The Magnification of a Microscope Given Distances and Focal Lengths

[PhysicsFishy]

Alright, I was hoping that someone could help me out with this... The problem is:

"The distance between the eyepiece and the objective lens in a certain compound microscope is 21.0 cm. The focal length of the eyepiece is 2.60 cm, and that of the objective is 0.400 cm. What is the overall magnification of the microscope?"

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I know that for a telescope that:

M=(-)fo/fe

M=magnification
fo=focal length of objective lens
fe=focal length of eyepiece lens
(-)=negative inserted if image is inverted

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If I apply this equation to the problem I get "-.15" which is incorrect. The distance between the lenses is a variable, so I know that I somehow need to apply that but am unsure how to do it. Is there another equation I can use? Thanks for any help.

 

[jbsteele]

Yes, you can use a slightly different equation to calculate the magnification of a compound microscope. It is: M = (fo x D) / (fe x d)Where M = magnification, D = distance between the lenses, fo = focal length of the objective lens, fe = focal length of the eyepiece lens, and d = distance from the object to the objective lens.In this particular problem, we are given the distance between the lenses (D) and the focal lengths of both the objective and eyepiece lenses (fo and fe). We also know that the distance from the object to the objective lens (d) is typically very small and can be assumed to be zero. Therefore, the equation simplifies to: M = (fo x D) / fePlugging in the given values, we get:M = (0.400 cm x 21.0 cm) / 2.60 cm M = 8.08x