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In geometrical optics, suppose radiation has a nonnegative radiance $L(\mathbf{x},\mathbf{s})$ whose directional contributions add as power. Here $\mathbf{s}$ is the actual propagation unit vector, $d\Omega$ is a solid angle element, and the collecting oriented surface $A$ has chosen unit normal $\mathbf{n}(\mathbf{x})$. Assume the following integrals exist. The outward and inward radiant fluxes areBoth quantities are nonnegative; their difference is the signed net radiant flux. The dot product supplies the cosine of the angle with the unit normal, hence the projected area. For incoming radiation the outward unit normal gives a negative dot product; the minus sign makes the collected power positive. Measuring the incoming angle from a source-facing unit normal instead gives the familiar positive $\cos \theta$. The units check is $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\times {\mathrm{m}}^2\times \mathrm{sr}=\mathrm{W}$.

Equivalently, the incoming surface integral is ${\Phi }_{-}={\int }_{A}EdA$ with local irradiance $E$; the emitted part of ${\Phi }_{+}$ is ${\int }_{A}MdA$ with radiant exitance $M$. For a small flat patch and a narrow accepted cone, nearly constant radiance and angle give $\Phi \simeq LA\cos \theta \Delta \Omega$. This approximation requires uniformity across the patch and cone. Direction-independent outgoing radiance over a whole hemisphere gives the different exact result ${\Phi }_{+}=\pi LA$, by Lambert's cosine law and the spherical coordinates calculation in radiant exitance. Replacing the cosine-weighted integral by the hemisphere's $2\pi$ solid angle would double the answer.^[ref]

These radiance formulas presume a description in additive directional power; they do not reconstruct arbitrary coherent wave interference from separate ray contributions. For general electromagnetic fields, the signed instantaneous power crossing an oriented surface is ${\int }_A\mathbf{S}\cdot \mathbf{n}dA$, using the Poynting vector $\mathbf{S}$; optical measurements commonly average over many wave periods. A zero net radiant flux can conceal equal nonzero incoming and outgoing radiant fluxes.

For normal incidence in vacuum, a uniform beam delivers energy $IA\Delta t$ to an illuminated surface area $A$ during time $\Delta t$. A perfect absorber receives momentum $IA\Delta t/c$, so force divided by surface area gives $p_{\mathrm{abs}}=I/c$. A perfect stationary mirror sends the photons back with reversed momentum, so momentum conservation gives twice the normal transfer, $p_{\mathrm{refl}}=2I/c$. These are time-averaged pressures; the microscopic arrivals need not be uniform.

For a stipulated model light intensity $I=1000\mathrm{W}{\mathrm{m}}^{-2}$ and the exact SI speed of light $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$,These are computed predictions, not measured data or a universal sunlight value. If the beam fully covers a normally oriented surface area $A=0.010{\mathrm{m}}^2$, the intercepted radiant flux is $\Phi =IA=10\mathrm{W}$ and the absorber's force is $\Phi /c\simeq 33.356\mathrm{nN}$. Using the total source power would be wrong if only part of its beam reaches the receiver. The absorption prediction assumes complete photon absorption, while the reflection prediction assumes complete specular reflection; transmission, scattering of light and directed thermal photon emission require their own outgoing momentum accounting.

Measure the angle of incidence $\theta$ between the incoming path and the surface's normal line, with $0\le \theta <\pi /2$. There are two separate cosine factors. A fixed actual surface area $A$ intercepts only its projected area $A\cos \theta$, so $\Phi =IA\cos \theta$. Each incoming photon supplies normal momentum $(E_{\gamma }/c)\cos \theta$ on absorption, or twice that on specular reflection. Thus, for radiometric absorptance $a$ and reflected power fraction $R$,This model assumes transmitted light continues straight in the same medium and carries away its original momentum. It omits thermal recoil and assumes a stationary receiver. At $\theta =60^{\circ }$, fixed beam light intensity therefore gives one quarter of the normal-incidence pressure. For fixed intercepted radiant flux $\Phi$, however, the normal force has only one cosine factor. Equivalently, if the given quantity is receiving-surface irradiance $E_{\mathrm{inc}}=I\cos \theta$, then $p_{\perp }=(a+2R)E_{\mathrm{inc}}\cos \theta /c$. The distinction matters when interpreting an experiment's angle scan. ^[ref]

Absorption also transfers a tangential momentum component at oblique incidence. It contributes a tangential force, rather than normal pressure. Ideal specular reflection preserves the tangential momentum component, so it transfers only normal momentum.

