Hello aspirant,

In a compound microscope if a magnified virtual image is produced at a distance of 25 cm from the eye-piece and the focal length of its objective lens is 1 cm and the magnification is 100 with the tube length of the microscope as 20 cm, the focal length of the eyepiece lens (in cm) will be:-

Given,

L = 20, f0 = 1cm, M = 100

We know,

M= (v0/u0)(1+(D/f))

or, M= (L/f0)(1+(D/f))......[as, v0 ≈L, u0≈f0]

or, 100= (20/1)(1+(25/f))

on solving we get

f = 6.25 cm...........(Ans.)

Hope this helps,

All the Best!!