{title2=$n_1^2A_1\sin^2\theta_1=n_2^2A_2\sin^2\theta_2$}

When both endpoint bundles have the filled circular-cone geometry of <etendue of a circular aperture and cone>, <conservation of etendue> gives
$$
n_1^2A_1\sin^2\theta_1=n_2^2A_2\sin^2\theta_2.
$$
For 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 $\theta_2\approx14.4775^\circ$. 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.

\Image[/etendue-area-angle-exchange.png]
{title=Equal <etendue>: four times the spatial <area> exchanges for one quarter of the directional <area>}
{description=The circles are cross-sections of the four-dimensional <phase space>: spatial disks in $(x,y)$ and direction disks in $(p_x,p_y)$. Scales agree within each row. The displayed transformation preserves their product volume; it is a schematic of endpoint bundles, not a claim that a particular <optical lens> exactly realizes the transformation at every <angle>.}

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.


Area-angle tradeoff ( $n_{1}^{2} A_{1} sin^{2} θ_{1} = n_{2}^{2} A_{2} sin^{2} θ_{2}$ )

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Area-angle tradeoff (n12​A1​sin2θ1​=n22​A2​sin2θ2​)

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Area-angle tradeoff ( $n_{1}^{2} A_{1} sin^{2} θ_{1} = n_{2}^{2} A_{2} sin^{2} θ_{2}$ )

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