Aperture truncation of etendue

Area-angle tradeoff

{title2=$G=n^2A\pi\sin^2\theta_{\max}$}

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> is
$$
G=n^2A\int_0^{2\pi}\!\int_0^{\theta_{\max}}
\cos\theta\sin\theta\,d\theta\,d\phi
=n^2A\pi\sin^2\theta_{\max}=\pi A\,\mathrm{NA}^2,
$$
where $\mathrm{NA}=n\sin\theta_{\max}$ 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 $G=n^2A\pi$, rather than $2\pi n^2A$. <Radiative Transfer lecture notes by John E. Greivenkamp>[Section 22, slides 22-3–22-4, solid-angle element and exact cone solid angle; slide 22-7, projected-area hemisphere factor]{ref}

For an original model example, choose $n=1$, $D=10\,\mathrm{mm}$, and $\theta_{\max}=30^\circ$. Then
$$
A=25\pi\,\mathrm{mm^2},\qquad
G=\frac{25\pi^2}{4}\,\mathrm{mm^2\,sr}
\approx61.685\,\mathrm{mm^2\,sr}.
$$
If the <radiance> is uniform across this filled bundle and equals $L=100\,\mathrm{W\,m^{-2}\,sr^{-1}}$, its <radiant flux> is $\Phi=LG/n^2\approx6.1685\,\mathrm{mW}$. Neither the geometry nor the chosen <radiance> is an experimental measurement.


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 Incoming links: Etendue of a circular aperture and cone