A thin equiconvex glass lens of refractive index 1.5 has both surfaces of radius of curvature 30 cm. Its right surface is silvered, so the lens…

Light passes through the left surface (water to glass), reflects from the silvered right surface (a concave mirror of radius $R=30\ \mathrm{cm}$ seen from inside glass), and passes again through the left surface (glass to water). Powers (in the reduced form $n/f$): each refraction $(n_g-n_w)/R=(1.5-\tfrac43)/30$; the mirror $2n_g/R=3/30$. Total $P=2\cdot\dfrac{1/6}{30}+\dfrac{3}{30}=\dfrac{10/3}{30}=\dfrac19\ \mathrm{cm^{-1}}$. The equivalent mirror has focal length (in water) $f=n_w/P=\dfrac{4/3}{1/9}=12\ \mathrm{cm}$. The image coincides with the object when the object is at the centre of curvature of this equivalent mirror, $u=2f=24\ \mathrm{cm}$. A direct check with the paraxial ray transfer gives the returning ray crossing the axis at distance $uf/(u-f)$, which equals $u$ only for $u=2f$. Using $f=12$ cm itself would put the image at infinity, and the other values mix up the equivalent power.