You are given two converging lenses of focal length 1.25 cm and 5 cm to design a compound microscope. If it is desired to have a magnification of 30, then find out the separation between the objective and eyepiece.

Given $f_{ o }$ = 1.25 cm $f_{ e }$ = -5 cm

Magnification, m= 30,

D= 25 cm

If the object is very close to the principal focus of the objective and the image formed by the objective is very close to eyepiece, then magnifying power of a microscope is given by

m = -L/$f_{ o }$ . D / $f_{ e }$

====> 30 = L/1.25 . 25/5

===> L = 125x30x5/25x100

====> L = 25 x 30/100

===> L = 30/4

===> L = 7.5 cm

This is a required separation between the objective and the eyepiece.

The answer is wrong it should be 6.25cm as the formula used is only used when the final image is formed at infinity. The formula to be used is ,m=L/fo(1+D/fe) Uploadingâ€¦