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The instantaneous rightward energy flux along a string is the transverse force times the transverse velocity,For a rightgoing time-harmonic wave of peak displacement wave amplitude $|A|$, averaging over a wave period givesA leftgoing time-harmonic wave has the opposite signed energy flux. The positive ratios of reflected and transmitted power to incident power are thereforeAll explicit angular frequency factors cancel. In the ideal wave equation on a string, $T$ and the two linear mass densities are constant, so the mechanical impedances have no frequency or wavelength dependence. Consequently $\mathcal{T}$ is independent of wavelength even though the two string wavelengths generally differ.

Finally,This is conservation of energy at a lossless, massless junction. The transmitted displacement wave amplitude can exceed the incident one when $Z_R<Z_L$; the factor $Z_R/Z_L$ is essential when comparing power.

Put $\delta =k_{M}l$. Eliminating the middle-segment complex amplitudes using the right-hand boundary conditions givesHence the displacement reflection coefficient isFor finite positive real mechanical impedances, its denominator cannot vanish at real $\delta$. The real and imaginary parts of the numerator must therefore vanish separately:This gives all three possibilities.

If the exterior strings differ, $Z_L\ne Z_R$, then $\cos \delta =0$, so $\sin \delta \ne 0$ andThe middle segment is a quarter-wave impedance transformer. Its two reflected contributions cancel at these odd quarter-wavelength thicknesses. Here ${\lambda }_M=2\pi /k_M$ is specifically the wavelength in the middle string; the matching is not generally valid at every frequency.

If $Z_L=Z_R$ but $Z_M$ differs, the second condition requires $\sin \delta =0$, givingA whole number of half-wavelengths reproduces the exterior mechanical impedance at the input. The formal $n=0$ limit is zero frequency or zero segment length, rather than a finite-wavelength resonance.

Finally, if $Z_L=Z_M=Z_R$, both conditions hold identically: $\boxed{r=0\text{ at every wavelength}}$. Since the system is lossless, each of these cases also gives unit transmitted power.

The expression is a scalar diffraction theory integral for a monochromatic wave. The complex quantity ${\psi }_P$ represents a field complex amplitude, with physical field obtained by taking the real part after multiplication by $e^{-i\omega t}$. The light intensity is proportional to $|{\psi }_P{|}^2$. The coordinates $x,y$ label transverse positions in the aperture plane, $\Sigma$ is its transmitting region, and $h(x,y)$ is the incident field multiplied by the aperture's complex transmission. The wavenumber is $k=2\pi /\lambda$ in the propagation medium, and $R$ is the axial propagation distance if $h$ already includes the incident field.

To derive the Fresnel diffraction, start from Huygens-Fresnel principle contributions with propagation factor $e^{iks}/s$. With the coordinate origin at the projection of $P$ into the aperture plane,The paraxial approximation requires $\rho /R\ll 1$, where ${\rho }^2=x^2+y^2$. Replace the slowly varying $1/s$ and the obliquity factor by their axial values, but retain the quadratic variation of wave phase. Neglect of the next wave phase term requires $k{\rho }^4/(8R^3)\ll 1$ over the relevant aperture. Thus, apart from a common constant,The quadratic wave phase may be of order one or much larger; setting it to zero would require the stronger conditions for Fraunhofer diffraction. The scalar treatment also neglects the detailed vector field at aperture edges and assumes dimensions large enough compared with wavelength for this approximation to be useful.

For a finite-distance axial point source at distance $p$ and an observer at distance $q$, the incident field contributes its own quadratic wave phase. If $h$ instead denotes only the aperture transmission, the combined curvature is described by an effective distanceThese are two equivalent conventions for distributing the incident wave phase between $h$ and the exponential. Both distances must be included in the later calculation.

The Debye model approximates a three-dimensional crystal as an elastic continuum supporting quantized normal modes, or phonons. At long wavelengths, the acoustic phonon branches have approximately linear dispersion relations ${\omega }_s(k)=v_{s}k$: one longitudinal branch and two transverse branches. Replace their complicated actual spectra by these linear forms, then introduce a cutoff chosen to preserve the total $3N$ vibrational degrees of freedom for $N$ atoms. This is an approximation to the spectrum, not a claim that a real crystal has no non-acoustic branches.

