A telescope has an objective of focal length `30 cm` and an eye piece of focal length `3.0 cm`. It is focussed on a scale distant `2.0 m`. For seeing with relaxed aye, calculate the separation between the objective and eye piece.

Answer

`37.97 cm`

Explanation

Here, `f_(0) = 30 cm, f_(e) = 3 cm, u_(0) = - 200 cm`.
Separation between the lenses `= (v_(0) + |u_(0)|)`
From `(1)/(v_(0)) - (1)/(u_(0)) = (1)/(f_(0))`
`(1)/(v_(0)) = (1)/(f_(0)) + (1)/(u_(0)) = (1)/(30) - (1)/(200) = (20 - 3)/(600) = (17)/(600)`
`v_(0) = (600)/(17)cm = 35.29 cm`
From `(1)/(v_(e)) - (1)/(u_(e)) = (1)/(f_(e))`
`(1)/(u_(e)) = (1)/(v_(e)) - (1)/(f_(e)) = (1)/(-25) - (1)/(3) = (-28)/(75)`
`u_(e) = (-75)/(28) = - 2.68 cm`
`:.` Separation between the lenses
`= 35.29 + 2.68 = 37.97 cm`