In aerial mapping a camera uses a lens with a 100 cm focal length. The height at which the airplane must fly, so that the photograph of a 1 km long strip on the ground fits exactly on the 20 cm long filmstrip of the camera, is

200 km

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20 km

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5 km

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1 km

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Solution

The correct option is C 5 km
Focal length, f = 1 m,
size of object, $h_{1} = 1000$ m, size of image, $h_{2} = 0.2$ m
image is real, so
$- \frac{h_{2}}{h_{1}} = \frac{v}{u} \Rightarrow v = - 2 \times 10^{- 4} u$
$∵ \frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$1 = - \frac{1}{2 \times 10^{- 4} u} - \frac{1}{u} \Rightarrow u = - 5000 m$ or $u = - 5 k m$

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In aerial mapping a camera uses a lens with a 100 cm focal length. The height at which the airplane must fly, so that the photograph of a 1 km long strip on the ground fits exactly on the 20 cm long filmstrip of the camera, is

The correct option is C 5 kmFocal length, f = 1 m,size of object, h1=1000 m, size of image, h2=0.2 mimage is real, so−h2h1=vu⇒v=−2×10−4u∵1f=1v−1u1=−12×10−4u−1u⇒u=−5000m or u=−5km