Compound Microscope: Solving for Objective Focal Length

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To solve for the focal length of the objective in a compound microscope, the user applied the equations for image distance and magnification but encountered difficulties due to unknown variables. The total length of the microscope tube is 18.0 cm, which is crucial for determining the image distance. The focal length of the eyepiece is given as 2.08 cm, and the near-point distance is 25.0 cm. The user attempted to substitute variables to isolate the objective focal length but questioned whether their assumption about the tube length equating to the object distance was correct. Clarifying these assumptions and correctly applying the formulas is essential for accurately calculating the objective's focal length.
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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?
 
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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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