In astromical telescope ,`f_(0)=2m`" and "`f_(e)=5cm`" .Initially it is set for normal adjustment and then for near vision.The corresponding angular magnifications (given "D=25cm

An astronomical telescope has an eyepiece of focal-length 5cm. If the angular magnification in normal adjustment is 10, when final image is at least distance of distinct vision (25cm) from eye piece, then angular magnification will be :

An astronomical telescope consists of two convex lenses of focal lengths 20 cm and 4 cm. For the nomal adjustment of the telescope, what is the angular magnification produced by the telescope ?

The focal lengths of the objective and of the eye-piece of a compound microsope are f_(0) and f_(e) respectively. If L is the tube length and D ,the least distance of distinct vision, then its angular magnification, when the image is formed at infinity, is

A 20 D lens is used as a magnifier. Where should the object be placed to obtain maximum angular magnification? (Given, D=25 cm).

An astronomical telescope has f_(o)= 200 cm and f_(epsilon) = 4cm . Telescope is focused to see an object 10 km from the objective. Final image is formed at infinity. Find length L of the tube and the angular magnification of telescope.

Magnification produced by astronominal telescope for normal adjustment is 10 and length of telescope is 1.1m. The magnification when the image is formed at least distance of distinct vision (D = 25cm) is-

The focal lengths of objective and eye piece of a microscope are 1.25 cm and 5 cm respectively. Find the position of the object relative to the objective in order to obtain an angular magnification of 30 in normal adjustment.

The magnifying power of terrestrial telescope is 25 when it is in normal adjustment and the length of the telescope is 124 cm. If the focal length of the erecting lens is 5 cm, the focal lengths of the objective and the eye-piece are respectively.