Question:
A person with a normal near point (25cm) 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 9.0mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope,
Answer:
Focal length of the objective lense (f0bj) = 8 mm = 0.8 cm
Focal length of the eyepiece (feye) = 2.5 cm
Object distance for the objective lens (Uobj) = -.9 mm = -0.9 cm
Distance of vision (d) = 25 cm
Image distance for the eyepiece (veye) = -d = -25 cm
Object distance for the eyepiece (ueye) = ?
Using lens formula
1/ veye - 1/ ueye = 1/feye
1/ ueye =1/ veye -1/feye
1/ ueye = 1/-25 - 1/2,5 = -1-10/25 = -11/25
ueye = -25/11 = -2.27 cm
Value of the image distance for objective lens (f0bj) can be obtained by using lens formula
1/ vobj - 1/uobj = 1/fobj
1/ vobj =1/fobj + 1/ uobj = 1/0.8 - 1/0.9 = 0.9-0.8/0.72 = 0.1/0.72
vobj = 7.2 cm
So, sum of the distance for objective lens and the eyepiece = ueye + vobj = 2.27 + 7.2 = 9.47 cm
The magnifying power of the microscope is = vobj (1+ d/ fobj)/uobj
= 7.2(1 + 25/2.5)/0.9 = 1(1+10) = 88
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