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(B) The magnifying power $m$ of an astronomical telescope in normal adjustment is given by the formula:$m = \frac{f_o}{f_e}$where $f_o$ is the focal l

Vedclass · Accepted Answer

$200$

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(B) The magnifying power $m$ of an astronomical telescope in normal adjustment is given by the formula:$m = \frac{f_o}{f_e}$where $f_o$ is the focal length of the objective lens and $f_e$ is the focal length of the eyepiece.Given:Magnifying power $m = 10$Focal length of eyepiece $f_e = 20 \, cm$Substituting the values into the formula:$10 = \frac{f_o}{20 \, cm}$$f_o = 10 	imes 20 \, cm$$f_o = 200 \, cm$Therefore,the focal length of the objective lens is $200 \, cm$.

An astronomical telescope has a magnifying power $10$. The focal length of the eyepiece is $20 \, cm$. The focal length of the objective is.....$cm$.

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