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`.