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

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(B) For a telescope adjusted for parallel rays (final image at infinity),the magnifying power $m$ is given by:$m = \frac{f_o}{f_e} = 9$where $f_o$ is

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$18 \ cm, 2 \ cm$

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(B) For a telescope adjusted for parallel rays (final image at infinity),the magnifying power $m$ 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 $f_o = 9f_e$ ... $(i)$The distance between the objective and the eyepiece for parallel rays is given by:$L = f_o + f_e = 20 \ cm$Substituting equation $(i)$ into this expression:$9f_e + f_e = 20 \ cm$$10f_e = 20 \ cm$$f_e = 2 \ cm$Now,calculating $f_o$:$f_o = 9 	imes 2 \ cm = 18 \ cm$Therefore,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 objective and eyepiece are respectively:

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