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 :

Question

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 : (A) + 11.2 (B) -11.2 (C) -8.8 (D) + 8.8

Tardigrade Admin · Accepted Answer

f0 = 30 cm fe = 3 cm M=(f0/fc)(1-(fC/D)) =(30/3)(1-(3/25)) =(22×30/3×25)=(44/5)=+8.8

+ 11.2

-11.2

-8.8

+ 8.8

f0​=30cmfe​=3cm M=fc​f0​​(1−DfC​​) =330​(1−253​) =3×2522×30​=544​=+8.8

$f_{0} = 30 c m f_{e} = 3 c m$
$M = \frac{f _{0}}{f _{c}} (1 - \frac{f _{C}}{D})$
$= \frac{30}{3} (1 - \frac{3}{25})$
$= \frac{22 \times 30}{3 \times 25} = \frac{44}{5} = + 8.8$