Question

An astronomical telescope has a magnifying power 10 in normal adjustment distance between the objective lens and eye piece is 22cm calculate focal length of objective lens

Answer 1

Answer:

Explanation:plz see the attachment

Answer 2

The focal length of objective lens is 18.33 cm.

Explanation:

The magnification for the astronomical telescope is given by

$m=\frac{f_{o} }{f_{e} }$

Here, $f_{o}$ is the focal length of the objective lens and $f_{e}$ is the focal length of the eye piece.

Given m = 5.

Therefore,

      $5=\frac{f_{o} }{f_{e} }$

or    $f_{o} }=5{f_{e} }$                                            (1)

The distance between objective lens and the eye piece is equal to the length of the telescope and it is given as

$L=f_{o}+f_{e}$

Given, L = 22 cm.

So,  

$22cm=f_{o}+f_{e}$

From equation (1), we get

$22cm=f_{o}+\frac{f_{o}}{5}$

$f_{o}=18.33cm$

The focal length of objective lens is 18.33 cm.

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