A compound microscope has lenses of focal length `10 mm and 30 mm`. An object placed at `1.2 cm` from the first lens is seen through the second lens at `0.25 m` from the eye lens. Calculate
(i) magnifying power
(ii) distance between the two lenses.

Answer

`-46.7 ; 8.7 cm`

Explanation

Here, `f_0 = 10 mm = 1 cm, f_e = 30 mm = 3 cm`
`u_0 = -1.2 cm, v_e = -0.25 m = -25 cm`
`M = ?`
From `(1)/(v_0)-(1)/(u_0)=(1)/(f_0)`
`(1)/(v_0)=(1)/(f_0)+(1)/(u_0)=(1)/(1)-(1)/(1.2)=(0.2)/(1.2)=(1)/(6)`
`v_0 = 6 cm`
From `(1)/(v_e)-(1)/(u_e)=(1)/(f_e)`
`(1)/(u_e)=(1)/(v_e)-(1)/(f_e)=(1)/(-25)-(1)/(3)=(-28)/(75)`
`u_e = (-75)/(28) = -2.7 cm`
`M = (v_0)/(-u_0) (1 + (d)/(f_e)) = (6)/(-1.2) (1+(25)/(3)) = -46.7`
`L = v_0 + |u_e| = 6 + 2.7 = 8.7 cm`.