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 $$30$$, find out the separation between the objective and the eyepiece.
Correct option is A. 7.5cm
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
$$m=-\dfrac{L}{f_0}(\dfrac{D}{f_e})$$
Given, magnifying power is to be $$m = 30$$, the focal length of objective lens $$f_o= 1.25 cm$$, the focal length of eyepiece $$f_e =5cm$$
We know, $$D$$ is the least distance for distinct vision, i.e., $$D = 25cm$$.
Then, we have, $$30 =\dfrac{L}{1.25}\times(\dfrac{25}{5}) = 5L/1.25 = 4L$$
$$\therefore L = 7.5 cm$$
Thus, the separation between the objective and the eyepiece is $$7.5cm$$.
So, the correct option is (A).