Question

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

Answer 1

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.

Answer 2

if the object is very close to the principal focus of the objective lens and the image formed by the objective lens is very close to the eyepiece and final image in a microscope is formed at infinity, then magnifying power of microscope is given by
                   $\bold{m=-\frac{L}{f_0}(\frac{D}{f_e})}$
Here, m  is magnifying power e.g., m = 30
           fo is the  focal length of objective lens e.g., fo = 1.25 cm
           fe is the focal length of eyepiece e.g., fe = -5cm
           D is the least distance of distrinct vission, e.g.,  D = 25cm
   so, 30 = -L/1.25(25/-5) = 5L/1.25 = 4L
         L = 7.5 cm

hence, the seperation between objectives and eyepiece is 7.5cm