A photograph of a landscape is captured by a drone camera at a height of 18 km . The size of the camera film is $$ 2 cm \times 2 cm $$ and the area of the landscape photographed is $$ 400 km^{2} $$. The focal length of the lens in the drone camera is :

The photograph shows an area of $$400\;{\rm km^{2}}$$ on the ground. If this area is square, its side is the square-root of its area.

Side of landscape, $$L = \sqrt{400\;{\rm km^{2}}} = 20\;{\rm km}$$.

The film is a square of side $$l = 2\;{\rm cm} = 0.02\;{\rm m}$$.

Linear magnification of a lens is $$m = \frac{\text{image size}}{\text{object size}} = \frac{l}{L}$$

Substituting the sizes, $$m = \frac{0.02\;{\rm m}}{20\,000\;{\rm m}} = 1 \times 10^{-6}$$.

Magnification is also $$m = \frac{v}{u}$$, where   $$u$$ = object distance = height of drone $$= 18\;{\rm km}=18\,000\;{\rm m}$$,   $$v$$ = image distance (distance of film from lens).

Therefore $$v = m\,u = 1\times10^{-6}\times 18\,000 = 0.018\;{\rm m}=1.8\;{\rm cm}$$.

For a very distant object ($$u \gg f$$), the lens formula $$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$ gives $$v \approx f$$ because $$\frac{1}{u}$$ is negligible compared with $$\frac{1}{v}$$.

Hence focal length, $$f \approx v = 1.8\;{\rm cm}$$.

Answer: $$1.8\;{\rm cm}$$  (Option A)