A telescope has an objective lens of focal le | vedclass

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(A) The height of the image $h'$ formed by the objective lens can be calculated using the magnification formula for a lens: $m = \frac{h'}{h} = \frac{

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$5$

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(A) The height of the image $h'$ formed by the objective lens can be calculated using the magnification formula for a lens: $m = \frac{h'}{h} = \frac{v}{u}$.Given:Height of the object $h = 50 \ m = 5000 \ cm$.Distance of the object $u = -2 \ km = -2000 \ m = -200000 \ cm$.Focal length of the objective lens $f = 200 \ cm$.Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:$\frac{1}{v} - \frac{1}{-200000} = \frac{1}{200}$$\frac{1}{v} = \frac{1}{200} - \frac{1}{200000} = \frac{1000 - 1}{200000} = \frac{999}{200000}$$v = \frac{200000}{999} \approx 200.2 \ cm$.Now,$h' = h 	imes \frac{v}{|u|} = 5000 	imes \frac{200000/999}{200000} = \frac{5000}{999} \approx 5.005 \ cm$.Rounding to the nearest integer,the height of the image is $5 \ cm$.

$A$ telescope has an objective lens of focal length $200 \ cm$ and an eyepiece with focal length $2 \ cm$. If this telescope is used to see a $50 \ m$ tall building at a distance of $2 \ km$,what is the height of the image of the building formed by the objective lens in $cm$?

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