= Solid angle of a disk
{title2=$\Omega=2\pi(1-d/\sqrt{d^2+a^2})$}
Consider a flat circular <disk> of <radius> $a>0$, with its center a perpendicular distance $d>0$ from the viewpoint $O$. The line from $O$ to its center is normal to the <disk>. Its rim defines a <right circular cone> with $\tan\theta=a/d$ and $\cos\theta=d/\sqrt{d^2+a^2}$. The <solid angle of a circular cone> immediately gives
$$
\boxed{\Omega=2\pi\left(1-\frac{d}{\sqrt{d^2+a^2}}\right).}
$$
For $a=d$, this is the same $45^\circ$ half-opening example: about $1.8403$ <steradians>. Both the <radius> and distance enter; doubling both leaves the <solid angle> unchanged.
A useful local version applies to a small surface patch of <area> $dA$, at distance $r$ from $O$. If $\psi$ is the <angle> between its <unit normal> and the line to $O$, its unsigned geometric contribution is
$$
d\Omega=\frac{|\cos\psi|\,dA}{r^2}.
$$
The numerator is the <projected area>. For a visible patch, perpendicular projection followed by radial projection onto a small piece of a <sphere> gives this formula. Integrating counts the directions once provided no direction encounters multiple counted patches. For an opaque detector, use its unoccluded face. At grazing view, the <projected area> tends to zero.
For this on-axis <disk>, a ring of <radius> $\rho$ has $r=\sqrt{d^2+\rho^2}$ and $|\cos\psi|=d/r$. Its <polar coordinates> gives $dA=\rho\,d\rho\,d\phi$, so the same answer follows directly from the <surface integral>:
$$
\Omega=2\pi\int_0^a\frac{d\,\rho}{(d^2+\rho^2)^{3/2}}\,d\rho
=2\pi\left(1-\frac{d}{\sqrt{d^2+a^2}}\right).
$$
The changing distance and obliquity are both already accounted for; an extra detector-wide cosine factor would count the projection twice.
\Image[/solid-angle-disk-detector-and-distance.png]
{title=The exact <solid angle> of an on-axis circular detector approaches a hemisphere at close range}
{description=Left: a meridian section, drawn with $a=d$. Right: the geometric acceptance fraction $\Omega/(4\pi)$ for a <point emitter> with <isotropy>, compared with the distant-source approximation $a^2/(4d^2)$. These are calculated geometric curves, not measured detector efficiencies. The dashed approximation is displayed outside its accurate range to show how it overestimates acceptance.}
For $d\gg a$, the <Taylor series> of $(1+a^2/d^2)^{-1/2}$ gives
$$
\Omega=\frac{\pi a^2}{d^2}+O\!\left(\frac{a^4}{d^4}\right).
$$
Thus the accepted fraction of an ideal <point emitter>'s isotropic output is approximately $A/(4\pi d^2)$, where $A=\pi a^2$. This is the geometry behind the <inverse-square law> for a small detector. At fixed $a$, however, taking $d\to0^+$ gives $\Omega\to2\pi$, so the accepted fraction approaches $1/2$ rather than diverging.
For a chosen numerical example, let $a=1\,\mathrm{cm}$ and $d=5\,\mathrm{cm}$. The exact <solid angle> is approximately $0.122015$ <steradians>, giving an acceptance fraction $0.00970966$. An ideal <point emitter> with <isotropy> and total <radiant flux> $1\,\mathrm{mW}$ would therefore send \b[about $9.71\,\mathrm{\mu W}$ onto the disk]. The distant-source approximation gives $10.0\,\mathrm{\mu W}$, about $3.0\%$ high. This calculation assumes unobstructed propagation and counts incident <radiant flux>; <photon absorption>, detector response, and any <optical attenuator> require separate factors.
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