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Question

A compound microscope consists of an objective of focal length 2 cm and an eye-piece of focal length 5 cm. When an object is kept 2.4 cm from the objective, final image formed is virtual and is at 25 cm from the eye-piece. Determine magnifying power of this compound microscope in this set up, i.e., in normal use.

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Solution

Given
fo=2cm=Focal length of the objective
fe=5cm=Focal length of the eye piece
Me=Magnification of the eye piece
Mo=Magnification of the objective
Me=1+Dfo
Me=1+255
Me=6
For the objective
u=2.4cm,v=?fo=2cm
1v12.4=12
1v=1212.4
1v=12241024
1v=112
v=12cm
Mo=vu=122.4=5
Magnifying power of the compound microscope
M=Mo×Me
M=5×6
M=30

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