The focal lengths of the objective and the eyepiece of a compound microscope are 1 cm and 2 cm respectively and the distance between them is 12 cm . If the least distance of distinct vision of the observer is 25 cm , what should a small object distance be placed to see it ?

Explanation

For the eyepiece ,`f_e= 2 cm , v=-25 cm`. So, from the equation of the lens we have,
`1/(-25)-1/u=1/2 or, 1/u=-(1/(25)+1/2)=-(27)/(50) or, u=-(50)/(27) cm`
So, the image formed by the objective is formed in front of the eyepiece at a distance `(50)/(27) cm ` . Hence, the image is formed behind the objective at a distance `(12-(50)/(27)) or, (274)/(27) cm` i.e ., the image distance of the object relative to the objective, v`=(274)/(27) cm ` .
The focal length of the objective , `f_o=1 cm ` from the equation of the lens we have ,
`1/(274/27)-1/u=1/1 or, (27)/(274)-1/u=1`
or, `-1/u=1-(27)/(274)=(247)/(274) or, u=-(274)/(247) =-1.11 cm `
`therefore` The required object distance =1.11 cm .