A radiative-convective climate column applies a radiative-convective equilibrium model to the vertical thermal structure relevant to climate change. Radiative transfer determines heating and cooling; convective adjustment represents unresolved vertical redistribution when the chosen atmospheric lapse rate is too steep. Radiative equilibrium alone need not give a physically admissible lower-atmospheric profile.

In 1967, Syukuro Manabe and Richard T. Wetherald computed equilibrium with prescribed relative humidity, allowing atmospheric water vapor to change with temperature. Their average-cloud case warmed by $2.36\mathrm{K}$ for carbon dioxide increasing from 300 to 600 parts per million, versus $1.33\mathrm{K}$ with fixed absolute humidity. These are model outputs, not measured warming. The comparison isolates an important water-vapor feedback.^[ref]

An original two-temperature calculation exposes the bookkeeping. Let an opaque surface absorb solar irradiance $Q$ and emit blackbody thermal radiation $\sigma T_s^4$. A single isothermal atmospheric layer transmits all sunlight and absorbs a fraction $ϵ$ of surface infrared emission, emitting $ϵ\sigma T_a^4$ both upward and downward. This is a grey thermal-band emissivity, with $0<ϵ\le 1$. Let $H$ be upward nonradiative heat flux, and let $C_s,C_a>0$ be heat capacities per unit area. Then conservation of energy givesThe two equations sum toHere $L$ is outgoing longwave radiation. Both $H$ and downward atmospheric emission cancel internally: they redistribute energy without creating it. The factor two in atmospheric emission counts two faces. Dropping the downward face would spoil the surface budget. Each right-hand term has units $\mathrm{W}{\mathrm{m}}^{-2}$; each $C$ has units $\mathrm{J}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$.

Choose $Q=240\mathrm{W}{\mathrm{m}}^{-2}$, $H=80\mathrm{W}{\mathrm{m}}^{-2}$ and $ϵ=1$. At equilibrium, $L=\sigma T_a^4=240\mathrm{W}{\mathrm{m}}^{-2}$ and $\sigma T_s^4=400\mathrm{W}{\mathrm{m}}^{-2}$, giving $T_a\simeq 255.06\mathrm{K}$ and $T_s\simeq 289.81\mathrm{K}$. Direct substitution gives surface balance $240+240-400-80=0$ and atmospheric balance $400-480+80=0$. These deliberately chosen fluxes illustrate the model; they are not an observed global budget or a reconstruction of the 1967 calculation. Prescribing $H$ illustrates heat transfer accounting but does not close a realistic convection law.

For fixed $Q,H,ϵ$, the positive-temperature equilibrium is locally stable. Put $a=4\sigma T_s^3>0$ and $b=4ϵ\sigma T_a^3>0$. The Jacobian matrix of the temperature equations has trace $-a/C_s-2b/C_a<0$ and determinant $ab(2-ϵ)/(C_s{C}_a)>0$. Thus both eigenvalues have negative real part. Temperature-dependent climate feedbacks alter this test. A single grey layer also omits the optical spectrum, cloud structure and the independent thermal behavior of the troposphere and stratosphere.

A black surface receiving absorbed stellar irradiance $Q$, beneath a single non-solar-absorbing cloud with thermal emissivity $e>0$, satisfies $T_c^4=T_g^4/2$ when both bodies are in pure radiative equilibrium. The surface balance is $\sigma T_g^4=Q+e\sigma T_c^4$, giving the displayed result. If instead convection pins $T_c=q{T}_g$, the whole-column balance gives $T_g^4=Q/[\sigma (1-e+e{q}^4)]$; the nonradiative energy budget cannot then be discarded.

Radiometry measures the energy and power carried by electromagnetic radiation, including light. It asks how much radiation a source supplies, where it travels, and how much reaches a receiving surface. The measurements are physical rather than weighted by the sensitivity of the human eye: equal radiant fluxes at different wavelengths need not look equally bright. ^[ref].

