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?
Given:
Focal length of the objective lens, fo = 144 cm
Focal length of the eyepiece, fe = 6.0 cm
The magnifying power of the telescope is given as,
…(1)
Where,
f0 = focal length of objective lens
fe = focal length of eyepiece
By putting the values in equation (1) we get,
⇒ m = 144/6
⇒ m = 24
The separation between the objective lens and the eyepiece is calculated as:
d = fo + fe
⇒ d = 144 + 6
⇒ d = 150cm
Hence 24 is the magnifying power of the telescope and the separation between the objective lens and the eyepiece is 150cm.
NOTE: Separation of lenses or length of telescope in case of refractive telescopes is always equal to the sum of focal lengths of the two lenses because one lens converges the rays from infinity at its focus and the eyepiece magnifies the image formed( virtual object) at its focus.