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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.

When both endpoint bundles have the filled circular-cone geometry of etendue of a circular aperture and cone, conservation of etendue givesFor the same refractive index, quadrupling the area halves the sine of the cone half-angle. In the $10\mathrm{mm}$, $30^{\circ }$ example, an ideal output of diameter $20\mathrm{mm}$ must have $\sin {\theta }_2=1/4$, so ${\theta }_2\approx 14.4775^{\circ }$. The exact result differs from simply halving $30^{\circ }$. Conversely, reducing the diameter to $5\mathrm{mm}$ requires a complete forward hemisphere, ${\theta }_2=90^{\circ }$, as a limiting case.

Equal bounding area and solid angle do not alone prove equal etendue: an actual output can have unoccupied directions or position-direction correlations. The filled-cone equation is an exact model of the specified bundles and a capacity constraint on an optical instrument, not a guarantee of an aberration-free implementation.

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]

Lossless reversible propagation preserves the etendue of the same bundle of light rays. Here the model uses stationary isotropic media, specular reflection and refraction, with every selected light ray reaching the output and with no optical gain or scattering of light. An aperture that blocks some incident light rays changes which bundle is counted; the full incident and transmitted etendue need not be equal. ^[ref]

The underlying phase space has four coordinates. In a planar section with constant refractive index, let $(x,y)$ locate a light ray and setThe Jacobian determinant from spherical coordinates givesThus etendue is a four-dimensional volume, preserved by the optical counterpart of Liouville theorem in Hamiltonian mechanics. Individual position and direction extents may change, and correlations between them matter. The Lagrange optical invariant is a related paraxial approximation invariant for pairs of light rays, rather than the full four-dimensional measure.

Etendue measures how much position-and-direction space a bundle of light rays occupies. A small spot accepting a wide range of directions can have the same etendue as a large spot accepting a narrow range. This makes etendue useful for deciding whether an optical instrument can deliver an extended source's entire admitted radiant flux to a small receiver. For forward crossings of a surface, the geometrical optics definition iswhere $\mathcal{B}$ is the occupied set of surface positions and directions, $n$ is the local refractive index, $dA$ is actual surface area, $d\Omega$ is solid angle, and $\theta$ is the angle to the forward unit normal. The cosine supplies the projected area. Opposite crossing directions must be counted separately with positive projected area, rather than cancelling them in a signed integral. ^[ref]

The shorthand $G\simeq n^{2}A\Omega$ assumes constant refractive index, a narrow nearly normal bundle, and the same admitted directions across the whole area $A$. At larger angles, use the cosine-weighted integral; a position-dependent angular acceptance cannot generally be replaced by the product of two independent extents.

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 fundamental angular floor is the diffraction limit of a telescope, of order $\lambda /D$. The ideal diffraction grating has $R_g=mW/d$, whereas its slit-limited resolving power has $R\theta D=mW\lambda /d$. Requiring $R\le R_g$ givesThe coefficient depends on the adopted line-resolution and aperture conventions. Below this scale, a narrower spaxel oversamples the same spatial mode rather than creating a new independently resolved element. Equivalently a spatial optical mode occupies etendue of order ${\lambda }^2$.

There is also an engineering floor before arbitrary shrinkage: the optical camera focal length, photodetector size and support clearances do not shrink, while the number of photodetectors increases as ${\theta }^{-2}$. Thus the shrinking-beam model's constant combined optical camera volume is eventually overtaken by fixed per-camera overheads. For fixed brightness, smaller spaxels receive fewer photons, so read noise and photon shot noise can constrain useful sampling before the formal optical floor. Diffraction limits independent spatial information; fixed photodetector and mechanical dimensions limit practical volume reduction.

The number of spaxels is $M=(\beta /\theta {)}^2$. To obtain definite scaling exponents, keep the spectral resolving power, wavelength interval, diffraction grating angles and groove spacing, input focal ratio, and photodetector sampling of a spectral resolution element fixed. A fixed photodetector then accommodates a fixed number $M_0$ of spectra, so $N\simeq M/M_0$ (rounded up in an actual instrument). The corresponding etendue per spaxel scales as ${\theta }^2$. Angular ratios are independent of whether both angles are in arcseconds; optical invariant equations use radians.

Write $b$ for the collimated beam diameter. The diffraction grating result $R\theta D=W(\sin \alpha +\sin \beta )$ gives $b\propto W\propto \theta$. A fixed collimator focal ratio gives $f_{\mathrm{coll}}\propto b\propto \theta$. With a fixed image width $p$ in photodetector detector pixels, the invariant $\theta D=p{A}_{\mathrm{cam}}/f_{\mathrm{cam}}$ and $A_{\mathrm{cam}}\propto \theta$ instead imply $f_{\mathrm{cam}}$ is constant: the optical camera focal ratio increases as $1/\theta$.

In a simple on-axis beam-envelope model, each collimator and collimated disperser space has area proportional to $b^2$ and length proportional to $b$, whereas the optical camera cone has area proportional to $b^2$ and fixed length. HenceThe constants include $1/M_0$ and fixed design parameters. At fixed field size, the combined collimator and diffraction grating volumes scale as $M{\theta }^3\propto \theta \propto M^{-1/2}$, while the combined optical camera beam-cone volume scales as $M{\theta }^2\propto 1$.

