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(A) The magnification produced by the objective lens of a telescope is given by the ratio of the height of the image $(I)$ to the height of the object

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) The magnification produced by the objective lens of a telescope is given by the ratio of the height of the image $(I)$ to the height of the object $(O)$,which is also equal to the ratio of the focal length of the objective $(f_0)$ to the object distance $(u_0)$: $m = \frac{I}{O} = \frac{f_0}{u_0}$ Given: Height of the building $(O)$ = $50 \, m = 5000 \, cm$ Focal length of objective $(f_0)$ = $200 \, cm$ Distance of the building $(u_0)$ = $2 \, km = 2000 \, m = 200,000 \, cm$ Substituting the values into the formula: $\frac{I}{5000} = \frac{200}{200000}$ $I = 5000 	imes \frac{200}{200000}$ $I = 5000 	imes \frac{1}{1000} = 5 \, cm$ Therefore,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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