Question

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 1

The focal length of the objective lens,fo=144cm
Focal length of the eyepiece,fe=6.0cm
The magnifying power of the telescope is given as m=$\frac{f_o}{f_e}$=$\frac{144}{6}$=24
The separation between the objective lens and the eyepiece is calculated as:ƒº+ƒe=144+6=150cm
Hence, the magnifying power of the telescope is 24 and the separation between the objective lens and the eyepiece is 150cm.