For a camera with focal length \( 150\ \text{mm} \) and a \( 30\ \text{cm} \times 30\ \text{cm} \) format size, what height above ground (in meters) is necessary for a vertical photograph to cover an area of \( 9\ \text{km}^2 \)?
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Use the vertical-photograph scale relation S = f/H together with S = photo side / ground side, after converting the 9 km² area to a ground square side.
Step 1: Note the given quantities. Focal length $f = 150$ mm $= 0.15$ m. Photo format $= 30$ cm $\times 30$ cm, so photo area $A_p = 0.3 \times 0.3 = 0.09\ \text{m}^2$. Ground area covered $A_g = 9\ \text{km}^2 = 9 \times 10^6\ \text{m}^2$. Step 2: Relate photo scale to the ratio of areas. For a vertical photograph the linear scale $S$ satisfies $S^2 = \dfrac{A_p}{A_g}$, since area scales as the square of the linear scale. Step 3: Compute $S$. \[ S^2 = \frac{0.09}{9 \times 10^6} = 1 \times 10^{-8} \]\[ S = \sqrt{1 \times 10^{-8}} = 1 \times 10^{-4} = \frac{1}{10000} \] Step 4: Convert the scale into a flying height using $S = f/H$. \[ H = \frac{f}{S} = 0.15 \times 10000 = 1500\ \text{m} \] Step 5: Cross check using the linear side-ratio method. The ground side is $\sqrt{9\ \text{km}^2} = 3000$ m and the photo side is $0.3$ m, giving the same $S = 0.3/3000 = 1/10000$, confirming the area-based computation.\[ \boxed{H = 1500\ \text{m}} \]
