= Solution
Use the original diagram's positive distances: $u$ from the object to the vertex, $d$ from the vertex to the image, and $R$ from the vertex toward the centre of curvature on the outgoing side. Let a paraxial <light ray> strike at height $Y$. Its incoming inclination is $\alpha\simeq Y/u$ above the <optical axis>, its outgoing inclination is $\beta\simeq Y/d$ below it, and the inward surface normal is inclined by $\theta\simeq Y/R$ below it. Thus the incidence and refraction angles relative to that normal are respectively $\alpha+\theta$ and $\theta-\beta$.
The local tangent-plane form of <Snell's law>, using the <small-angle approximation>, gives
$$
n_1(\alpha+\theta)=n_2(\theta-\beta),\qquad n_1\alpha+n_2\beta=(n_2-n_1)\theta.
$$
Substitute the three height-to-distance ratios and cancel $Y$:
$$
\boxed{\frac{n_1}{u}+\frac{n_2}{d}=\frac{n_2-n_1}{R}.}
$$
This <refraction at a spherical surface> result keeps only first-order ray angles. The surface sag shifts the height's axial reference only at higher order. Distances for other object, image or curvature orientations must be signed consistently; the displayed positive-distance form follows the particular orientation in the paper.
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