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The bulk photon mode density per volume per angular frequency is ${\omega }^2/({\pi }^2{c}^3)$, including two transverse polarizations. Multiply it by the thermal oscillator energy $\hslash \omega /(e^{\hslash \omega /(k_{B}T)}-1)$ to obtain the displayed spectrum. Its integral is $u={\pi }^2{k}_B^4{T}^4/(15{\hslash }^3{c}^3)$. This is volume energy density rather than radiance or surface exitance; isotropic equilibrium radiation has outward black-surface flux $cu/4$. zero-point energy is excluded from this thermal spectrum.

Insert a centred opaque circular stop of diameter $d=6\mathrm{mm}$ directly in the plane of the $D=10\mathrm{mm}$ bundle. It transmits the same $30^{\circ }$ directions at every surviving position, but discards the surrounding annulus. ThereforeWith the uniform radiance of etendue of a circular aperture and cone, the transmitted radiant flux is $2.2207\mathrm{mW}$ and the blocked fraction is $64\%$. Surviving light rays retain their radiance; the stop has not concentrated them. Lossless propagation after the stop preserves the etendue of this surviving subset, not that of the original incident bundle. A stop placed elsewhere can select positions and directions together, so its physical area ratio alone need not give the transmitted radiant flux fraction.

The rate per unit path length at which light is scattered into the backward hemisphere. It influences water-leaving radiance and must be distinguished from the absorption coefficient.

Basic radiance is radiance divided by the square of the refractive index. For radiance $L$ at a fixed wavelength,Lossless propagation conserves reduced radiance along each light ray; $L$ itself is conserved only when the endpoint refractive indices agree. A larger irradiance at a smaller image can result from admitting more directions without increasing reduced radiance. Uniform wave absorption can lower radiant flux while leaving the occupied geometric etendue unchanged; an opaque aperture removes part of that occupied set. ^[ref]

A function relating reflected radiance in one direction to incident irradiance from another direction at a surface location. A local nonemissive, nonfluorescent optical reflection model integrates $f_r{L}_i\cos {\theta }_i$ over incident solid angle to obtain outgoing radiance. Transmission, wavelength conversion and nonlocal transport need additional models.

A diffuse thermal emitter has emitted radiance independent of outgoing direction over the outward hemisphere. Its emitted radiant intensity from a planar patch still varies with viewing angle through projected area, as described by Lambert's cosine law. Diffuse emission does not mean equal power per solid angle from the patch. This approximation does not by itself impose a grey body spectrum or specify how incident radiation is reflected.

Directional emissivity is the ratio of emitted radiance to blackbody radiance at the same temperature in a specified outgoing direction. It can be total over wavelength or additionally spectrally resolved. Normal-direction emissivity is not automatically total hemispherical emissivity: the latter integrates all directions with the projected area weighting. A diffuse thermal emitter removes this directional dependence of emitted radiance.

A proposed optical bench test uses a steady illuminated diffuse patch, a relay made from two optical lens elements, interchangeable centred circular stops, and a calibrated optical power meter large enough to collect the whole output bundle. Start with small cone angles so an ordinary relay approximates the area-angle tradeoff. Measure the patch's image diameter and the accepted angular extent at both conjugate planes; the transverse magnification gives the spatial scale, but does not by itself measure etendue.

Map the local radiance with a baffled photodiode on translation and rotation stages, using two small aligned holes to select a narrow known solid angle. Keep the sampled position fixed while scanning direction, calibrate the sampling footprint and angular response, and repeat at several positions. Numerically integrate $n^2\cos \theta dAd\Omega$ over the occupied samples, and integrate $(L/n^2)dG$ to predict the collected radiant flux. This checks the filled-cone assumption instead of inferring it from two outer dimensions.

Compare the relay without a stop and with a stop of $0.6$ times the illuminated diameter in the same plane. The uniform-bundle prediction is a radiant flux ratio of $0.36$, with unchanged local radiance in the surviving samples. Cover the input for a dark reading, monitor source drift, repeat the open-stop reading, check detector linearity with calibrated optical attenuators, and enlarge the collecting detector or scan it to check for missed light rays. Lens losses, angular response, illumination gradients, alignment errors and diffraction can spoil the ideal prediction; their effects must be measured separately. This is a procedure to try, with predicted readouts, not purported experimental data.

Suppose an extended source fills every position of the entrance aperture with a cone of half-angle ${\theta }_{\mathrm{in}}$, and all these light rays must reach a one-sided planar receiver. If the receiver accepts at most a cone of half-angle ${\theta }_{\mathrm{out}}$, its available etendue bounds the admitted etendue:ConsequentlyThis original derivation combines conservation of etendue with the exact cone integral; the second inequality allows the complete output hemisphere. It bounds capacity and does not prove that a chosen shape attains the bound.