The five quantities below answer different questions about the same radiation. The International System of Units makes the distinctions visible: a watt is a joule per second, while a steradian measures solid angle. ^[ref].

Resolving wavelength or frequency adds a further “per unit interval”, as in spectral radiance. The notation is consistent with NIST's table of spectral quantities. In particular, radiant flux is power, whereas the existing astronomical radiative flux is power per area. ^[ref].

This bookkeeping makes an optical camera's collected signal calculable from a source's radiance and the geometry of the receiving aperture. It also lets radiation measurements from different instruments be compared. For example, NIST describes separate calibrations of spectral radiance and spectral irradiance for instruments used to observe Earth. ^[ref].

The Rayleigh range of a fundamental Gaussian beam in vacuum is the positive axial distance from the beam waist at which its radius grows from $w_0$ to $\sqrt{2}{w}_0$. Thus its contour area doubles and its peak irradiance halves. Within $|z|\le z_R$, the radius stays at most $\sqrt{2}{w}_0$; for $|z|\gg z_R$, beam divergence dominates. At fixed vacuum wavelength, the Rayleigh range scales quadratically with the beam waist radius. The formula here uses the vacuum wavelength, not an unlabelled wavelength in a material.

The ratio of water-leaving radiance to incident downwelling irradiance just above the surface. It depends on both absorption and backscattering and has units of inverse steradian.

Consider an original, explicitly normalized model of primary coma, rather than a prescription for a particular optical lens. Let $(u,v)$ be Cartesian coordinates in a circular optical pupil divided by its radius $a$, with $u^2+v^2\le 1$. Let $h=\alpha /{\alpha }_0$ be a signed small field angle divided by an arbitrary positive reference angle. With polar coordinates $u=\rho \cos \theta$, $v=\rho \sin \theta$, prescribe the wavefront errorHere $W_0$ is an arbitrary positive optical path difference scale, not a measured error. The cubic optical pupil dependence and linear field dependence are the classical Seidel form.^[ref]

In the paraxial approximation, light ray direction errors follow the transverse gradient of wavefront error. For a reference sphere a distance $L$ from the exit optical pupil in a uniform image medium of refractive index $n$, the corresponding image-error scale is $s=L{W}_0/(na)$. Choose image-error coordinate signs so the normalized displacement is $\nabla w$; reversing the convention reverses the plotted orientation. Image coordinates below are errors relative to the ideal image of this particular object point, rather than positions of different objects:The partial derivatives explain the asymmetry: opposite optical pupil points $(u,v)$ and $(-u,-v)$ land together, rather than on opposite sides of the ideal image. For $h>0$, every $X$ is nonnegative, while $Y$ can have either sign. The figures stipulate this single wavefront error, omit other optical aberrations and diffraction, and assign no instrument dimensions.

For a ring of fixed optical pupil radius $\rho$, the double-angle formulas for trigonometric functions giveThus each ring maps into a circle of radius $|h|{\rho }^2$ whose centre is displaced by $2h{\rho }^2$. Doubling $\rho$ quadruples both sizes; for $h=1$, the edge ring has centre $(2,0)$ and radius $1$, while the half-radius ring has centre $(1/2,0)$ and radius $1/4$. Superposing these displaced circles produces the comet-like envelope.

For uniform irradiance over the circular optical pupil, equal area samples give an equal-weight geometrical image. Its centroid is $(h,0)$ because $⟨u^2⟩=⟨v^2⟩=1/4$ and $⟨uv⟩=0$. Moving the coordinate origin to that centroid translates the image; it does not remove its asymmetric shape. Adding a linear term to $w$ likewise translates all light rays equally. These origin choices must be distinguished from changing the optical aberration itself.

The model's full transverse extent is $3|h|$ along $X$ and $2|h|$ along $Y$. Halving the field angle halves the blur, and reversing it reverses the pattern about the ideal image. This isolates why coma matters for an optical camera's usable field of view: collecting more light does not ensure that off-axis points remain compact. Near diffraction-limited resolution, a physical optics calculation of the point spread function is needed instead of interpreting the geometrical tip as an infinitely sharp image.^[ref]

A single-layer greenhouse model represents an atmosphere by one uniform layer that absorbs and emits a fraction of thermal radiation. With infrared emissivity $ϵ$, the layer emits $ϵ\sigma T_a^4$ both upward and downward.