For periodic boundary conditions in volume $V$, each allowed wavevector occupies $(2\pi {)}^3/V$ in the space of wavevectors. Counting a spherical shell for each branch gives the density of states per unit angular frequencyAn effective speed of sound $v_D$ can be defined by $3/v_D^3=1/v_l^3+2/v_t^3$. In the single-speed Debye model, choose ${\omega }_D=v_D{k}_D$ so thatThus the Debye wavevector, Debye frequency and Debye temperature satisfyHere $g$ has units of seconds when angular frequency is measured in ${\mathrm{s}}^{-1}$; its integral is the dimensionless number of normal modes. The spherical cutoff replaces the real Brillouin zone by an equal-count continuum construction.

Each phonon mode is a quantum harmonic oscillator. Since phonon number is not conserved at thermal equilibrium, the chemical potential is zero and its Bose-Einstein distribution gives mean thermal energy $\hslash \omega /[e^{\hslash \omega /(k_{B}T)}-1]$. The zero-point energy is independent of temperature in the harmonic, fixed-volume approximation and does not contribute to the heat capacity at constant volume. Excluding that constant term,Differentiating the first integral at fixed volume avoids losing the temperature-dependent upper limit in the second one. The resulting Debye heat capacity isFor $T\ll {\Theta }_D$, extend the upper limit to infinity. Expanding $1/(e^x-1)={\sum }_{r\ge 1}{e}^{-rx}$ and integrating term by term gives ${\int }_0^{\infty }{x}^3/(e^x-1)dx=6\zeta (4)={\pi }^4/15$. ThereforeThis $T^3$ law follows from the three-dimensional ${\omega }^2$ density of states and the availability only of low-frequency modes at low temperature. It is a central success of the Debye model.

For $T\gg {\Theta }_D$, the Bose-Einstein distribution reduces to the classical equipartition theorem: every mode contributes thermal energy $k_{B}T$. Consequentlythe Dulong-Petit law, or $3R$ per mole of atoms. The simple model omits detailed phonon dispersion relation, distinct optical phonon branches, and anharmonic effects such as thermal expansion. In a metal, the low-temperature total heat capacity additionally contains a term approximately proportional to $T$ from Electrons, so the lattice $T^3$ term should not be mistaken for the whole measurement.

Let the infinite outer walls be at $x=0,D$, the finite steps at $x=a,b$, and the three potential energy values be $V_L,V_B,V_R$. The original drawing specifies the orderingThe energy eigenstates can be chosen real. The Dirichlet boundary conditions are $\psi (0)=\psi (D)=0$; wavefunction matching at a finite potential step requires both $\psi$ and ${\psi }^{\prime }$ to be continuous at $a,b$. In particular, neither finite step acts as an additional hard wall.

The Time-independent Schrödinger equation givesThus a wavefunction is sinusoidal in a classically allowed region, with wavenumber $k=\sqrt{2m(E-V)}/\hslash$, and a linear combination of exponential functions in a classically forbidden region, with inverse decay length $q=\sqrt{2m(V-E)}/\hslash$. Its curvature changes sign relative to $\psi$ at a step crossing $E$; its slope does not jump.

By the Sturm oscillation theorem, the three energy eigenstates have exactly zero, one, and two interior nodes, respectively. Choose their global phases so that their slopes at the left wall are positive:

For $E_0,E_1$, the right-hand piece is proportional to $\sinh [q_R(D-x)]$, rather than a pure decaying exponential: this enforces the finite-distance hard wall exactly. The barrier pieces likewise contain both exponential solutions. Their small tails are an instance of quantum tunnelling.

This is an original illustrative calculation, with vertical lines marking only the region boundaries. It uses $D=1$, $a=0.4$, $b=0.6$, $V_L=0$, $V_R=0.62$, $V_B=0.82$, and ${\hslash }^2/(2m{D}^2)=0.002$ in arbitrary energy units. Its lowest energy eigenvalues are approximately $0.098,0.384,0.696$, so it has exactly the required ordering. Here both nodes of the third wavefunction happen to lie towards the left; its dominant lobe is on the right. These precise positions and relative amplitudes are illustrative: the original drawing supplies no numerical lengths, barrier heights, or mass, so it does not determine a unique quantitative wavefunction. The required sketch features are the matching, curvature, hard-wall zeros, and ordered node counts.

