A small telescope has an objective lens of focal length `144 cm` and an eye-piece of focal length `6.0 cm`. What is the magnifying power of the telescope ? What is the separation between the objective and the eye-piece ?

Explanation

Given, focal length of objective, `f_(o)=144cm`
Focal length of eyepiece, `f_(e)=6cm`
For normal adjustment,
Magnifying power of telescope for normal adjustment,
`m=(f_(o))/(f_(e))=(144)/(6)=24`
(b) The length of the telescope in normal adjustment
`L=f_(o)+f_(e)=144+6=150cm`.