Past exam of the physics course of the year 4 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / Section A / A6 / Solution

The quoted angular diameter gives half-angle $\theta =0.005$. At distance $r=10^{13}\mathrm{m}$, the uniform beam covers area $A\simeq \pi (r\theta {)}^2=7.85\times 10^{21}{\mathrm{m}}^2$. Its intensity is $I=P/A\simeq 2.93\times 10^{-21}\mathrm{W}{\mathrm{m}}^{-2}$. For a vacuum electromagnetic wave, $I={ϵ}_{0}c{E}_{\mathrm{rms}}^2$, soNo extra factor two belongs in this rms relation. The frequency fixes the wavelength but is unnecessary once the total power and stipulated beam divergence are specified. This is the ideal directed-beam estimate, before propagation or receiver losses.