= Aperture truncation of etendue
{title2=$G_{\rm transmitted}/G_{\rm incident}=(d/D)^2$}
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. Therefore
$$
\frac{G_{\rm transmitted}}{G_{\rm incident}}=\frac{d^2}{D^2}=0.36,
\qquad
G_{\rm transmitted}=\frac{9\pi^2}{4}\,\mathrm{mm^2\,sr}
\approx22.207\,\mathrm{mm^2\,sr}.
$$
With 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.
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