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

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(C) The magnifying power 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

Vedclass · Accepted Answer

$18\; cm$ and $2\; cm$

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(C) The magnifying power 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 $f_o = 9f_e$ ..... $(i)$The length of the telescope tube in normal adjustment is the sum of the focal lengths of the objective and eyepiece: $L = f_o + f_e = 20\; cm$ ..... $(ii)$Substituting equation $(i)$ into equation $(ii)$:$9f_e + f_e = 20\; cm$$10f_e = 20\; cm$$f_e = 2\; cm$Now,substituting $f_e = 2\; cm$ back into equation $(i)$:$f_o = 9 	imes 2\; cm = 18\; cm$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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