A person with a normal near point `(25 cm)` using a compound microscope with an objective of focal length `8.0 mm` and eye piece of focal length `2.5 cm` can bring an object placed `9.0 cm` from the objective in sharp focus. What is the separation between the two lenses ? Calculate the magnifying power of the microscope ?

Explanation

Focal length of the objective lens, fo=8mm = 0.8cm
Focal length of the eyepiece, fe=2.5 cm
Object distance for the objective lens, u0=-9.0 mm=-0.9cm
Least distance of distant vision, d=25 cm
Image distance for the eyepiece, ve=-d=-25cm
Object distance for the eyepiece = `mu_(c)`
using the lens formula, we can obtain the valueof `mu_(e)` as:
`1/v_(e)-1/u_(c)=1/f_(e)`
`1/mu_(c)=1/v_(c)-1/f_(c)`
`=1/-25 -1/2.5=(-1-10)/25=-11/25`
`therefore mu_(c)= -25/11 = -2.27` cm
We can also obtain the value of the image distance for the objective lens `v_(0)` using the lens formula.
`1/v_(0)-1/u_(0) = 1/f_(0)`
`1/v_(0)=1/(f_(0))+1/u_(0)`
`=1/0.8 -1/0.9 = (0.9-0.8)/(0.72) = 0.1/0.72`
`therefore v_(0)=7.2cm`
=9.47 cm
the magnifying power of the microscope is calculate as:
`v_(0)/|mu_(0)|(1+d/f_(c))`
`=7.2/0.9(1+25/2.5)= 8(1+10) = 88`
Hence,the magnifying power of the microscope is 88.