Question

The total magnification produced by a compound microscope is `20`. The magnification produced by the eye piece is `5`. The microscope is focussed on a certain object. The distance between the objective and eye piece is observed to be `14 cm`. If least distance of distinct vision is `20 cm`, calculate the focal length of objective and eye piece.

Answer 1

Correct Answer - `2 cm` ; `5 cm`
Here, `m = - 20, m_(e) = 5`
`:. m_(0) = (m)/(m_(e)) = (-20)/(5) = -4`
As `m_(e) 1 + (d)/(f_(e))`
`5 = 1 + (20)/(f_(e)) :. F_(e) = (20)/(4) = 5 cm`
As `m_(e) = (f_(e))/(u_(e) + f_(e))`
`:. 5 = (5)/(u_(e) + 5) u_(e) = (5 - 25)/(5) = - 4 cm`
Distance between objective and eye piece
`= v_(0) + |u_(e)| = 14`
`v_(0) = 14 - |u_(e)| = 14 - 4 = 10 cm`
magnification produced by objective lens
`m_(0) = (v_(0))/(u_(0))`
`-4 = (10)/(u_(0)) :. u_(0) = -2.5 cm`
From `(1)/(f_(0)) = (1)/(v_(0)) - (1)/(u_(0))`
`= (1)/(10) - (1)/(-2.5) = (1 + 4)/(10) = (5)/(10)`
`f_(0) = (10)/(5) = 2 cm`