Question

In an astronomical telescope, focal length of objective lens is `75 cm` and that of eye piece is `5 cm` . Calculate the magnifying power and the distance between the two lenses, when final image of distant object is seen at a distance of `25 cm` from the eye.

Answer 1

Here, `f_(0) = 75cm, f_(e) = 5 cm, m = ?`,
`L = ? D = 25 cm`
`m = -(f_(0))/(f_(e))(1 + (f_(e))/(d))`
`= -(75)/(5)(1 + (5)/(25)) = - 15 xx (6)/(5) = -18`
From `(1)/(v_(e)) - (1)/(u_(e)) = (1)/(f_(e))`,
`-(1)/(u_(e)) = (1)/(f_(e)) - (1)/(v_(e)) = (1)/(5) - (1)/(-25) = (6)/(25)`
`u_(e) = (-25)/(6) = - 4.17 cm`
Distance between the two lenses, `L = f_(0) + |u_(e)|`
`= 75 + 4.17 = 79.17 cm`