Solution
= Solution
The <astronomical magnitude> convention is $m_1-m_2=-2.5\log_{10}(F_1/F_2)$, with both <fluxes> measured through the same <photometric passband>. A five-unit <magnitude difference> therefore corresponds to
$$
\boxed{\frac{F_b}{F_f}=10^{(6-1)/2.5}=100.}
$$
The brighter endpoint has one hundred times the visual <flux> of the fainter endpoint. Absolute <fluxes> require a <magnitude zero point>, which is not given.