Single-slit diffraction is the spreading of light transmitted by one slit. Consider an ideal rectangular slit of width $a$, illuminated normally by a uniform plane wave with a single wavelength $\lambda$ in the surrounding medium. This monochromatic wave has uniform wave amplitude and equal incident wave phases across the slit, providing the required optical coherence. In the scalar Fraunhofer diffraction model, the normalized angular wave intensity isHere $\theta$ is measured from the forward normal in the plane across the slit width. The sinc function uses the unnormalized convention $\mathrm{sinc}u=\sin u/u$. The limit makes the central value one. This ideal envelope neglects additional direction-dependent propagation and detector factors; near the forward direction it describes the normalized screen irradiance. ^[ref].

The noncentral zeros satisfyThe restriction follows from the range of the sine of a real angle. The first minima have $\sin \theta =\pm \lambda /a$ when $\lambda /a\le 1$. If $a<\lambda$, this model has no noncentral zero in the forward hemisphere; equality puts the first zeros at grazing directions, outside the small-angle approximation and off any finite perpendicular screen. ^[ref].

For $0<\lambda /a<1$, define the central width as the separation between the two first minima. Its angular width and its width on a perpendicular screen areThe angular approximation uses radians and requires $\lambda /a\ll 1$; the screen approximation uses $y\approx L\theta$. The tangent relation is exact straight-line geometry, even though the underlying Fraunhofer diffraction field is approximate. This width is distinct from the full width at half maximum.

For $\lambda =500\mathrm{nm}$, $a=0.1\mathrm{mm}$ and $L=1\mathrm{m}$, the positive first minimum isThe dimensionless far-field parameter is $a^2/(\lambda L)=0.02$. Doubling the slit width to $a=0.2\mathrm{mm}$ gives $W\approx 5.00002\mathrm{mm}$ and parameter $0.08$; both curves below use the Fraunhofer diffraction approximation. Choosing $L=2\mathrm{m}$ halves those parameters and doubles both screen widths.

Each curve is normalized to its own $I_0$; the arrows measure between first minima. The calculation uses $\theta =\arctan (y/L)$ before evaluating the sinc function, rather than replacing $\sin \theta$ by $y/L$. The lower diagram shows the exact screen geometry with the vertical scale enlarged. Normalization removes the change in peak brightness when the slit width changes.

The wave interference explanation is especially simple: divide the slit into equal narrow strips and add their complex amplitudes, rather than their wave intensities. With wavenumber $k=2\pi /\lambda$, the continuous slit givesIntegrating the complex exponential gives the endpoint difference; its continuous value at zero is one. Taking the squared complex modulus gives the stated wave intensity. This is the Fourier transform of an interval indicator expressed in angular coordinates. Feynman's continuous-source argument obtains the same pattern by letting the number of sources tend to infinity. ^[ref].

At the positive first minimum, the wave phases span exactly $2\pi$ across the slit. Every strip in one half has a partner $a/2$ away, with equal wave amplitude and phase difference $\pi$, so the pairs cancel. At $\theta =0$, all the wave phases align instead. ^[ref].

The drawing uses $N=8$ equally weighted midpoint samples $x_j/a=-1/2+(j+1/2)/8$, with $j=0,\dots ,7$. At the first minimum their complex amplitudes are a common rotation of the eight roots of unity, so their head-to-tail polygon closes exactly. This finite sum illustrates cancellation; the continuous integral, rather than eight point sources, defines the slit pattern at general angles. The circle shows the continuous range of wave phases; coloured arrows show the eight samples.

A narrower aperture requires a larger angle to accumulate the cancelling phase difference, so narrowing the slit widens the beam. This is a useful limit on geometrical optics: confining light more tightly at an opening increases its subsequent angular spread. At fixed incident irradiance and slit height, reducing $a$ also reduces transmitted power, exposing a practical balance between diffraction-limited resolution and light throughput. In a diffraction grating, this single-slit diffraction envelope modulates the wave interference between different slits.

