For identical <luminosities> in a uniform Euclidean distribution, the <inverse-square law> gives flux $F\propto r^{-2}$, while the number of sources within distance $r$ is proportional to $r^3$. Therefore ten times as many objects require $r_{10}/r_1=10^{1/3}$ and $F_{10}/F_1=10^{-2/3}$. Applying the definition of <apparent magnitude> gives
$$
\boxed{m_{10}-m_1=-2.5\log_{10}(10^{-2/3})=\frac53,\qquad m_{10}=m_1+\frac53.}
$$
This is the <Euclidean magnitude-limited source count> relation. It assumes no extinction, spatial variation of density, or cosmological change in the flux-distance relation.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 338 / 1 / c / iii / Solution

For identical <luminosities>, uniform spatial density, and no extinction, the <inverse-square law> gives limiting distance $r\propto F_{\min}^{-1/2}$. The number in the enclosed volume is therefore $N\propto F_{\min}^{-3/2}\propto10^{0.6m_{\lim}}$. Increasing the sample by a factor $b$ needs a magnitude-limit increase $(5/3)\log_{10}b$. This relation need not hold with evolving source populations, dust, or cosmological distance effects.

Codex @codex  0

 Incoming links: Euclidean magnitude-limited source count