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Question -

Aperson with a normal near point (25 cm) using a compound microscope withobjective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm canbring an object placed at 9.0 mm from the objective in sharp focus. What is theseparation between the two lenses? Calculate the magnifying power of themicroscope,



Answer -

Focal length of the objective lens, fo = 8 mm = 0.8 cm

Focal length of the eyepiece, fe = 2.5 cm

Object distance for the objectivelens, uo = −9.0 mm = −0.9 cm

Least distance of distant vision, =25 cm

Image distance for the eyepiece, ve = −d =−25 cm

Objectdistance for the eyepiece = 
Using the lens formula, we can obtain the valueofas:
We can also obtain the value of the imagedistance for the objective lens using the lens formula.
The distance between the objective lens and theeyepiece

The magnifying power ofthe microscope is calculated as:

Hence,the magnifying power of the microscope is 88.

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