Past exam of the astrophysics course of the University of Oxford / 2021 / 5 / b / Solution

TODO: measure the relative-distance offset from an intact source figure or its numerical photometry.

For corresponding populations whose intrinsic absolute magnitudes are calibrated to match, the distance modulus in the same filter iswhere $A$ is interstellar extinction in that filter. Subtracting the two equations yieldsHere $\Delta m=m_2-m_1$, $\Delta M=M_2-M_1$, and $\Delta A=A_2-A_1$. For a matched horizontal branch after a metallicity correction, or matched unevolved main sequences after correcting colour and composition, $\Delta M\simeq 0$. Neglecting differential interstellar extinction then gives $D_2/D_1\simeq 10^{\Delta m/5}$. A one-astronomical magnitude shift would mean a factor $1.585$ in distance, illustrating the calibration rather than measuring these panels.

A defensible numerical ratio cannot be extracted from the surviving figure. The corresponding sequences and most of the vertical scales are hidden by the corrupted image. Different bottom-axis labels or limiting astronomical magnitudes do not establish a common-population vertical offset: exposure depth and intrinsic population differences can also change them. An intact figure or numerical photometry is needed to measure $\Delta m$, then correct $\Delta M$ and $\Delta A$. The equation above is the quantitative inference supported by the available source; inventing an offset would not solve the missing-data problem.