JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Geometrical Optics

Treat the photographed landscape as a square patch of area $400{\text{km}}^2$, so its side is


The image on the film is a square of side $y=2\text{cm}=2\times 10^{-2}\text{m}$.

The magnification of the lens is


For a thin lens, $m=\frac{v}{u}$ in magnitude, where $u=18\text{km}=1.8\times 10^4\text{m}$ is the object distance (drone height). So


Since the object distance is enormously larger than any reasonable focal length ($u\gg f$), the image forms essentially at the focal plane, so $v\approx f$. The focal length of the lens is therefore $1.8$ cm.