord joins $F(-w_{\ast })$ to $F(w_{\ast })$, where the first maximum radius satisfies $C(w_{\ast })\cos (\pi w_{\ast }^2/2)+S(w_{\ast })\sin (\pi w_{\ast }^2/2)=0$. This gives $w_{\ast }=1.2094$ andHere x is signed displacement along the fence from the grazing optical axis. The pointwise Fresnel gain versus integrated transmitted power distinction explains why an aperture can concentrate more power at one receiver than the fully open field through constructive interference; this is not an increase of integrated transmitted power.

The existing-opening constraint in Fresnel aperture optimization matters literally: the original gap already occupies $[0,0.5]\mathrm{m}$. The symmetric optimum would require filling its far end. If only further removal is permitted and the final opening must be connected, maximize the chord subject to $w_{-}\le 0$ and $w_{+}\ge 1.768$. The constrained optimum is approximatelyThese are different optimization assumptions; both are stated so the original gap is not silently replaced. The model treats the fence as an ideal opaque screen and ignores ground reflections or additional propagation paths. ^[ref]

The grazing-gap signal is the chord from $F(0)=0$ to $F(1.768)$, approximately $(0.3245,0.4818)$ in the (C,S) plane. Its squared length determines received power after applying the same propagation normalization used for the open field. Reversing which gap edge grazes the ray reflects both limits and leaves the power unchanged.

With the whole fence removed, the vector joins $F(-\infty )=(-1/2,-1/2)$ to $F(+\infty )=(1/2,1/2)$. It is $(1,1)$, with squared length2. The normalized power ratio must compare the gap chord to this full chord, not to the half-plane chord from the origin to one limiting point.

Write the lattice reciprocal vector as $G=2\pi /a$ in one dimension. The nearly-free electron model starts from ${\epsilon }_k={\hslash }^2{k}^2/(2m_e)$ and couples plane waves separated by G. Near a zone boundary, the two-wave Hamiltonian isAt $k=G/2$ the free waves are degenerate and the periodic potential opens a gap $2|V_G|$. A reduced-zone drawing folds k modulo G into $[-G/2,G/2]$, displaying multiple bands at each reduced wavevector. The original plot shows free folded parabolas, avoided crossings and an allowed vertical optical transition. ^[ref]

Including two spin states, filling a free Fermi sphere gives $n=2(4\pi k_F^3/3)/(2\pi {)}^3=k_F^3/(3{\pi }^2)$. A monovalent BCC cell contributes two electrons, soThe sphere therefore lies inside every first-zone face, since the supplied distance is the nearest boundary.

For the shortest BCC reciprocal vector, $G=2\sqrt{2}\pi /a$. In the zero-potential folded-band limit, the energy of a vertical transition from k along G to the neighbouring free branch is ${\hslash }^2(G^2-2Gk)/(2m_e)$. It decreases over the occupied interval and is smallest at $k=k_F$. Thus the monovalent BCC direct optical threshold isThe final state is empty because $G-k_F>k_F$. Finite weak $V_G$ changes the energies and makes optical matrix elements nonzero; an exactly free electron gas by itself has no interband absorption. The expression is a leading weak-potential threshold, not a claim that every peak equals the onset.

Using the supplied lattice constants and bare-electron free bands gives $E_F=(3.24,2.12,1.85)\mathrm{eV}$ and thresholds $(2.06,1.35,1.18)\mathrm{eV}$ for Na,K,Rb. The source plot's broad higher-frequency features shift toward lower energy in the same order as a increases, qualitatively consistent with the predicted $a^{-2}$ scale. Their peak shapes, heights and exact positions need matrix elements, band corrections and scattering; low-frequency intraband conductivity is a separate contribution. No measured trace is reproduced.

With a positive charge magnitude e, the relaxation-time equation for electrons is $m_e^{\ast }\dot{\mathbf{v}}=-e\mathbf{E}-m_e^{\ast }\mathbf{v}/\tau$. Steady drift is opposite the field, ${\mathbf{v}}_d=-e\tau \mathbf{E}/m_e^{\ast }$, so the positive electron mobility and electrical conductivity areThis is the weak-field Drude model; the mobility is a magnitude, not the signed velocity-to-field ratio.

