1. Given data

• Height of drone camera above landscape: $u = 18\,\text{km}$
• Film size: $2\,\text{cm} \times 2\,\text{cm}$ so film area is $4\,\text{cm}^2$
• Area of landscape photographed: $400\,\text{km}^2$

2. Find the linear size of the landscape photographed

Since the photographed area is square (because the film is square), let the side of the landscape be $L$.

$L^2 = 400\,\text{km}^2$ $L = 20\,\text{km}$

So the landscape dimensions are: $20\,\text{km} \times 20\,\text{km}$

3. Use linear magnification

For a camera, $m = \frac{\text{image size}}{\text{object size}} = \frac{v}{u}$

Here,

• image side $= 2\,\text{cm}$
• object side $= 20\,\text{km}$

Convert $20\,\text{km}$ to cm: $20\,\text{km} = 20 \times 10^5\,\text{cm} = 2 \times 10^6\,\text{cm}$

Thus, $m = \frac{2}{2\times 10^6} = 10^{-6}$

Hence, $\frac{v}{u} = 10^{-6}$

Now convert $u = 18\,\text{km}$ to cm: $u = 18 \times 10^5\,\text{cm} = 1.8\times 10^6\,\text{cm}$

Therefore, $v = 10^{-6} \times 1.8\times 10^6 = 1.8\,\text{cm}$

4. Use lens formula

For a distant object, $u \gg v$, so focal length is approximately equal to image distance: $f \approx v$

More exactly, $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

Since $u = 1.8\times 10^6\,\text{cm}$ is extremely large, $\frac{1}{u} \text{ is negligible}$

So, $f \approx 1.8\,\text{cm}$

5. Check options

• A: $2.8\,\text{cm}$ ❌
• B: $0.9\,\text{cm}$ ❌
• C: $2.5\,\text{cm}$ ❌
• D: $1.8\,\text{cm}$ ✅

Therefore, the correct answer is: $\boxed{1.8\,\text{cm}}$