For the $30^{\circ }$ input cone with $n_{\mathrm{in}}=n_{\mathrm{out}}=1$, $C\le 4$: a $10\mathrm{mm}$ entrance cannot deliver the entire filled bundle to a circular receiver smaller than $5\mathrm{mm}$ in diameter. If the output half-angle is restricted to $60^{\circ }$, $C\le 3$ and the receiver diameter must be at least $10/\sqrt{3}\approx 5.7735\mathrm{mm}$. For a receiver medium with model refractive index $n_{\mathrm{out}}=1.5$, the full-hemisphere capacity bound instead gives $C\le 9$; that larger value also depends on transmitting the bundle into that medium.

Attempting $C=10$ in the air-to-air example forces rejection of some input. For the uniform radiance already specified, even filling the entire output hemisphere givesThe maximum mean irradiance gain remains $\eta C\le 4$. With nonuniform radiance, rejecting a given fraction of geometric etendue need not reject the same fraction of radiant flux.

These bounds use a filled extended-source bundle in passive reversible geometrical optics, comparing the same wavelength and the same radiation channels. An ideal optical point source does not fill that four-dimensional set. Diffraction, optical coherence, or coupling a Gaussian beam into an optical fiber require a physical optics calculation; the cone bound is not a universal guarantee for a single wave mode. Likewise, optical gain or fluorescence changes the propagation assumptions.

Consider a planar circular aperture of diameter $D$ in constant refractive index $n$. Suppose every position in its area $A=\pi D^2/4$ is filled with directions inside the same rotationally symmetric cone about the unit normal, with half-angle ${\theta }_{\max }$. Using spherical coordinates, the exact etendue iswhere $\mathrm{NA}=n\sin {\theta }_{\max }$ is the appropriate side's numerical aperture. The ordinary cone solid angle is $2\pi (1-\cos {\theta }_{\max })$; its cosine-weighted value is $\pi {\sin }^2{\theta }_{\max }$. A full forward hemisphere therefore gives $G=n^{2}A\pi$, rather than $2\pi n^{2}A$. ^[ref]

For an original model example, choose $n=1$, $D=10\mathrm{mm}$, and ${\theta }_{\max }=30^{\circ }$. ThenIf the radiance is uniform across this filled bundle and equals $L=100\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$, its radiant flux is $\Phi =LG/n^2\approx 6.1685\mathrm{mW}$. Neither the geometry nor the chosen radiance is an experimental measurement.

The focal ratio, or f-number, of a converging optical instrument focused at infinity is the dimensionless quantitywhere $f>0$ is effective focal length and $D$ is the on-axis entrance pupil diameter. The notation f/4 means $N=4$, not $f=4$ in a particular unit. A $100\mathrm{mm}$ focal length with $D=25\mathrm{mm}$ gives f/4. Closing the aperture stop until $D=12.5\mathrm{mm}$ gives f/8 while retaining the same focal length. The mechanical opening need not equal $D$. ^[ref].

For a thin lens in air with a circular aperture stop at the thin lens, the converging bundle from a distant on-axis point has half-angle $\alpha$ satisfyingThe last step uses the small-angle approximation. For a compound optical lens, the cone is set by the image-side optical pupil and its distance from the receiving plane; the simple triangle is not an exact high-angle rule. At finite object distance a thin lens instead gives $v=f(1+|m|)$, by its imaging equation and transverse magnification, so its working cone corresponds to $v/D=N(1+|m|)$. Nominal focal ratio therefore does not alone specify close-range image irradiance. ^[ref].

Two useful scalings follow under stated conditions. For a fixed uniform incident beam and unchanged transmission, collected radiant flux scales as $D^2$. For an extended source with fixed radiance, the central image irradiance scales approximately as $N^{-2}$ in the paraxial approximation: the solid angle of the cone scales as $(D/f{)}^2$. For an ideal uniformly illuminated unobscured circular entrance pupil, the first dark ring of the point spread function has angular radius and focal plane radiusHere $\lambda$ is wavelength. These are first-zero radii, not diameters or a universal rule for resolving arbitrary objects. A proof outline evaluates the circular aperture's Fraunhofer diffraction using its Fourier transform, then uses the first zero of the resulting Bessel function. ^[ref].