For an intrinsic nondegenerate semiconductor, $n_i=p_i=\sqrt{N_c{N}_v}{e}^{-E_g/(2k_{B}T)}$. Equal electron and hole masses give $N_c=N_v$ and intrinsic chemical potential midway between the band edges. At300K, the exponential is $e^{-19.34}=3.98\times 10^{-9}$. Using the later supplied mass $0.015m_e$ givesThis is many orders below a typical metal's carrier density. Since ${\sigma }_i=e{n}_i({\mu }_e+{\mu }_h)$, ordinary finite carrier mobilities leave its intrinsic conduction negligible on a metallic scale; the gap alone does not specify an exact conductivity without the mobilities.

Under the explicitly stipulated full donor ionization, $n\simeq N_d=10^{24}{\mathrm{m}}^{-3}$ and the supplied mobility is $0.0100{\mathrm{m}}^2{\mathrm{V}}^{-1}{\mathrm{s}}^{-1}$. ThereforeThe band diagram places the intrinsic chemical potential at midgap and a shallow isolated donor level just below the conduction edge. The doped chemical potential moves upward. Quantitatively, the degeneracy test for a fully ionized light-mass donor gas gives $N_d/N_c=21.7$, so a nondegenerate logarithmic chemical-potential formula is inconsistent. The parabolic-band zero-temperature estimate is $E_{F,d}-E_c={\hslash }^2(3{\pi }^2{N}_d{)}^{2/3}/(2m^{\ast })=0.243\mathrm{eV}$; its300K value is about0.241eV. The diagram accordingly shows the doped chemical potential inside the conduction band, not between an isolated donor level and the edge.

The hydrogenic donor state scaling using the supplied hydrogen values isThe binding is much smaller than $k_{B}T=25.9\mathrm{meV}$, indicating a shallow donor on the isolated-state estimate. However, the mean donor spacing is only $N_d^{-1/3}=10\mathrm{nm}$, well below the orbit radius. The donor-state overlap estimate invalidates an independent isolated-level ionization model at this concentration. Shallow binding supports abundant mobile carriers, but the full ionization versus isolated donor occupation distinction means full ionization cannot be justified solely by comparing binding with temperature while simultaneously retaining isolated levels and a degenerate chemical potential. The conductivity above follows the question's carrier-density assumption; an overlapping impurity/conduction-band description is needed for a self-consistent microscopic account. ^[ref]

One instrument is a diffraction grating spectrometer. Its principal maxima satisfy $d\sin \theta =m\lambda$ at normal incidence. N coherently illuminated grooves give a narrow principal peak; applying the Rayleigh criterion to the next wavelength yields the resolving power of a diffraction gratingHigher order and more illuminated grooves improve resolution, but finite entrance-slit width, aberrations, detector sampling and overlap between spectral orders limit useful resolution. Angular dispersion $d\theta /d\lambda =m/(d\cos \theta )$ is distinct from the ability to resolve two finite-width peaks.

A second instrument is a Fabry–Pérot interferometer. For plate spacing L, refractive index n and internal angle theta, transmission resonances satisfy $2nL\cos \theta =m\lambda$. Multiple coherent reflections produce an Airy transmission profile. For identical lossless mirrors of intensity reflectivity $\mathcal{R}$, the high-finesse limit givesThe frequency free spectral range is approximately $c/(2nL\cos \theta )$ for nondispersive n. A larger optical spacing narrows a line but also shrinks the unambiguous frequency interval; nearby orders need sorting. Mirror flatness, absorption, alignment and aperture broadening lower the effective finesse. This resolving-power formula assumes separated narrow lines and a stated line-separation convention, rather than equating transmission peak spacing with peak width.

The transient is the homogeneous part of a damped harmonic oscillator, $m\ddot{x}+b\dot{x}+kx=0$. Define $\alpha =b/(2m)$ and ${\omega }_0=\sqrt{k/m}$. Its characteristic roots are $-\alpha \pm \sqrt{{\alpha }^2-{\omega }_0^2}$.

Initial position and velocity determine A,B. For an imposed force or support motion, the full solution is a particular response plus this transient, which decays for b>0 and k>0. The mechanical energy obeys $d(K+U)/dt=-b{\dot{x}}^2$. In light damping its cycle-averaged envelope decays as $e^{-2\alpha t}$, with $Q\simeq {\omega }_0/(2\alpha )$. Initial conditions can still produce an overshoot in a nonoscillatory regime; the damping category describes the roots rather than every possible trajectory.

