The focal length of the objective and eyepiece of a compound microscope are 4 cm and 6 cm respectively. If an object is placed at a distance of 6cm from the objective, what is the magnification produced by the microscope ?

Explanation

Given that `f_(0)=4cm, f_(e)=6cm, u_(0)=6`
`(1)/(f_(0))=(1)/(upsilon_(0))+(1)/(u_(0))`
`(1)/(upsilon_(0))=(1)/(f_(0))-(1)/(u_(0))=(1)/(4)-(1)/(6)`
`(1)/(upsilon_(0))=(2)/(24)`
`upsilon_(0)=12cm.`
Magnifying power `m=(v_(0))/(u_(0))(1+(D)/(f_(e)))`
`=(12)/(6)[1+(25)/(6)]=2[(31)/(6)]=(62)/(6)`
`m = 10.33`