A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5 cm. What is the magnifying power of the telescope for viewing distant objects when

(i) the telescope is in normal adjustment (i.e. when the final image is at infinity)?

(ii) the final image is formed at the least distance of distinct vision (25 cm)?

Given focal length of objective lens,$f_{ o }$ = 140cm

and focal length of eye lens, $f_{ e }$ = 5 cm

(i) For normal adjustment, the magnification, m = $f_{ o }$/$f_{ e }$ = -140/5 =-28

(ii) For least distance of distinct vision, the magnification,

m= -$f_{ o }$/$f_{ e }$ (1 + D/$f_{ e }$) = -140/5 (1 + 5/25)

m = -28 (1+0.2) =-33.6