For the normally incident plane wave, the incident complex amplitude and phase are uniform across the aperture. Treat the field using scalar diffraction, with the aperture represented by its prescribed amplitude transmittance $h(x,y)$. Let $R$ be the distance from the aperture centre to the observation point and let ${\ell }_x,{\ell }_y$ be the transverse direction cosines of that point.

Across an aperture of characteristic size $a$, the Fraunhofer diffraction assumptions are that $a/R\ll 1$, so $1/r\simeq 1/R$ and the obliquity factor is nearly constant, and that the quadratic contribution to the propagation phase is negligible:The remaining path-length dependence is linear:A coherent monochromatic wave and an aperture large enough relative to wavelength for the scalar aperture model complete the usual physical assumptions. The complex amplitude becomeswhere $C$ includes the common propagation phase, distance factor and nearly constant obliquity factor. For the central part of a planar screen in the paraxial approximation, ${\ell }_x\simeq X/R$ and ${\ell }_y\simeq Y/R$. Small observation angles are useful for that screen-coordinate formula; a general far-field Fourier transform can instead use the exact direction cosines. A lens can also produce this pattern in its focal plane without requiring a physically very distant screen.

For $N$ equally spaced illuminated slits, with $N\simeq W/D$, summing a geometric series of complex amplitudes givesHere $I_1(\theta )$ is the one-slit single-slit diffraction wave envelope. The grating equation for principal maxima iswith height $N^2{I}_1({\theta }_m)$ and much weaker subsidiary maxima between orders. The first zeros around order $m$ are $\beta =m\pi \pm \pi /N$, so the angular distance from a maximum to its first minimum is, for large $N$,Changing wavelength by $\delta \lambda$ moves the same diffraction order byThe Rayleigh criterion places the maximum for one wavelength at the first zero for its neighbour. Equating these separations givesThis is the ideal resolving power of a diffraction grating, for an illuminated and transmitted order with sufficiently narrow source and instrumental broadening. The zeroth order has no wavelength separation and therefore no useful spectral resolving power. The distinction between $W=ND$ and the span $(N-1)D$ of the outer slit centres is a small edge convention; the exact state count in the resolving-power formula is $N$.

The required resolving power of a diffraction grating isFor $N=260$ illuminated slits, $\mathcal{R}=260|m|$. In orders $1,2,3,4$, the ideal resolvable wavelength differences are respectively $2.267$, $1.133$, $0.756$ and $0.567\mathrm{nm}$. ThusFourth order gives $\mathcal{R}=1040>982.2$, only a modest margin. It must also exist: for normal incidence the grating equation requires $D\ge 4\lambda \simeq 2.36\mu \mathrm{m}$, with $D>4\lambda$ needed to avoid an exactly grazing order. The relevant order must transmit appreciable light intensity rather than lie at a single-slit diffraction zero, and the source and detector widths must preserve this resolution.

The original question gives only the slit count, not the line spacing, slit width, usable diffraction order or instrumental broadening. It therefore supports the conditional conclusion “yes in fourth or higher usable order”, but not an unconditional historical claim that this particular instrument resolved the sodium absorption line doublet in its observed configuration.

To determine a crystal structure, scatter probes whose wavelength is comparable to atomic spacings, typically $0.1\mathrm{nm}$. X-ray diffraction, neutron diffraction and electron diffraction probe periodic order through coherent elastic scattering. With incident and outgoing wavevectors ${\mathbf{k}}_i,{\mathbf{k}}_f$, define $\mathbf{Q}={\mathbf{k}}_i-{\mathbf{k}}_f$. Elastic scattering has $|{\mathbf{k}}_i|=|{\mathbf{k}}_f|$ and constructive interference occurs when $\mathbf{Q}=\mathbf{G}$, a reciprocal lattice vector. The equivalent Bragg's law iswhere the detector scattering angle is $2\theta$, not $\theta$. Peak positions reveal plane spacings and the unit cell; systematic absent reflections constrain centring and the atomic basis.

For X-ray diffraction, a laboratory tube or a synchrotron supplies a collimated beam. In a monochromatic single-crystal experiment a goniometer rotates the specimen and an area detector records reflection positions and integrated wave intensities. A powder experiment uses many randomly oriented crystallites; the reflections form cones recorded as rings or peaks versus $2\theta$, permitting indexing and determination of lattice constants without selecting one crystallite. For a cubic cell, for example, $d_{hkl}=a/\sqrt{h^2+k^2+l^2}$. The white-beam Laue method is especially useful for crystal orientation.

