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Compound microscope

  1. Apr 9, 2007 #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 2.08 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 -4350, what is the focal length of the objective? (Include the sign.)

    I tried using the equations

    d i =d- fo

    M = (d i * N)/(fo*fe)

    since I could not find d i b/c I didn't know fo, I substituted d-fo for d i in the second equation. When I isolated fo, I got

    fo=(dN)/(Mfe +N)

    I thought that I was doing this right...what did I mess up?
     
  2. jcsd
  3. Apr 9, 2007 #2
    did I do an algebraic mistake or...I assumed that the 18 cm = d....was that a wrong assumption to make?
     
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