An object is placed at a distance of 5 cm from the objective of a compound microscope . If the final image is formed at the least distance of distinct vision and coincides with the object then calculate the focal lengths of the objective and the eyepiece . given that the least distance of distinct vision =25 cm and the magnifying power of the instrument = 15 cm.

Explanation

For the objecttive ,
image distance =v,object distance =u
For the eyepiece , image distance `=v_1`, object distance =`u_1`
`therefore` According to the problem,
u=5 cm and `v_1=25 cm `
Since the object and the final image coincide, so the distance between the objective and the eyepiece, L =25 -5 =20 cm .
`therefore v+u_1=20 " "...(1)`
Again , total magnification of the instrument = magnification by the objective `xx` magnification by the eyepiece.
` therefore 15 =v/uxx(v_1)/(u_1) or, 15 =v/5 xx(25)/(u_1) or, v/(u_1)=3 " " .....(2)`
Solving equations (1) and (2) we have , v=15 cm and `u_1`= 5cm
For the objective ,
u=-5 cm and v=+15 cm [`because` image is real]
According to the equation of lens we have,
`1/(15) -1/(-5) =1/(f_e) or, f_o =3.75 cm`
For the eyepiece,
`u_1=-5 cm and v_1 =-25 cm `[as image is virtual ]
`1/(-25)-1/(-5)=1/f_e or, f_e =6.25 cm `
So, focal lengths of the objective and eyepiece are 3.75 cm and 6.25 cm respectively.