Incident radiant flux per unit area and unit wavelength or frequency. Integrating $E_{\lambda }$ over a wavelength interval gives the contribution to irradiance in that interval. Conversion between $E_{\lambda }$ and $E_{\nu }$ requires the absolute Jacobian determinant associated with $\nu =c/\lambda$, as for spectral radiance.

For collected spectral radiant flux $f(\lambda )$ and a linear photodiode with spectral responsivity $r(\lambda )$, the dark-subtracted electric current is $i=\int rfd\lambda$. A reference optical spectrum $f_A$ sets the conversion $C_A$ from electric current to collected luminous flux. Applying it to another optical spectrum $f_T$ givesAssume nonzero calibration and test signals. Then $F{\Phi }_{v,\mathrm{indicated}}$ is the corrected luminous flux. Multiplying either optical spectrum by an overall positive scale leaves $F$ unchanged. If $r$ is proportional to $V$, $F=1$ for every optical spectrum. Uniformly overfilled aperture measurements use wavelength-resolved irradiance instead, multiplying the electric current expression by aperture area; the same ratio cancellation gives the illuminance correction. This is the algebra behind source-dependent calibration, rather than a new measured correction curve. ^[ref]

A proposed bench test can isolate the mechanism without a gas burner. Illuminate a blackened and a polished sample of matched material and geometry, record both temperatures and their relative displacement, then block the beam. Measure incident irradiance separately. Swap the surface treatments, hold the ambient temperature fixed for the illumination test, and change it with the beam blocked for a common-heating control. This is a model-based experiment proposal, not a report of measurements.

A larger response of the dark sample would support a difference in absorbed power; the displacement and cooling curves test the assumed linear thermal expansion coefficient and time constant. A stable illumination step that never reaches the switching gap challenges the chosen parameter model rather than the existence of thermal expansion. Nonuniform sample temperature, unequal losses and mechanical friction can separate the physical device from the ideal equations.

Ultraviolet is electromagnetic radiation with wavelengths shorter than the violet end of ordinarily visible light. The CIE convention uses $100$–$400\mathrm{nm}$, subdivided into UV-A at $315$–$400\mathrm{nm}$, UV-B at $280$–$315\mathrm{nm}$ and UV-C at $100$–$280\mathrm{nm}$. These divisions name intervals; they do not mark universal changes in every material's response. The boundary with visible light is not sharp, and other fields use near-, far-, vacuum- and extreme-ultraviolet divisions. ^[ref]. For example, NASA's Solar Dynamics Observatory observes extreme-ultraviolet bands below $100\mathrm{nm}$; the CIE interval is therefore not the full extent of astronomical ultraviolet usage. ^[ref].

The CIE ultraviolet interval corresponds to frequencies about $7.49\times 10^{14}$–$3.00\times 10^{15}\mathrm{Hz}$ and photon energies $4.97\times 10^{-19}$–$1.99\times 10^{-18}\mathrm{J}$, using vacuum wavelength and $E=hc/\lambda$ with Planck constant $h$ and speed of light $c$. Shorter wavelength means larger energy per photon; larger irradiance means more incident power per area. These are different comparisons. At the same incident power, a $254\mathrm{nm}$ ultraviolet beam has fewer photons per second than a $940\mathrm{nm}$ infrared beam, even though each ultraviolet photon has more energy.

A common ultraviolet mechanism is photon absorption promoting an electron from a ground state to an excited state of a chemical atom or molecule. By contrast, many infrared absorption bands involve stretching or bending chemical bonds. For either mechanism, both the energy difference and the coupling to the incident electromagnetic field matter. One concrete example is the $217\mathrm{nm}$ absorption discussed for butadiene: an electron moves between molecular orbitals, and the position of the absorption reveals their energy separation. Ultraviolet absorption can therefore probe molecular orbitals, while infrared spectroscopy probes many molecular motions. ^[ref].

Subsequent chemical reactions depend on the absorbing chemical substance and its environment; the label ultraviolet alone does not predict a reaction rate. At sufficiently short wavelength, a photon can cause photoionization if its energy exceeds the relevant ionization energy. There is no single wavelength that separates all ionizing from all nonionizing behavior for every chemical substance. Likewise, eventual heating after photon absorption is possible in both ultraviolet and infrared. The distinction is useful because different optical spectra expose different processes, not because the two ranges obey different laws of electromagnetism.

