If for an astronomical telescope focal length of eyepiece is 50 m and that of objective is 10 m . So, What must be the distance between them( in m) to view an object 250 m away from the objective ?

Answer

`72.50`

Explanation

Given,
`f_(0) = 50m, f_(e) = 10 m`
For Objective,
`1/V_(0) - 1/u_(0) = 1/f_(0)`
`therefore 1/v_(0) - 1/(-250) = 1/50`
`therefore 1/v_(0) = 1/50 - 1/250 = (5-1)/250 = 1/62.5`
`v_(0) = 62.5` m
For normal adjustment, the image formed by objective will be at the focus of the eyepiece.
`therefore L = v_(0) + f_( e) = 72.5` m