A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.
Focal length of the objective lens,fo=8mm=0.8cm Focal length of the eyepiece, fe=2.5cm
Object distance for the objective lens,μo=−9mm=−0.9cm
Least distance of distant vision d=25cm
Image distance for the eyepiece,ve=−d=−25cm
Object distance for the eyepiece =ue
Using the lens formula, we can obtain the value of ue
1ve−1ue=1fe
1−25−1ue=12.5
So, ue=−2.27cm
We can also obtain the value of the image distance for the objective lens vo using the lens formula.
1vo−1uo=1fo
1vo−1−0.9=10.8
vo=7.2 m
Distance between the objective lens and the eye piece is |ue|+vo
so separation is 2.27+7.2=9.47cm
The magnifying power of the microscope is calculated as
m=vo|uo|(1+d/fe)
⇒7.20.9(1+252.5)=88