 Solution

The wavelength is $\lambda =c/f\simeq 0.50\mathrm{m}$, 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 $\sin \theta \simeq 1.22\lambda /D$, while its half-power point has $\sin \theta \simeq 0.514\lambda /D$. ThusThe 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 $G_d\simeq 1.64$. 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,For ideal aperture efficiency this is 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: $V_{\mathrm{rms}}^2=P_{\mathrm{available}}{R}_L$ for the respective matched loads. A short-dipole rather than half-wave convention changes the order-one gain factor, not the main estimate.