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

For object, object distance, `u_(o)=-0.9cm`,
Focal length, `f_(o)=0.8cm`
Using than lens formula,
`(1)/(v_(o))-(1)/(u_(o))=(1)/(f_(o))`
`rArr" "(1)/(v_(o))=(1)/(0.8)-(1)/(0.9)=(0.9-0.8)/(0.72)`
`=(0.1)/(0.72)=(10)/(72)`
`rArr" "v_(o)=7.2cm`
`therefore" "` Objective makes real image 7.2 cm from it one other side of the object.
For eyepiece, image distance, `v_(e)=-25cm`
Focal length, `f_(e)=2.5cm`
Again,
`(1)/(v_(e))-(1)/(u_(e))=(1)/(f_(e))`
`-(1)/(u_(e))=(1)/(f_(e))-(1)/(v_(e))`
`=(1)/(2.5)-(1)/(-25)=(11)/(25)`
`rArr" "u_(e)=-(25)/(11)cm =-2.27cm`
`therefore" Eyepiece lies ahead by 2.27 cm from the real image formed by the objective."`
Distance between objective and eyepiece
`=v_(o)+|u_(e)|=7.2+2.27=9.47 cm`
Magnifying power of microscope is
`M=(v_(o))/(|u_(o)|)(1+(D)/(f_(e)))`
`=(7.2)/(0.9)(1+(25)/(2.5))=88`