At distance $L$, the <small-angle approximation> gives $\ell=L\theta$. Thus \b[the transverse resolution is approximately $2.5\times10^{10}\,\mathrm m$], or $3.0\times10^{10}\,\mathrm m$ with the circular-aperture <Rayleigh criterion>. This is a transverse separation at the source, rather than a determination of its distance along the line of sight.

Solution

Past exam of the physics A course of the University of Cambridge / 2015 / Paper 2 / Section A / 1 / b / Solution

 Solution (source code)