The crystal basis structure factor isafter geometric and thermal corrections, including the Debye–Waller factor. For X-rays, $f_j$ describes scattering from the atomic electron density. Thus positions provide lattice geometry while relative wave intensities constrain atomic positions and species. Wave intensities do not directly give the phases of $F_{\mathbf{G}}$: structure refinement or other constraints must address the phase problem.

Neutron diffraction uses a reactor or spallation source, wavelength selection and a detector bank or time-of-flight detection. Neutrons interact with nuclei through nuclear scattering lengths that vary between elements and isotopes rather than simply with atomic number. This makes them useful for light atoms, isotope contrast and distinguishing some neighbours in the periodic table; their magnetic interaction also probes magnetic order. Electron diffraction in a transmission electron microscope gives strong signals from very small thin crystallites, although strong multiple scattering often complicates quantitative wave intensity analysis. These techniques complement one another.

A phonon dispersion relation requires both momentum and energy transfer, so elastic diffraction alone is insufficient. In coherent inelastic neutron scattering, the measured transfers for phonon creation areHere $\mathbf{q}$ is the phonon wavevector reduced into a Brillouin zone, and $s$ labels an acoustic phonon branch or optical phonon branch. Negative energy transfer corresponds to annihilating an already occupied phonon. For neutrons $E={\hslash }^2{k}^2/(2m_n)$, so wavelength and energy selection are linked. Typical thermal-neutron energies of a few to tens of millielectronvolts ($1\mathrm{meV}=10^{-3}$ electronvolts) are well matched to many lattice vibrations.

A triple-axis spectrometer uses a monochromator crystal to select ${\mathbf{k}}_i$, the sample as the second scattering axis, and an analyser crystal to select ${\mathbf{k}}_f$ before detection. Rotate these elements and the specimen to hold $\mathbf{Q}$ fixed while scanning $E_i-E_f$; inelastic peaks locate ${\omega }_s(\mathbf{q})$. Repeat at different $\mathbf{q}$ along symmetry directions to map the phonon dispersion relation. The dynamic structure factor governs peak weights, including the selection of phonon polarization by the scattering geometry. A time-of-flight spectrometer instead infers outgoing neutron energy from travel time over a known flight path and records many directions simultaneously.

Inelastic X-ray scattering can also resolve phonons when very high energy resolution is available. Optical Raman scattering and Brillouin scattering measure vibrations near the zone centre because photon momentum transfers are small compared with a typical reciprocal lattice spacing; they complement, rather than generally map the entire zone like, momentum-resolved neutron measurements. Ultrasound measurements give the long-wavelength sound speed, the initial slope $d\omega /dq$ of an acoustic phonon branch.

Assume a hollow square electromagnetic waveguide with perfectly conducting walls, filled with vacuum or air. For side $a$, the waveguide cutoff frequency isA transverse electric mode can have one zero index, but not both. A transverse magnetic mode requires both indices positive. Thus the lowest cutoff belongs to the two degenerate transverse electric modes ${\mathrm{TE}}_{10}$ and ${\mathrm{TE}}_{01}$, withThe corresponding cutoff vacuum wavelength is $2a=2.0\mathrm{cm}$. Strictly, a travelling mode carrying power along the guide needs $f>f_c$: at the cutoff itself, the axial wavenumber and group velocity are zero. The number above is therefore the lower frequency threshold, not a propagating mode with nonzero longitudinal energy flux exactly at that frequency. A dielectric filling would replace $c$ by the wave speed in that medium.

Equilibrium radiation and cavity frequencies. A blackbody absorbs all incident radiation at every frequency and direction. In thermal equilibrium its emission spectrum depends only on its temperature. Radiation inside an isothermal absorbing cavity, sampled through a small aperture that hardly disturbs equilibrium, realizes black-body radiation to a good approximation. Microscopic absorption and emission establish the equilibrium distribution; a perfectly reflecting wall is the convenient idealization used to count the modes once that distribution is specified.

