= Solution
\b[<antenna> quantities.] <radiation resistance> represents emitted <power> by $P_{\rm rad}=I_{\rm rms}^2R_{\rm rad}$ at the feed. It is not <Joule heating> in the <antenna>; a separate loss resistance accounts for that. <antenna gain> is the directional far-field intensity relative to an isotropic radiator supplied with the same input <power>:
$$
G(\theta,\phi)=\frac{4\pi r^2S_r(\theta,\phi)}{P_{\rm in}}.
$$
It includes <radiation efficiency> and equals <antenna directivity> only for a lossless <antenna>. <antenna effective area> is defined by the <power> available to a matched load, $P_{\rm av}=A_{\rm eff}S_{\rm inc}$, for incidence from a specified direction and matched polarization. It need not equal a literal geometric collecting area.
\b[Gain–area relation.] To derive the <antenna gain-effective-area relation>, take a passive reciprocal <antenna>. Its <electromagnetic reciprocity> makes the receive pattern proportional to the transmit pattern; write $A_{\rm eff}(\Omega)=K G(\Omega)$. The normalization follows from <thermal equilibrium>. In the <Rayleigh-Jeans law> limit, <blackbody radiation> radiance per unit <frequency> per unit <solid angle> is $2k_BT/\lambda^2$ for two polarizations. A single <antenna> polarization couples to one half, so the available incident noise <power> is
$$
dP_{\rm inc}=\frac{k_BT}{\lambda^2}\int A_{\rm eff}(\Omega)\,d\Omega\,d\nu.
$$
The open-circuit Johnson voltage spectrum of the <antenna> resistance is $4k_BTR_{\rm rad}\,d\nu$. A <matched load> receives half the open-circuit voltage, giving available <power> $k_BT\,d\nu$. In a lossless <antenna> at <thermal equilibrium> these <powers> agree; hence $\int A_{\rm eff}\,d\Omega=\lambda^2$. Since $\int G\,d\Omega=4\pi$,
$$
\boxed{A_{\rm eff}(\theta,\phi)=\frac{\lambda^2}{4\pi}G(\theta,\phi)}.
$$
The same efficiency factor multiplies receiving area and gain if reciprocal dissipative losses are included consistently. Impedance mismatch or polarization mismatch reduces delivered <power> further. The factor of two for thermal polarization modes and the factor of four between open-circuit voltage variance and matched-load <power> are separate effects.
\b[Normalize the <radar> beam.] Interpret the azimuth as signed about the beam centre at $\phi=0$, so both sides of the narrow lobe are included. Put $u=\theta-\pi/2$. To leading order in the small angular width, $\sin\theta\simeq1$ and the angular integrals can be extended over the real line:
$$
P=\int N\,d\Omega\simeq C\int e^{-u^2/(2\alpha^2)}du
\int e^{-\phi^2/(2\alpha^2)}d\phi=2\pi\alpha^2C.
$$
The <Gaussian antenna beam solid angle> gives the peak on the positive $x$-axis consequently has
$$
C\simeq\frac{P}{2\pi\alpha^2},\qquad
G_0\simeq\frac2{\alpha^2},\qquad
A_0\simeq\frac{\lambda^2}{2\pi\alpha^2},\qquad \lambda=\frac c\nu.
$$
Here $N$ is <power> per <solid angle>, equal to $r^2S_r$ in the far field. The leading neglected solid-angle correction is relative order $\alpha^2$.
\b[Return using the supplied electric-dipole response.] The <electric-dipole radar cross section of a small conducting sphere> determines this return. At target range $D$, the incident intensity is $S_t=C/D^2$. Write its <root mean square> <electric field> as $E_t$, with $S_t=\epsilon_0cE_t^2$. The induced <electric dipole> is $p_{\rm rms}=4\pi\epsilon_0a^3E_t$. A dipole's angular radiation pattern, normalized by $\int\sin^2\chi\,d\Omega=8\pi/3$, gives $dP_{\rm sc}/d\Omega=(3P_{\rm sc}/8\pi)\sin^2\chi$. The backward direction is perpendicular to the induced dipole, so the supplied total dipole <power> gives
$$
\left.\frac{dP_{\rm sc}}{d\Omega}\right|_{\rm back}
=\frac{3}{8\pi}\frac{8\pi^3\nu^4p_{\rm rms}^2}{3\epsilon_0c^3}
=k^4a^6S_t,\qquad k=\frac{2\pi\nu}{c}.
$$
The received intensity is this angular <power> divided by $D^2$. Using the <monostatic radar equation>, or directly multiplying by the receive area, gives
$$
P_R=\frac{A_0}{D^2}k^4a^6\frac{C}{D^2}
=\frac{4\pi^2Pa^6}{\lambda^2\alpha^4D^4}.
$$
The receiver threshold is a voltage across the <matched load>, so the detectable <power> is $V_m^2/R$, without another matching factor:
$$
\boxed{D_{\rm electric}=
\left(\frac{4\pi^2PR\nu^2a^6}{c^2\alpha^4V_m^2}\right)^{1/4}}.
$$
This is the electric-dipole result following from the response data provided. It assumes $ka\ll1$, far-field propagation on each leg, alignment at the beam maximum and no attenuation or additional receiver losses.
\b[Perfect-conductor qualification.] <perfect-conductor sphere dipole interference> arises because the sphere also excludes the oscillating <magnetic field>. Solving the magnetic boundary problem gives $\mathbf m=-2\pi a^3\mathbf H_t$ in addition to the <electric dipole>. The <magnetic polarizability of a perfectly conducting sphere> follows from its induced surface currents.<Electromagnetic Fields and Energy by Hermann A. Haus and James R. Melcher>[section 8.4, Example 8.4.3: magnetic dipole induced in a perfectly conducting sphere]{ref} In the backward direction the magnetic contribution has half the electric scattering <amplitude> and the same sign, so the returned intensity is $9/4$ of the electric-only value. Equivalently, the monostatic <radar cross section> is $\sigma_{\rm back}=9\pi k^4a^6$, rather than $4\pi k^4a^6$. Including both leading dipoles therefore gives
$$
\boxed{D_{\rm PEC}=
\left(\frac{9\pi^2PR\nu^2a^6}{c^2\alpha^4V_m^2}\right)^{1/4}
=\sqrt{\frac32}\,D_{\rm electric}}.
$$
Both results have been distinguished because the supplied auxiliary data describe only the <electric dipole>, while the literal perfect-conductor boundary conditions also generate the magnetic one. The backward interference is a leading-order effect, not a higher-order correction in $ka$.<Scattering of EM Waves by Michael Fowler>[section Scattering by a Small Perfectly Conducting Sphere: electric and magnetic dipoles and backward interference]{ref}
\Image[/past-exam-of-the-optics-and-electrodynamics-course-of-the-university-of-cambridge/2015/paper-1-radar-scattering.png]
{title=Original electric-only and electric-plus-magnetic small-sphere scattering comparison}
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