An equiconvex lens made of glass of refractive index 1.5 has both surfaces of radius 30 cm. One of its faces is silvered, and a small object is…

The focal length of the lens is $\dfrac1f=(1.5-1)\left(\dfrac1{30}+\dfrac1{30}\right)$, so $f=30\ \mathrm{cm}$. The silvered face is a concave mirror of focal length $15\ \mathrm{cm}$. First refraction: $\dfrac1v=\dfrac1{30}-\dfrac1{10}$ gives $v=-15\ \mathrm{cm}$, a virtual image $15\ \mathrm{cm}$ in front, which acts as a real object for the mirror at a distance equal to the mirror's focal length $15\ \mathrm{cm}$, so the reflected rays are parallel. They refract again through the lens and converge at its focal point, $30\ \mathrm{cm}$ in front of the lens. Equivalent single mirror: $\dfrac1F=\dfrac2{30}+\dfrac1{15}$, $F=7.5\ \mathrm{cm}$; object at $10\ \mathrm{cm}$ gives $v=30\ \mathrm{cm}$, matching.