 Solution

The phrase “image” does not specify whether a real image or a virtual image is required. The later mention of an image plane makes a projected real image a natural interpretation. Let $u>0$ be object distance and $v>0$ real-image distance from a converging thin lens. A transverse magnification of magnitude $15$ gives $v=15u$, and the object-to-image separation gives $u+v=200\mathrm{mm}$. HencePlace the optical lens $12.5\mathrm{mm}$ beyond the object along the illuminating beam. The real image is inverted and has transverse magnification $-15$.

For the alternative single-optical lens magnifying glass interpretation, the virtual image is upright and lies on the object's side, so signed $v=-15u$. Its separation from the object is $|v|-u=14u=200\mathrm{mm}$. The thin lens equation then givesThe source does not explicitly select between these two geometries; they should not be conflated with a $200\mathrm{mm}$ lens-to-image distance.