Q. A small telescope has an objective lens of focal length 144cm and an eyepiece of focal length 6.0cm.What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?

Answer

A Focal length of the objective lens f_o = 144cm

A Focal length of the eyepiece f_e = 6.0cm

A magnifying power of the telescope is given as:

M = f₀ /f_e

=144/6 = 24

These parathion among the objective lens and the eyepiece is calculated as

F₀ + f_e

=144 + 6 = 150 cm

Therefore the magnifying powers of the telescope is 24 and the separation between the objective lens and the eye piece is 150cm.