Ritter's 1801 observation used a chemical readout rather than a thermometer. In Caltech/IPAC's account, sunlight passed through a glass optical prism and silver chloride was placed in different parts of the optical spectrum. Darkening was weak toward red, stronger toward violet, and pronounced beyond the visible violet edge. The measured evidence was a change in the chemical substance's appearance at a spatial location with no visible color, establishing radiation beyond the eye's ordinary range. It was not a calibrated measurement of a universal ultraviolet optical spectrum. ^[ref].

A controlled modern interpretation of such a chemical test would compare equal exposure durations and matched fresh samples, keep an unexposed sample in darkness, monitor temperature, and distinguish scattered visible light from the intended short-wavelength beam with characterized filtering. These are proposed controls, not an assertion that Ritter performed them. Unequal sample preparation or heating can change a reaction rate. The optical prism, terrestrial atmosphere and sample response restrict which wavelengths the test detects. Darkening is a detector-weighted response, so its spatial maximum is not automatically the maximum of incoming radiant flux or of photon energy.

Thomas Larason and Jeanne Houston's NIST report documents a quantitative ultraviolet detector comparison. An argon source feeds a double-diffraction grating spectrometer; a shutter and mirrors deliver the selected band to a working-standard or test photodiode, successively at the same position. A beam splitter supplies a monitor photodiode. Amplifiers and voltmeters record detector and monitor simultaneously; shutter-closed readings provide dark baselines. The $200$–$500\mathrm{nm}$ scan, typically with $4\mathrm{nm}$ bandpass, measures spectral responsivity. Monitor normalization suppresses source drift; known spectral lines calibrate wavelength. The report assesses temperature effects and unwanted light outside the band, weighted by detector response. The latter need not cancel between different spectral responsivities. ^[ref].

This determines electrical response without visual perception. Bandpass and spatial nonuniformity limit the calibration; aging and changed temperature can affect later use. The scan includes visible light and ultraviolet; no response below $200\mathrm{nm}$ follows from it.

For a narrow beam, the idealized substitution calculation isHere $R$ is spectral responsivity; $I$ and $M$ are dark-subtracted detector and monitor electric currents. This follows by writing $I=RP$ for each detector and using the monitor to track changes in radiant flux. It assumes linear response, unchanged beam splitting and collection geometry, and a sufficiently narrow band. For an illustrative standard with $R_{\mathrm{std}}=0.15\mathrm{A}{\mathrm{W}}^{-1}$, ratios $I_{\mathrm{std}}/M_{\mathrm{std}}=2.0$ and $I_{\mathrm{test}}/M_{\mathrm{test}}=1.6$ give $R_{\mathrm{test}}=0.12\mathrm{A}{\mathrm{W}}^{-1}$. These invented arithmetic inputs illustrate the method; they are not NIST measurements.

An independent photon-counting model connects spectral responsivity to quantum efficiency. If $\eta (\lambda )$ is the number of collected electrons per incident photon, there is no internal multiplication, and $q$ is the elementary charge, thenAt illustrative constant $\eta =0.60$ and $P=1\mu \mathrm{W}$, this gives about $0.123\mu \mathrm{A}$ at $254\mathrm{nm}$ and $0.455\mu \mathrm{A}$ at $940\mathrm{nm}$. The ultraviolet readout is smaller in this model because there are fewer incident photons, not because their individual energies are smaller. Real quantum efficiency depends on wavelength, windows, absorption and carrier collection; a detector's band gap and construction determine its usable range. Constant $\eta$ across these two wavelengths is an analytical comparison, not a claimed specification of a real photodiode.

The curve is the calculated relation $E=hc/\lambda$, and the shaded ranges use the stated CIE conventions. The lower diagram abstracts the detector comparison rather than reproducing an instrument drawing. Together with the infrared heat-balance illustration, it shows why a calibrated detector response, rather than what looks bright to an observer, is the bridge between invisible radiation and a quantitative measurement.

