= Solution
The <wavelength> is $\lambda=c/f\simeq0.50\,\mathrm m$, and the aperture diameter is $D=1.0\,\mathrm m$. A uniformly illuminated circular aperture has normalized <power> pattern $[2J_1(x)/x]^2$, where $x=\pi D\sin\theta/\lambda$. Its first null is at $\sin\theta\simeq1.22\lambda/D$, while its half-power point has $\sin\theta\simeq0.514\lambda/D$. Thus
$$
\boxed{\text{Point to within roughly }0.25\,\mathrm{rad}\ (15^\circ)\text{ for modest loss}.}
$$
The first-null small-angle estimate is $0.61\,\mathrm{rad}$, or roughly $35^\circ$; using the sine gives about $38^\circ$. This is a broad beam because the diameter is only about two <wavelengths>. Real illumination and aperture efficiency alter the exact beam shape; a required allowable gain loss would specify the pointing tolerance more precisely.
For a matched, correctly polarized half-wave dipole, $A_{e,d}=G_d\lambda^2/(4\pi)$ with $G_d\simeq1.64$. For an efficient dish, $A_{e,s}\simeq\eta\pi(D/2)^2$. Received available <powers> are proportional to these areas. With both <voltages> referred to the same matched receiver resistance,
$$
\boxed{\frac{V_{\rm dipole}}{V_{\rm dish}}\simeq
\sqrt{\frac{G_d\lambda^2}{\eta\pi^2D^2}}
=\frac{0.204}{\sqrt\eta}.}
$$
For ideal aperture efficiency this is \b[about one fifth], or five times greater <voltage> from the dish. Terminal <voltages> without a common impedance or matching convention are not fixed by aperture alone: $V_{\rm rms}^2=P_{\rm available}R_L$ for the respective <matched loads>. A short-dipole rather than half-wave convention changes the order-one gain factor, not the main estimate.
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