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 2.5cm 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 ve1−ue1=fe1 −251−ue1=2.51 So, ue=−2.27cm We can also obtain the value of the image distance for the objective lens vo using the lens formula. vo1−uo1=fo1 vo1−−0.91=0.81 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=∣uo∣vo(1+d/fe) ⇒0.97.2(1+2.525)=88