In a rectangular electromagnetic cavity, the wave equation and perfect conductor wall boundary conditions select component wavenumbers $k_x=\pi l/L_x$, $k_y=\pi m/L_y$, $k_z=\pi n/L_z$. Combining the vacuum dispersion relation $\omega =c|\mathbf{k}|$ and $\nu =\omega /(2\pi )$ givesFor example, electric-field components can have the spatial dependenceFor three positive indices, Gauss's law requires $k_x{A}_x+k_y{A}_y+k_z{A}_z=0$, leaving two independent polarizations of an electromagnetic wave.

Index qualification. The original paper describes the indices as positive integers. That describes the three-nonzero-index family but omits valid electromagnetic modes with exactly one zero index, which have one polarization of an electromagnetic wave. For instance, with $n=0$, $l,m\ge 1$, only $E_z$ above remains; it is normal to the top and bottom faces and vanishes at the side walls. Two zero indices give no nontrivial cavity field. The omitted one-zero-index family does not affect the leading three-dimensional mode count, but is essential to the thin-cavity limit below. Retaining the paper's strict positivity restriction for every mode would instead give a thickness gap and would not yield its requested two-dimensional power law.

Three-dimensional mode count. A lattice cell in positive $\mathbf{k}$ space has volume ${\pi }^3/V$, where $V=L_x{L}_y{L}_z$. In a continuum approximation, a shell of radius $k$ and thickness $dk$ contains its positive octant, multiplied by two polarizations:Using $k=2\pi \nu /c$ givesFor a finite cavity the exact density of states is discrete; this formula is its bulk continuum limit, appropriate when many modes contribute and wall corrections are small.

Mean photon number. A mode is a quantum harmonic oscillator with thermal excitation energies $sh\nu$, $s=0,1,2,\dots$. The zero-point energy does not enter its thermal occupation. Put $q=e^{-h\nu /(k_{B}T)}$. Its partition function and mean occupation areThis is the Bose-Einstein distribution with zero photon chemical potential: equilibrium absorption and emission do not conserve photon number.

Thermal energy. Multiply the density of states by the mean excitation energy $h\nu ⟨s⟩$ and integrate:The substitution $x=h\nu /(k_{B}T)$ isolates all the dependence on cavity dimensions and temperature:Expanding $1/(e^x-1)={\sum }_{r=1}^{\infty }{e}^{-rx}$ gives an integral ${\sum }_{r}6/r^4=6\zeta (4)={\pi }^4/15$. HenceThis is the thermal energy only, excluding the zero-point contribution. The three-dimensional continuum regime requires $k_{B}T$ large compared with the geometric level spacings in all three directions.

Spectral sketch. The spectral energy density per unit ordinary frequency is Planck's law,At small frequency it rises as $8\pi k_{B}T{\nu }^2/c^3$, while at large frequency it decays exponentially. Raising temperature increases the spectrum at every fixed positive frequency, moves the peak to larger frequency in proportion to $T$, and increases the integrated area in proportion to $T^4$. A plot per unit wavelength has a different peak position because the integration variable changes.

Thin-cavity radiation. Let $A=L_x{L}_y$. The relevant low-temperature window isNonzero thickness indices are then thermally suppressed, while many in-plane modes are populated. The $n=0$ family has one polarization of an electromagnetic wave. Its allowed in-plane wavenumbers form a lattice of cell area ${\pi }^2/A$; a quarter of a circular shell therefore containsThus the dimensional crossover of thermal cavity radiation givesThe integral is $2\zeta (3)$ by the same geometric-series argument. A model with a different number of independent polarizations of an electromagnetic wave changes the prefactor but not the $T^3$ scaling. For a cavity with fixed finite $L_x,L_y$, cooling still further eventually freezes its lowest in-plane mode as well: the strict $T\to 0$ thermal energy is exponentially suppressed. The requested area law is the intermediate quasi-two-dimensional regime, not an arbitrarily low-temperature continuum limit.

A transmission line must be treated as a distributed circuit when its propagation time is significant on the signal timescale. Let $R^{\prime },L^{\prime },G^{\prime },C^{\prime }$ denote resistance, inductance, leakage conductance and capacitance per unit length. Applying Kirchhoff's voltage law and Kirchhoff's current law to a short length $dx$ gives the telegrapher's equationsFor a lossless line, $R^{\prime }=G^{\prime }=0$. Differentiating again gives wave equations withThe travelling solutions can be written $V=V_{+}(t-x/v)+V_{-}(t+x/v)$ and $I=[V_{+}(t-x/v)-V_{-}(t+x/v)]/Z_0$. The minus sign for the backward wave is essential to power transport.