Small sound disturbances in a gas couple density, pressure and longitudinal velocity. Linearizing mass conservation and the pressure-gradient force about a uniform state givesEliminating velocity yields ${\partial }_t^2{\rho }^{\prime }=c_s^2{\nabla }^2{\rho }^{\prime }$. When local heat exchange is negligible over an oscillation, $c_s^2=(\partial p/\partial \rho {)}_s$. For an ideal gas,The adiabatic index is $\gamma =C_p/C_V$; an isothermal disturbance instead has $c_T^2=p_0/{\rho }_0$. Pressure and velocity in a progressive plane wave are in phase and obey $p^{\prime }={\rho }_0{c}_{s}u$, with average energy flux $⟨p^{\prime }u⟩=p_{\mathrm{peak}}^{\prime 2}/(2{\rho }_0{c}_s)$. Particle oscillation speed is not the sound propagation speed. Viscosity, heat conduction and molecular relaxation dissipate sound; changes in active internal degrees of freedom can cause dispersion. The continuum acoustic model requires wavelength large compared with the molecular mean free path and small fractional perturbations.

A solid's atoms do not travel freely through it, but its phonons and mobile electronic quasiparticles carry energy between scattering events. In a near-equilibrium isotropic diffusive regime, their transport can therefore use a kinetic thermal-conductivity approximation. Carriers moving along the local temperature gradient sample energies characteristic of their last scattering position. Linearizing that temperature difference and averaging squared directional cosines givesHere C is the contributing carrier heat capacity per unit volume, c its group speed and ell a transport mean free path. More accurately, the formula is a mode/energy-weighted sum of heat capacities times $v_g^2\tau /3$, not a product of arbitrary separately averaged values. Anisotropic crystals require a tensor, and ballistic samples or strongly nonquasiparticle regimes need a different description.

Lattice energy is transported by phonons; long-wavelength acoustic modes usually dominate low-temperature conduction. Their speed is set by elastic wave propagation. In the Debye model their volumetric heat capacity grows as $T^3$ at low temperature and approaches the classical lattice limit at high temperature. With boundary-limited mean free path, this gives a low-temperature $T^3$ conductivity; increasing phonon scattering can instead produce a peak and a declining high-temperature trend.

In metals, electrons near the Fermi surface carry both heat and charge. Their heat capacity is linear in T and their typical transport speed is the Fermi velocity, not a classical gas thermal speed. Degeneracy restricts the active thermal fraction but the large speed and long mean free path can make electronic heat conduction dominant. Insulators have few mobile electrons, so phonon conduction usually dominates. The approximate parallel-channel sum is $\kappa ={\kappa }_e+{\kappa }_{\mathrm{ph}}$; which channel dominates is material- and temperature-dependent. ^[ref]

Phonon paths are limited by boundaries, defects, isotopes and phonon–phonon interactions; electrons scatter from impurities, defects, boundaries and lattice vibrations. At low temperature a clean sample can be boundary-limited, while residual electronic impurity scattering survives as phonons freeze out. In a quasiparticle relaxation-time approximation, independent resistive rates are sometimes added, ${\tau }^{-1}\simeq {\sum }_j{\tau }_j^{-1}$, but this is not an exact rule for every coupled process.

Umklapp scattering transfers crystal momentum by a reciprocal lattice vector and can relax a heat current. The normal phonon collisions and heat-current relaxation distinction is that a normal phonon collision conserves the total crystal momentum of the phonon distribution; it is not automatically a resistive thermal-scattering event in the same way. Normal collisions can redistribute populations or establish hydrodynamic flow, with boundaries and other momentum-relaxing processes then setting resistance. Likewise, different angular scattering kernels distinguish a single-particle lifetime from a transport lifetime. These qualifications explain why estimating ell requires physical scattering information, not simply a collision count.

The Wiedemann-Franz law relates the electronic part of thermal conductivity to electrical conductivity:For degenerate electrons with elastic scattering and a transport function smooth over $k_{B}T$, a Sommerfeld expansion makes the same Fermi-level transport factor appear in charge conduction and in heat conduction. Their ratio cancels that factor and leaves the displayed Lorenz number. Heat conductivity is defined at zero electric current, so thermoelectric subtraction is part of the precise transport expression.

The law does not relate total solid heat conductivity to charge conductivity when phonons contribute substantially. Strong inelastic scattering, sharp energy dependence, nondegenerate carriers or nonstandard electronic regimes can change the ratio. Low-temperature residual elastic-scattering transport is a common setting where the standard value is recovered. The unit W·ohm/K² follows because ${\kappa }_e/(\sigma T)$ contains one extra inverse temperature beyond thermal conductivity itself.