A compound microscope has the objective and eyepiece mounted in a tube that is 18.9 cm long. The focal length of the eyepiece is 2.47 cm , and the near-point distance of the person using the microscope is 24.2 cm .
If the person can view the image produced by the microscope with a completely relaxed eye, and the magnification is -4476, what is the distance between the objective lens and the object to be examined?
Given data
Image distance is di = 2.47 cm -(18.9 cm)
=- 16.43 cm = -0.1643 m
Focal length of eyepiece is fe= 2.47 cm = .0247 m
Near point distance N= 0.242 m
Total Magnification is M=-4476
The expression for total magnification is,
Mtotal= -(di)(N)/((fO)(fe))
The Focal length of the objective eyepiece is,
fO= -(- 0.1643)( 0.242 )/(( -4476)( 0.0247))
= 3.596x10^-4 m
=0.359x10^-3 m
=0.359 mm
Therefore, the objective focal length is 0.359 mm.
A compound microscope has the objective and eyepiece mounted in a tube that is 18.9 cm...