In a compound microscope, the objective and eye piece have focal lengths `0.95 cm and 5 cm` respectively, and are kept at a distance of `20 cm`. The final image is formed at a distance of `25 cm` from the eye piece. Calculate the position of the object and the total magnification.

Answer

`(-95)/(94) cm` ; `- 94`

Explanation

Here, `f_(0) = 0.95 cm, f_(e) = 5 cm`.
`v_(0) + u_(e) = 20 cm, v_(e) = - 25 cm`.
Using `(1)/(v_(e)) - (1)/(u_(e)) = (1)/(f_(e))`, calculate `u_(e)`
`:. v_(0) = 20 - u_(e)`
Using `(1)/(v_(0)) - (1)/(u_(0)) = (1)/(f_(0))`, find `u_(0) = -(95)/(94) cm`
Total magnification, `m = (v_(0))/(u_(0))(1 + (d)/(f_(e))) = - 94`