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

Given, focal length of objective, `f_(o)=140cm`
`f_(e)=5cm`
(a) For normal adjustment of telescope,
Magnifying Power, `M=(f_(o))/(f_(e))=(140)/(5)-28`
(b) When image is at the least distance of distinct vision
`M=(f_(o))/(f_(e))(1+(f_(e))/(25))`
`=28(1+(5)/(25))`
`=28xx1.2=33.8`