Question

111. The focal lengths of objective lens and eye lens
of a Gallelian Telescope are respectively 30 cm and
3.0 cm. Telescope produces virtual, erect image of
an object situated far away from it at least distance
of distinct vision from the eye lens. In this condition,
the Magnifying Power of the Gallelian Telescope
should be:
1) +11.2
2) +8.8
3) -8.8
4) -11.2​

Answer 1

Answer:

Astronomical Telescope -

$m= \frac{-f_{o}}{f_{e}}\left ( 1+\frac{f_{e}}{D} \right )$

 

- wherein

$f_{o}$ = focal length of objective

$f_{e}$= focal length of eyepiece

 

Given focal length of  

Objective $f_0$= 30 cm

eyepiece $f_e$ = 3 cm

M = $\frac{f_0}{f_e}(1- \frac{f_e}{D})$

= $\frac{30}{3}(1- \frac{3}{25})= 10 \times \frac{22}{25}$ = +8.8

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