A telescope objective lens has a focal length of `100 cm` . When the final image is formed at the least distance of distinct vision, the distance between the lenses is `105 cm`. Calculate the focal length of eye piece and magnifying power of telescope.

Answer

`6.25 cm ; 20`

Explanation

Here, `f_(0) = 100 cm , f_(e) = ?, m = ?`
When the final image is formed at the least distance of distinct vision, distance between objective and eye lens `= f_(0) + |u_(e)| = 105 cm`
`:. u_(e) 105 - f_(0) = 105 - 100 = 5 cm`
For the eye piece, `u_(e) = - 5cm`,
`v_(e) = - d = - 25 cm`
`:. (1)/(f_(e)) = (1)/(v_(e)) - (1)/(u_(e)) = (-1)/(25) + (1)/(5) = (4)/(25)`
`f_(e) = (25)/(4) = 6.25 cm`
`m = (f_(0))/(f_(e))(1 + (f_(e))/(d)) = (100)/(6.25)(1 + (6.25)/(25)) = 20`