Question

Derive expression for magnifying power of compound microscope

Answer 1

The magnifying power of a compound microscope is the ratio of the angle subtended at the eye by the image to the angle subtended by the object at the unaided eye, when both are paced at the least distance of distinct vision It is denoted by 'M' and is given as
M = 1 + D/f
Where D is the least distance of the distinct vision and is equal to 'v' .
Since, D is constant so magnifying power depends upon the focal length of the lens. Smaller the focal length, greater will be the magnifying power of the lens.
Let,
f₍e₎ = Focal length of the eye piece
F₍o₎ = Focal length of the objective = u
M₍e₎ = Magnification of eye piece
M₍o₎ = Magnification of objective
v = L = Length of the microscope tube
M = M₍e₎ x M₍o₎ 
Hence,
M = v/u (1+D/f₍e₎)
M = L/F₍o₎ (1+D/f₍e₎)
Which is the required answer.

Answer 2

$\textbf{Answer is in Attachment !!}$