At $\lambda =0.55\mu \mathrm{m}$, the example's f/4 and f/8 first-ring radii are $2.684$ and $5.368\mu \mathrm{m}$. Closing the aperture stop halves $D$, quarters collected radiant flux and doubles this diffraction radius. It can also reduce optical aberrations by excluding outer light rays; actual sharpness reflects both effects. This explains why a smaller aperture need not produce a sharper image. ^[ref].

A proposed optical bench check puts a fixed collimator before a converging thin lens and a photodetector large enough to collect the entire image behind it. Compare two centred aperture stops of known diameters while fixing source output, detector gain and integration time. The integrated signal tests the $D^2$ collection law; a spatially resolved optical camera at the focal plane can separately measure the point spread function. Repeat a dark reading and the first aperture stop, and check that neither the incoming beam nor detector clips either bundle. A resolved pinhole source, imperfect collimation, optical aberrations and detector pixel size can dominate the apparent spot width; a single peak reading does not measure total collected radiant flux.

In vacuum, $\nu =c/\lambda$, where $c$ is the speed of light. Increasing wavelength decreases frequency, so the ordinary signed differentials satisfy $d\nu =-c{\lambda }^{-2}d\lambda$. A spectral slice's positive contribution to radiance must be unchanged. The change of variables formula therefore gives the absolute Jacobian determinantThese formulas use $L_{\lambda }$ per metre and $L_{\nu }$ per hertz. The factor $c/{\nu }^2$ has units $\mathrm{m}\mathrm{s}=\mathrm{m}\mathrm{H}{\mathrm{z}}^{-1}$, which converts the former density into the latter. If ${\ell }_{\mathrm{nm}}$ denotes the numerical density per nanometre in $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\mathrm{n}{\mathrm{m}}^{-1}$, its per-metre value is $10^9{\ell }_{\mathrm{nm}}$. Conversely, the per-nanometre value is $10^{-9}$ times the per-metre value. Likewise, a density per terahertz is $10^{12}$ times its per-hertz value because $1\mathrm{THz}=10^{12}\mathrm{Hz}$.

For a concrete stipulated spectrum, let $L_{\lambda }=C=10^9\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}{\mathrm{m}}^{-1}$ between ${\lambda }_1=500\mathrm{nm}$ and ${\lambda }_2=600\mathrm{nm}$, and zero outside. This is exactly $1\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\mathrm{n}{\mathrm{m}}^{-1}$ throughout the band. Using the exact International System of Units value $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$, the corresponding frequency bounds areThe transformed spectral radiance is $L_{\nu }=Cc/{\nu }^2$ on this interval. At ${\nu }_1$ it is approximately $1.200830743\times 10^{-12}\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\mathrm{H}{\mathrm{z}}^{-1}$, and at ${\nu }_2$ it is $8.339102380\times 10^{-13}$ in the same units. These are respectively $1.200830743$ and $0.833910238$ per terahertz. Thus a flat wavelength density becomes a decreasing frequency density; relabelling the horizontal axis alone would miss both the shape and the units. ^[ref].

The equality of band radiances has a short exact proof:The reversed bounds and the Jacobian determinant preserve the same spectral measure. Each half-band, $500$–$550\mathrm{nm}$ and $550$–$600\mathrm{nm}$, carries $50\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$, although the corresponding frequency widths are about $54.508$ and $45.423\mathrm{THz}$.

A spectrum's peak is consequently a statement about a particular density and coordinate. The frequency and wavelength peaks of Planck's law differ for precisely this reason. Integrated radiance over a specified physical band supplies a comparison that survives the change of variables.

The derivative of outgoing spectral radiance with respect to a pressure-resolved absorber abundance defines a kernel once the vertical measure is fixed. In an optically thin column with $k\propto p$ and $B/B_0=p/p_0$, $K_p=[k_0{B}_0/(p_{0}gM)]p(p/p_0-1)$. It is negative inside the atmosphere and zero at both endpoints. Its largest magnitude lies at $p_0/2$, whereas the logarithmic-pressure kernel $p{K}_p$ peaks in magnitude at $2p_0/3$. A uniformly perturbed abundance instead has one integrated derivative, not a pressure location.

Irradiance measures incident radiant flux per actual receiving area. Its units $\mathrm{W}{\mathrm{m}}^{-2}$ match radiant exitance, but it counts the incoming side of the radiation exchange. Measure incoming angles from the unit normal pointing toward the incident hemisphere. ThenFor $2\mathrm{W}$ distributed uniformly over $0.010{\mathrm{m}}^2$, $E=200\mathrm{W}{\mathrm{m}}^{-2}$. If the receiving plane is tilted $60^{\circ }$ in the same broad uniform parallel beam, the incident power per actual area is halved to $100\mathrm{W}{\mathrm{m}}^{-2}$. This receiving-surface cosine factor is geometric and does not assume Lambertian reflectance. Together, irradiance and radiant exitance describe how radiation supplies and removes energy at a surface, while radiance preserves the directional information.

