Past exam of the astrophysics course of the University of Oxford / 2015 / 8 / f / i / Solution

For a transverse comoving separation ${\ell }_c=100$ Mpc in a flat universe, the corresponding distance is the transverse comoving distance $D_M=D_C$, soEquivalently the physical separation is $100/3$ Mpc and dividing it by $D_A=5421/3$ Mpc gives the same angle. Dividing 100 comoving Mpc directly by $D_A$ would mix scale conventions. The density's $n_c^{-1/3}=100$ Mpc is a characteristic inter-object spacing; a particular nearest neighbour has a distribution of distances and orientations.