= 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 is
$$
m-M=5\log_{10}\frac D{10\ \mathrm{pc}}+A,
$$
where $A$ is <interstellar extinction> in that filter. Subtracting the two equations yields
$$
\boxed{\frac{D_2}{D_1}=10^{[\Delta m-\Delta M-\Delta A]/5}.}
$$
Here $\Delta m=m_2-m_1$, $\Delta M=M_2-M_1$, 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, $\Delta M\simeq0$. Neglecting differential <interstellar extinction> then gives $D_2/D_1\simeq10^{\Delta m/5}$. A one-<astronomical magnitude> shift would mean a factor $1.585$ in distance, illustrating the calibration rather than measuring these panels.
\b[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.
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