Compound microscope is an optical instrument which is used to obtain high magnification.
   It consists of two converging lenses:
    Eye piece
   The lens in front of object is called objective. Its focal length f1= fo is taken to be very small .The    objective forms a real, inverted, and magnified image of the object placed just beyond the focus of    objective.
Eye piece
   The lens towards the observer's eye is called piece .Focal length of eye piece is greater than the focal    length of objective. Eye piece works as a magnifying glass.
   The objective is so adjusted that the object is very closed to its focus. The objective forms a real,
   inverted and magnified image of the abject beyond 2fo on the right hand side. The eye piece is so    adjusted that it forms a virtual image at the least distance of distinct vision "d" .The final image is
   highly magnified.
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Magnifying power
   In order to determine the magnifying power of a compound microscope ,we consider an object oo' placed    in front of objective at a distance p1. Objective forms an inverted image II' at a distance of q1 from    objective.
   Magnification produced by the objective is given by:
Mo= size of image / size of object
Mo= q1/ p1--------------- (1)
   Eye piece works as a magnifying glass. It further magnifies the first image formed by objective.
   Magnification produced by the eye piece is given by:
Me= size of image / size of object
Me= q2/ p2
   We know that the eye piece behaves as a magnifying glass therefore the final image will be formed at    least distance of distinct vision i.e at 25 cm from the eye. Hence q2 = d
Me= d / p2--------------- (2)
   Using thin lens formula for eye piece :
1/f2 = 1/q2 + 1/p2
   Here f2 = fe, q2 = - d and p = p2
1/fe = 1/-d + 1/p2
1/fe = -1/d + 1/p2
Multiplying both sides by "d"
d/fe = -d/d + d/p2
d/fe = -1 + d/p2
1 + d/fe = d/p2   
                                d/p2 = 1 + d/fe----------------(3)

Comparing equation (2) and (3)
               Me = 1 + d/fe--------(4)
   Total magnification is equal to the product of the magnification produced by the objective and the eye    piece.
M =Mo X Me
M = (q1/p1)(1 + d/fe)
   In order to get maximum magnification, we must decrease p1 and increase q1 .Thus maximum possible
   value of p1 is fo i.e p = fo and maximum possible value of q1 is the length of microscope i.e q1 = L
   Therefore the magnification produced by a compound d microscope is given by:
M = (L/fO)(1 + d/fe)
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