A telescope has an objective lens of focal le | vedclass

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(A) The magnification $m$ produced by the objective lens is given by the ratio of the image height $I$ to the object height $O$,which is also equal to

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) The magnification $m$ produced by the objective lens is given by the ratio of the image height $I$ to the object height $O$,which is also equal to the ratio of the focal length $f_0$ to the object distance $u_0$ for distant objects.$m = \frac{I}{O} = \frac{f_0}{u_0}$Given:Object height $O = 50 \ m = 5000 \ cm$Focal length $f_0 = 200 \ cm$Object distance $u_0 = 2 \ km = 2000 \ m = 200000 \ cm$Substituting the values:$\frac{I}{5000} = \frac{200}{200000}$$I = 5000 	imes \frac{200}{200000}$$I = 5000 	imes \frac{1}{1000} = 5 \ cm$Thus,the height of the image formed by the objective lens 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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