Question

The total magnification produced by a compound microscope is 20. The magnification produced by the eye piece is 5. The microscope is focussed on a certain object. The distance between the objective and eyepiece is observed to be 14 cm. If least distance of distinct vision is 20 cm, calculate the focal length of the objective and the eye piece..

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