The focal length of objective and eye lens of a astronomical telescope are respectively `2m` and `5 cm` . Final image is formed at `(i)` least distance of distinct vision `(ii)` infinity. The magnifying power in both cases will be

Explanation

(i) When the final image is at infinity, then:
`MP = -(f_(0))/(f_(e)) = -(60)/(5) = -12` and length of the telescope is `L = f_(0) + f_(e) = 60 +5 = 65 "cm"`
(ii) For least distance of distinct vision, the magnifying power is :
`MP = -(f_(0))/(f_(e)) ( 1 + (f_(e))/(D)) = -(60)/(5)(1+ (5)/(25)) = - (12xx6)/(5) = -14.4`
Now `(1)/(f_(e)) = (1)/(v_(e)) - (1)/(u_(e)) implies (1)/(5) = -(1)/(25) (1)/(u_(e)) implies (-1)/(u_(e)) = (1)/(25) + (1)/(5) implies u_(e) = -4.17 "cm" implies |u_(e)| = 4.17"cm"`
The length of telescope in this position is `L = f_(0) + |u_(e)| = 60 + 4.17 = 64.17 "cm"`