A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. what is the magnifying power of the telescope for viewing distant objects when
(a) the telescope is in normal adjustment (i.e, when the final image is at infinity ),
(b) The final image is formed at the least distance of distinct vision (25 cm )

Explanation

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=f_(0)/f_(s)`
`=144/6=24`
The separation between the objective lens and the eyepiece is calculated as:
`f_(0)+f_(e)`
`=144 + 6=150 cm`
Hence, the magnifying power of the telescope is 24 and the seperation between the objective lens and the eyepiece is 150 cm.