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(A) Given:Height of the drone,$H = 18 \ 	ext{km} = 18 	imes 10^3 \ 	ext{m}$.Area of the landscape,$A_{	ext{landscape}} = 400 \ 	ext{km}^2 = (20 \

Vedclass · Accepted Answer

$1.8$

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(A) Given:Height of the drone,$H = 18 \ 	ext{km} = 18 	imes 10^3 \ 	ext{m}$.Area of the landscape,$A_{	ext{landscape}} = 400 \ 	ext{km}^2 = (20 \ 	ext{km}) 	imes (20 \ 	ext{km})$.Thus,the side length of the landscape is $x = 20 \ 	ext{km} = 20 	imes 10^3 \ 	ext{m}$.Size of the camera film is $2 \ 	ext{cm} 	imes 2 \ 	ext{cm}$,so the side length of the image is $y = 2 \ 	ext{cm} = 2 	imes 10^{-2} \ 	ext{m}$.Using the property of similar triangles,the ratio of the side length of the object to the side length of the image is equal to the ratio of the object distance to the focal length (since the image is formed at the focal plane for a distant object):$\frac{x}{y} = \frac{H}{f}$$\frac{20 \ 	ext{km}}{2 \ 	ext{cm}} = \frac{18 \ 	ext{km}}{f}$$\frac{20 	imes 10^3 \ 	ext{m}}{2 	imes 10^{-2} \ 	ext{m}} = \frac{18 \ 	ext{km}}{f}$$10^6 = \frac{18 \ 	ext{km}}{f}$$f = \frac{18 \ 	ext{km}}{10^6} = 18 	imes 10^{-6} \ 	ext{km} = 18 	imes 10^{-6} 	imes 10^5 \ 	ext{cm} = 1.8 \ 	ext{cm}$.Therefore,the focal length of the lens is $1.8 \ 	ext{cm}$.

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

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