These are integral-field spectrograph volume scaling laws for the shrinking beam, not a claim that the whole apparatus can shrink without a floor. A photodetector of fixed transverse size $d_0$ needs space for its field: an optical camera envelope interpolating between optical pupil width $b$ and photodetector width $d_0$ has volume proportional to $f_{\mathrm{cam}}(b^2+b{d}_0+d_0^2)$, rather than just $f_{\mathrm{cam}}{b}^2$. Replication then adds terms scaling as $M\theta$ and $M$, which can grow as the spaxels shrink. Other field clearances and mechanical margins also change the asymptote. Without the fixed spectral/design assumptions above, the information in the question does not determine unique physical-volume exponents. The role of fixed photodetector size and beam spread is discussed in Allington-Smith's instrument scaling model.

The etendue is the area–solid-angle acceptance of a ray bundle, more generally $\mathcal{G}=n_{mopt}^2\int \cos \theta dAd\Omega$. In an ideal lossless optical system it is conserved. Increasing pupil area increases accepted power only together with the appropriate field and detector cone; passive optics cannot arbitrarily increase radiance. For a small nadir pixel here, $A\Omega$ denotes entrance-pupil area times the solid angle of its surface footprint.

Choose $1.00$–$1.03\mu \mathrm{m}$ within the surface-sensitive window. A representative band-mean total radiance read from the original blue curve is ${\bar{L}}_{\lambda }\simeq 4\mu \mathrm{W}\mathrm{c}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\mu {\mathrm{m}}^{-1}$, uncertain at roughly the tens-of-per-cent level. Its band integral isThe provided SNR formula depends on $L\Delta \nu$ only through this band integral. If $\nu$ is wavenumber, the selected band spans about $291\mathrm{c}{\mathrm{m}}^{-1}$ and $L_{\nu }d\nu =L_{\lambda }|d\lambda |$; numerical radiance densities in the two coordinates must not be mixed. Convert the specific detectivity to SI: $D^{\ast }=5\times 10^9\mathrm{m}\mathrm{H}{\mathrm{z}}^{1/2}{\mathrm{W}}^{-1}$ and $\sqrt{a}=2\times 10^{-4}\mathrm{m}$. Using the footprint dwell time from part (c),A square $200\mathrm{m}$ footprint at altitude $259\mathrm{km}$ subtends $\Omega \simeq (200/259000{)}^2=5.96\times 10^{-7}\mathrm{sr}$. Thus the collecting area and circular aperture diameter areThe Rayleigh criterion independently requires only $D\gtrsim 1.22\lambda /(200/h)\simeq 1.6\mathrm{mm}$ at $1.015\mu \mathrm{m}$, so the noise requirement dominates diffraction. The $200\mu \mathrm{m}$ detector size corresponds to focal length about $0.259\mathrm{m}$ for this pixel field. This assumes one detector footprint per stated feature size; stricter spatial sampling would need smaller footprints and a revised photon budget.

The estimate assumes a clear, non-scattering atmosphere as in the graph, a night-side or otherwise isolated thermal signal, negligible extra stray-light and read-noise penalties beyond the supplied detectivity, and transmission $0.7$ for the instrument alone. Atmospheric attenuation is already included in the plotted incoming radiance. The selected band has little atmospheric-only signal, but actual hot-spot detection concerns contrast relative to the background: SNR $50$ on total radiance is not a guarantee for an arbitrarily small temperature anomaly. An illustrative radiance range $3$–$5$ in the plotted units would give diameters about $0.20$–$0.15\mathrm{m}$.

The optical telescope supplies collecting area and spatial selection, the spectral selector supplies a passband, and the photodetector converts received radiant flux into a measured signal. For an extended source, a useful schematic photon count iswhere $A\Omega$ is the etendue, $\tau$ the integration time, $\eta$ the net transmission and quantum efficiency, and $L_{\lambda }$ the spectral radiance. An illustrative independent-noise model givesHere $N_b$ and $N_d$ denote background and dark counts, and ${\sigma }_{\mathrm{read}}$ is readout standard deviation. This makes the roles of the three components explicit.

A larger optical telescope collects more photons at fixed angular pixel size and improves the diffraction-limited resolution. Its limits include optical aberrations, alignment, launch size, stray light and pointing. Conservation of etendue constrains simultaneous collecting area and angular acceptance. Increasing spatial resolution by narrowing a pixel's solid angle reduces its photons.

A narrower spectral passband or higher spectral resolving power separates diagnostic features and rejects unwanted background, but usually reduces photons per channel and hence signal-to-noise ratio. Selector transmission, leakage, slit width, detector format and the required time cadence limit the trade. A wider band helps only insofar as it adds useful signal rather than mostly background.

The photodetector's quantum efficiency, dark current, read noise, dynamic range and response time determine conversion losses and added noise. Cooling can suppress thermal background and dark current, but cannot remove photon shot noise. Longer integration improves a stationary signal's signal-to-noise ratio until motion, saturation, drift or required cadence becomes limiting.