The magnifying power of a telescope with tube | vedclass

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(D) The magnifying power $m$ of an astronomical telescope in normal adjustment is given by $m = \frac{f_o}{f_e} = 5$,where $f_o$ is the focal length o

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$10$

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(D) The magnifying power $m$ of an astronomical telescope in normal adjustment is given by $m = \frac{f_o}{f_e} = 5$,where $f_o$ is the focal length of the objective lens and $f_e$ is the focal length of the eyepiece.From this,we get $f_o = 5f_e$.The length of the telescope tube $L$ is given by $L = f_o + f_e = 60 \ cm$.Substituting $f_o = 5f_e$ into the length equation:$5f_e + f_e = 60 \ cm$$6f_e = 60 \ cm$$f_e = 10 \ cm$.Therefore,the focal length of the eyepiece is $10 \ cm$.

The magnifying power of a telescope with tube length $60 \ cm$ is $5$. Then the focal length of its eyepiece is: (in $cm$)

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