A compound microscope has an objective of focal length `2.0 cm` and an eye-piece of focal length `6.25cm` and distance between the objective and eye-piece is `15cm`. If the final image is formed at the least distance vision `(25 cm)`, the distance of the object form the objective is

Explanation

Here, `f_0 = 2.0 cm, f_e = 6.25 cm., u_0 = ?`
(a) `v_e = -25 cm`
As `(1)/(v_e)-(1)/(u_e)=(1)/(f_e) :. (1)/(u_e)=(1)/(v_e)-(1)/(f_e)=(1)/(-25)-(1)/(6.25) = (-1 - 4)/(25) = (-5)/(25)`
`u_e = -5 cm`.
As distance objective and eye piece `= 15 cm :. v_0 = (15 - u_e) = 15 - 5 = 10 cm`
As `(1)/(v_0)-(1)/(u_0)=(1)/(f_0)`
`(1)/(u_0)=(1)/(v_0)-(1)/(f_0)=(1)/(10)-(1)/(2)=(1 - 5)/(10)`
`u_0 = (-10)/(4) = -25 cm`
Magnifying power `=(v_0)/(|u_0|) (1+(d)/(f_e)) = (10)/(2.5)(1+(25)/(6.25)) = 20`
(b) As `v_e = oo , - u_e = f_e = 6.35 cm :. v_0 = 15 - 6.25 = 8.75 cm`
`(1)/(u_0)=(1)/(v_0)-(1)/(f_0)=(1)/(8.75)-(1)/(2.0)=(2 - 8.75)/(17.5)`
`u_0 = (-17.5)/(6.75) = -2.59`
Magnifying power `= (v_0)/(|u_0|) xx (v_e)/(|u_e|) = (8.75)/(2.59) xx (25)/(6.25) = 13.51`.