At a load $Z_L$, continuity of voltage and current gives the voltage reflection coefficientA matched termination $Z_L=Z_0$ absorbs the incident signal without reflection. An open circuit has ${\Gamma }_L=+1$ and a short circuit has ${\Gamma }_L=-1$. Superposed incident and reflected waves produce standing waves. A sinusoidal lossless line of length $\ell$ transforms the termination intofor the $e^{i\omega t}$ convention. At a quarter wavelength, $Z_{\mathrm{in}}=Z_0^2/Z_L$, allowing impedance transformation. A coaxial line supports a transverse electromagnetic mode whose ideal $L^{\prime },C^{\prime }$ are calculated in A4. Its energy travels in the fields between conductors; the characteristic impedance is not the wire's DC resistance. Losses introduce attenuation and make the characteristic impedance and propagation constant complex.

A cubic reflecting cavity supports standing waves with positive integer mode indices and wavevector components $k_i=\pi n_i/L$. A cell in mode space has volume $(\pi /L{)}^3$. In the continuum approximation, a spherical shell of radius $k$ and thickness $dk$ has one-eighth of its volume in the positive octant. There are two independent transverse photon polarizations for a generic wavevector. HenceThe photon energy is $ϵ=\hslash \omega =\hslash ck$, so $dk=dϵ/(\hslash c)$. ThereforeThis is the leading bulk density of states, valid when the cavity is large compared with the relevant wavelengths and many modes lie in the counting interval. A finite cavity has a discrete mode spectrum, with boundary corrections beyond this approximation. Modes with vanishing indices require separate boundary counting but do not change the leading $L^3$ term. Here $\hslash$ is the reduced Planck constant, as printed in the original PDF; replacing it by $h$ would introduce an erroneous factor of $(2\pi {)}^3$.

Kirchhoff's law of thermal radiation equates spectral absorptivity and spectral emissivity at the same wavelength and direction. A surface that is a poor absorber in the thermal infrared is also a poor emitter there; a coating cannot be assigned its solar absorption independently of its thermal emission spectrum.

(i) All-wavelength black paint. It absorbs the sunlight and also radiates maximally at the spacecraft's own temperature. With the isothermal-sphere assumptions of (d), the limiting black-surface distance isThe spacecraft can operate at $R\ge 50r$ in this model. A surface already assumed to be a blackbody is not improved by another black coating; relative to option (ii), however, this paint permits closer operation.

(ii) Black at short wavelengths and grey in the infrared. By Wien's displacement law, changing the black-body radiation temperature from $6000\mathrm{K}$ to $600\mathrm{K}$ shifts the wavelength peak from about $0.5\mu \mathrm{m}$ to about $5\mu \mathrm{m}$. Most solar power is therefore in the highly absorbing short-wavelength band, whereas most spacecraft emission lies beyond $2\mu \mathrm{m}$, where the spectral emissivity is reduced. The coating retains most of the heating but suppresses much of the cooling. Under the intended two-band grey body approximation, option (ii) makes the spacecraft hotter at the same distance and requires a larger minimum distance than black paint.

To express the comparison without inventing a grey absorptivity, let ${\bar{\alpha }}_S$ be radiometric absorptance averaged over the solar spectrum and ${\overline{ϵ}}_T$ be emissivity averaged over the spacecraft's own thermal spectrum. The radiative equilibrium equation becomesAt the failure threshold,For an ideal two-band coating take a constant radiometric absorptance $g<1$ beyond ${\lambda }_c=2\mu \mathrm{m}$, and unity below it. If $f(T)$ is the fraction of the power emitted by a blackbody below ${\lambda }_c$, Planck's law givesSince $f(T_S)>f(600\mathrm{K})$, this model has ${\bar{\alpha }}_S>{\overline{ϵ}}_{600}$ and hence $R_{\min }>50r$. The paper supplies neither a numerical value of $g$ nor a detailed spectrum, so no unique numerical distance for the grey paint follows. If its long-wavelength absorption is not approximately grey, the spectrum must be integrated explicitly rather than inferred from a peak alone. A coating designed to improve close approach would instead reflect much of the solar radiation while retaining strong thermal infrared emission.