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Use geometrized units and assume $M>0$, $|Q|\le M$, so the Reissner-Nordstrom spacetime contains a black hole. Its outer event horizon is $r_{+}=M+\sqrt{M^2-Q^2}$. The classical absorption cross-section is obtained in geometrical optics, by counting incident null geodesics captured from infinity. Spherical symmetry lets us use the equatorial plane. Write $f(r)=1-2M/r+Q^2/r^2$, and let a dot denote differentiation with respect to an affine parameter. The conserved Killing energy and angular momentum are $E=f\dot{t}$ and $L=r^2\dot{\varphi }$. Null normalization giveswhere the impact parameter is $b=L/E$ and $W=f/r^2$. An incoming ray turns around if the angular-momentum barrier reaches $E^2$ before the ray reaches the event horizon. The limiting captured ray approaches the exterior unstable circular null geodesic, at the maximum of $W$.

The stationary-point equation isThe outer root is the photon sphere radiusIt lies outside $r_{+}$: for $|Q|\le M$ it ranges from $3M$ down to $2M$, whereas $r_{+}$ ranges from $2M$ down to $M$. More precisely $r_{\mathrm{ph}}>2M$ except in the extremal case, and $r_{+}\le 2M$; even at extremality $r_{\mathrm{ph}}=2M>r_{+}=M$. At this root,so the circular null geodesic is unstable. The smaller stationary root lies at or inside the outer horizon and does not set capture from infinity. For $b<b_c$, the radial equation has no exterior turning point and the ray enters the event horizon; for $b>b_c$, it turns around. The critical impact parameter is thereforeThe stationary-point equation gives $Q^2=(3M{r}_{\mathrm{ph}}-r_{\mathrm{ph}}^2)/2$, hence $f(r_{\mathrm{ph}})=(r_{\mathrm{ph}}-M)/(2r_{\mathrm{ph}})$. Since a parallel incident beam is captured over a disk of radius $b_c$, the photon capture cross-section of a Reissner-Nordstrom black hole isThe equality case approaches the unstable orbit indefinitely, but it is a measure-zero boundary of the captured disk and does not change the cross-section.

For the suggested parameterization, choose $\psi$ with $Q=3M\sin \psi /\sqrt{8}$ and $\cos \psi \ge 0$. Then $\sqrt{9M^2-8Q^2}=3M\cos \psi$, so the same cross-section isThe black-hole range has $|\sin \psi |\le \sqrt{8}/3$, hence $\cos \psi \ge 1/3$. For $Q=0$, $r_{\mathrm{ph}}=3M$, $f(r_{\mathrm{ph}})=1/3$, and $b_c=3\sqrt{3}M$, givingthe classical Schwarzschild black hole light-capture result. As an additional limit check, $|Q|=M$ gives $r_{\mathrm{ph}}=2M$, $b_c=4M$, and ${\sigma }_{\mathrm{abs}}=16\pi M^2$. These are geometric capture areas, larger than the corresponding horizon areas; the relevant barrier is the photon sphere, not the horizon's apparent disk. The result applies to light in the short-wavelength limit, while a finite-frequency wave absorption cross-section depends on the greybody factors.