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Compound Microscopes

  1. Apr 8, 2007 #1
    1. A compound microscope has the objective and eyepiece mounted in a tube that is 18.0 cm long. The focal length of the eyepiece is 2.24 cm, and the near-point distance of the person using the microscope is 25.0 cm. If the person can view the image produced by the microscope with a completely relaxed eye, and the magnification is -4150, what is the focal length of the objective? (Include the sign.)



    2. M= (-di / f of objective) (N / f eyepiece)



    3. di = 18 - f objective - f eyepiece = 18-2.24- f obj = 15.76- fobjective
    4150 = - ( (15.76- fobjective) / f objective) (25.0/2.24)
    solved for f objective, and got -0.424 mm, but that's wrong.
     
  2. jcsd
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