The magnifying power of a telescope is 9. Whe | vedclass

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(A) The magnifying power $m$ of a telescope adjusted for parallel rays (normal adjustment) is given by $m = \frac{f_o}{f_e} = 9$,where $f_o$ is the fo

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$18 \ cm, 2 \ cm$

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(A) The magnifying power $m$ of a telescope adjusted for parallel rays (normal adjustment) is given by $m = \frac{f_o}{f_e} = 9$,where $f_o$ is the focal length of the objective and $f_e$ is the focal length of the eyepiece.From this,we get $f_o = 9f_e$.The length of the telescope tube in normal adjustment is $L = f_o + f_e = 20 \ cm$.Substituting $f_o = 9f_e$ into the length equation: $9f_e + f_e = 20 \ cm$.$10f_e = 20 \ cm \Rightarrow f_e = 2 \ cm$.Now,calculating $f_o$: $f_o = 9 	imes 2 \ cm = 18 \ cm$.Thus,the focal lengths are $18 \ cm$ and $2 \ cm$.

The magnifying power of a telescope is $9$. When it is adjusted for parallel rays,the distance between the objective and eyepiece is $20 \ cm$. The focal lengths of the lenses are:

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