Lambertian reflectance is the ideal diffuse-reflection model with direction-independent outgoing radiance, the frequency-integrated specific intensity. If a fraction $0\le \rho \le 1$ of incident radiant flux is reflected, a surface receiving irradiance $E$ has reflected radiance $L=\rho E/\pi$. The reflected outward power per area is $\pi L=\rho E$ by the same hemispherical integral used for radiant exitance. Its radiant intensity still acquires the projected area factor in Lambert's cosine law.

For $E=100\mathrm{W}{\mathrm{m}}^{-2}$ and $\rho =0.80$, the reflected radiance is $80/\pi \approx 25.4648\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$. A patch of area $0.020{\mathrm{m}}^2$ reflects $1.6\mathrm{W}$ in total, and its radiant intensities at $0^{\circ }$ and $60^{\circ }$ are respectively $1.6/\pi$ and $0.8/\pi \mathrm{W}\mathrm{s}{\mathrm{r}}^{-1}$. The radiance stays fixed as the view changes, provided the illumination stays fixed. Changing the incident angle can change irradiance and therefore the reflected radiance.

A spherical body of radius R observed from distance $d\gg R$ is illuminated and viewed from the same direction, with incident irradiance $F_{\star }$. For uniform Lambertian reflectance A, a patch at angle $\theta$ has outgoing radiance $A{F}_{\star }\cos \theta /\pi$. The observer receives this radiance times projected area divided by $d^2$. Since ${\int }_{\mathrm{hemisphere}}{\cos }^2\theta dS=2\pi R^2/3$, integration proves the displayed full-phase irradiance. Its geometric albedo is $2A/3$. Total reflected power alone does not specify the brightness in one direction for a non-Lambertian surface.

For an ideal diffuse emitting or reflecting planar surface with uniform direction-independent radiance, the radiant intensity in a direction at angle $\theta$ to its unit normal is $I(\theta )=I(0)\cos \theta$. This follows immediately from $I=LA\cos \theta$ and the projected area formula. The worked radiance example and diagram show the $0^{\circ }$ and $60^{\circ }$ cases. Lambertian reflectance supplies one physical idealization realizing the law; an ideal blackbody provides an emitting example.

The historical setting is Johann Heinrich Lambert's Photometria sive de mensura et gradibus luminis, colorum et umbrae, published in Augsburg in 1760. Its Part I, §139, explicitly compares normal and oblique illumination using the sine of the source's elevation, equivalently the cosine of its angle from the receiving normal vector. This is an early receiving-surface projection calculation; the modern outgoing-patch statement above keeps radiance, radiant intensity and irradiance distinct. ^{[ref] [ref]}.

The linear limb-darkening law parametrizes angular specific intensity by a stipulated limb-darkening coefficient $u$:Here $I$ denotes specific intensity at one frequency, or radiance integrated over a specified interval of frequency or wavelength; it does not denote radiant intensity. Within this stipulated range, $u=0$ gives a uniform disk and $I(0)/I_c=1-u$. The linear limb-darkening law is linear in $\mu$, while its radial profile $I(r)=I_c[1-u+u\sqrt{1-(r/R_{\star }{)}^2}]$ is curved. In the linear source function example above, $u=b_{\nu }/(a_{\nu }+b_{\nu })$ when $a_{\nu },b_{\nu }\ge 0$ and $a_{\nu }+b_{\nu }>0$.

For the illustrative choice $u=0.6$, the specific intensity at $r/R_{\star }=0$, $0.8$, and the limiting edge is respectively $I_c$, $0.76I_c$, and $0.4I_c$. These are synthetic values, not measurements of the Sun. Real optical spectra and stellar atmospheres need not obey a linear limb-darkening law; more flexible profiles can be necessary for accurate exoplanet transit photometry. ^[ref]

Noise-equivalent radiance is the incident radiance change producing signal-to-noise ratio one for a specified spectral and electrical bandwidth. A linear radiometer with voltage response $G=dV/dB$ and RMS voltage noise ${\sigma }_V$ has $\mathrm{NER}={\sigma }_V/|G|$. Its convention must distinguish per-sample noise from a noise amplitude per square-root bandwidth.

Outgoing longwave radiation is the outward thermal radiative flux at the top of the atmosphere, integrated over the specified thermal spectrum and outward directions. It differs from a directional spectral radiance.