devarshi-dt-logo

Question:

The focal lengths of objective lens and eye lens of a Galilean Telescope are respectively 30 cm and 3.0 cm. The telescope produces a virtual, erect image of an object situated far away from it at the least distance of distinct vision from the eye lens. In this condition, what is the magnifying power of the Galilean Telescope?

-11.2

8.8

11.2

-8.8

Solution:

Given:
fo = 30 cm
fe = 3.0 cm
uo = ∞
ve = D = least distance of distinct vision (assumed to be 25 cm)
Magnification M = fo/fe (1 - fe/D)
M = 30/3 (1 - 3/25)
M = 10 (1 - 0.12)
M = 10 (0.88)
M = 8.8