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(C) The focal length of the objective mirror $(f_o)$ is half of its radius of curvature $(R)$: $f_o = \frac{R}{2} = \frac{80 \, cm}{2} = 40 \, cm$. Gi

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$25$

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(C) The focal length of the objective mirror $(f_o)$ is half of its radius of curvature $(R)$: $f_o = \frac{R}{2} = \frac{80 \, cm}{2} = 40 \, cm$. Given the focal length of the eyepiece $(f_e)$ is $1.6 \, cm$. The magnifying power $(m)$ of a reflecting telescope is given by the ratio of the focal length of the objective to the focal length of the eyepiece: $m = \frac{f_o}{f_e} = \frac{40}{1.6} = 25$. Therefore,the magnifying power is $25$.

$A$ reflecting telescope has a large mirror as its objective with a radius of curvature equal to $80 \, cm$. What is the magnifying power of this telescope if the eyepiece used has a focal length of $1.6 \, cm$?

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