Question

the focal length of objective and eyepiece of an astronomical telescope are 20 cm and 5 cm respectively the final image is formed at the distance of 30 cm from the eyepiece magnifying power of the telescope will be​

Answer 1

Given that,

Focal length of eyepiece, $f_e=20\ cm$

Focal length of astronomical telescope, $f_o=5\ cm$

The final image is formed at the distance of 30 cm from the eyepiece

To find,

The magnifying power of the telescope.

Solution,

The formula that is used to find the mgnifying power of the telescope is given by :

$m=-\dfrac{f_o}{f_e}(1+\dfrac{f_e}{D})\\\\m=-\dfrac{20}{5}(1+\dfrac{5}{30})\\\\m=-4.67$

So, the magnifying power of the telescope will be​ -4.67.

Answer 2

Explanation:

Given that,

Focal length of eyepiece, $f_e=20\ cm$

Focal length of astronomical telescope, $f_o=5\ cm$

The final image is formed at the distance of 30 cm from the eyepiece

To find,

The magnifying power of the telescope.

Solution,

The formula that is used to find the mgnifying power of the telescope is given by :

$m=-\dfrac{f_o}{f_e}(1+\dfrac{f_e}{D})\\\\m=-\dfrac{20}{5}(1+\dfrac{5}{30})\\\\m=-4.67$

So, the magnifying power of the telescope will be -4.67.