The magnifying power of a telescope is nine. When it is adjusted for parallel rays, the distance between the objective and eyepiece is 20 cm. The focal length of objective and eyepiece are respectivel

Question

The magnifying power of a telescope is nine. When it is adjusted for parallel rays, the distance between the objective and eyepiece is 20 cm. The focal length of objective and eyepiece are respectively.

TestHub · Accepted Answer

For final image at infinity, magnifying power of a telescope is given by
$m = \frac{f_{o}}{f e} = 9$
where, m = magnification,
${a n d f}_{o}$ = focal length of eyepiece
$\Rightarrow f_{o} = 9 f_{e}$ …. (i)
Also, distance between objective and eyepiece
$= f_{o} + f_{e} = 20$ (given)
$\Rightarrow 9 f_{e} + f_{e} = 20 \Rightarrow f_{e} = 2 c m$
$f_{o} = 9 f_{e} = 18 c m$

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