 Etendue concentration limit

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.