In a transparent, stationary vacuum with Euclidean geometry and observers at rest in the same frame, the radiative transfer equation has zero absorption and emission, so $d{I}_{\nu }/ds=0$ along a light ray. The same specific intensity reaches the observer without geometric dilution. More generally, $I_{\nu }/{\nu }^3$ is a Lorentz invariant; collisionless propagation preserves this ratio along a ray even with gravitational or cosmological redshift, rather than preserving $I_{\nu }$ at shifted frequencies. ^[ref].

The elementary geometry is instructive. Two facing surface elements separated by distance $s$ exchange spectral radiant fluxThe receiver subtends $d{\Omega }_2=\cos {\theta }_{2}d{A}_2/s^2$ at the source. Describing that same traveling beam at the receiver instead uses $d{\Omega }_1=\cos {\theta }_{1}d{A}_1/s^2$. Both descriptions have identical geometric factors, so conservation of its radiant flux gives equal specific intensities at the two ends. This compares the same propagation direction at both ends, not emission in the opposite direction.

A uniform source remains equally bright per resolved direction under these assumptions, but occupies a smaller apparent solid angle farther away. For a small source of projected area $A_{\perp }$ at distance $d$ and a small receiver facing it, irradiance is approximately $E=L{\Omega }_{\mathrm{source}}\simeq L{A}_{\perp }/d^2$. The radiant flux intercepted by receiver area $A_d$ is approximately $E{A}_d$. With $L=10\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$, $A_{\perp }=0.020{\mathrm{m}}^2$ and $A_d=10^{-4}{\mathrm{m}}^2$, distances $10$ and $20\mathrm{m}$ give $E\simeq 0.002$ and $0.0005\mathrm{W}{\mathrm{m}}^{-2}$, and received radiant fluxes of $0.20$ and $0.050\mu \mathrm{W}$. The inverse-square law concerns collected radiation from the whole source; $L$ remains $10$ at either distance.

Consider an idealized planet that is a sphere of radius $R$, illuminated by parallel sunlight of irradiance $S$, measured on a plane perpendicular to the rays. Its intercepting projected area is $\pi R^2$, whereas its total surface area is $4\pi R^2$. Therefore the incoming power averaged per unit total surface area iswhere $\alpha$ is the broadband Bond albedo. The factor four is a sphere geometry result, not an atmospheric transmission factor. Local sunlight still varies over the illuminated hemisphere and vanishes on the dark side.

Now stipulate a single spatially uniform temperature $T$, no atmosphere, no internal heating, and emissivity one for outgoing thermal radiation. This is a thermal blackbody model using the Stefan–Boltzmann law; the prescribed solar-band reflection is treated separately, so the body is not a perfect absorber at all wavelengths. Let $C>0$ be the effective heat capacity per unit surface area, in $\mathrm{J}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$. Conservation of energy gives the ordinary differential equationHere $F$ is net heating per unit surface area, in $\mathrm{W}{\mathrm{m}}^{-2}$, and $\sigma$ is the Stefan-Boltzmann constant. Use $\sigma =5.670374419\times 10^{-8}\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}$, rounded from the exact constant reported by NIST. ^[ref] The uniform temperature is a model assumption: generally the surface average of $T^4$ differs from the fourth power of the surface-average temperature.

For fixed Bond albedo $\alpha$, radiative equilibrium requires $F(T_{\ast })=0$, givingWith the illustrative solar constant $S=1360\mathrm{W}{\mathrm{m}}^{-2}$, the global incoming average is $340\mathrm{W}{\mathrm{m}}^{-2}$:

Doubling albedo here lowers the equilibrium temperature by $33.231\mathrm{K}$, rather than halving it: the absorbed fraction changes from $0.7$ to $0.4$, and thermal radiation scales as $T^4$. These stipulated Bond albedos are whole-planet examples, not measurements of bare water and ice. The model predicts neither Earth's modern surface temperature nor its response to actual climate change; it omits the greenhouse effect, spatial heat transfer and seasons.

Each fixed-albedo equilibrium is a stable equilibrium: for $T_{\ast }>0$, the derivative is $F^{\prime }(T_{\ast })=-4\sigma T_{\ast }^3<0$. A small temperature increase produces net cooling, and a small decrease produces net heating. This is the simplest expression of the stabilizing Planck feedback. Allowing albedo to fall with temperature adds a competing term, as in ice-ocean albedo feedback.