Microscope objective and eyepiece focal lengths

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SUMMARY

The discussion focuses on calculating the focal lengths of a microscope's eyepiece and objective lens, specifically a 13.0x eyepiece and a 57.0x objective lens positioned 20.0 cm apart. The relevant equations include the magnification formula M = m(objective) * m(eyepiece) and the relationship between focal lengths, length = f(objective) + f(eyepiece). The user expresses difficulty in applying these formulas due to the presence of unknown variables and seeks clarification on concepts such as near points and relaxed vision.

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Homework Statement



A microscope has a 13.0 x eyepiece and a 57.0 x objective lens 20.0 cm apart. Calculate the focal length of each lens. Where the object must be for a normal relaxed eye to see it in focus?

Homework Equations



M=m(objective)*m(eyepiece)=f(objective)/f(eyepiece)

length=f(objective)+f(eyepiece)

with a relaxed eye m(eyepiece)=N/f(eyepiece) where N=near point (25 cm for normal naked human vision)

M≈(N*l)/[f(eyepiece)*f(objective)]

m(objective)=[l-f(eyepiece)]/do where do=distance of the object from the objective and l=length (distance between lenses)

The Attempt at a Solution



We had a similar problem where you could use f(objective)=length-f(eyepiece) to get M(tot)=[length-f(eyepiece)]/f(eyepiece) to determine the focal length of the eyepiece, but that does not work on this problem for some reason. All of the other equations I know of involving either focal length contain unknown variables and I can't find a way to get rid of those unknowns.

Also, I'm a little lost conceptually on the ideas of near points and relaxed vision. The problems we have had so far dealing with those concepts I have gotten lucky on but it would be nice to visualize and understand what I'm plugging in and why when I'm dealing with a "relaxed eye" situation.

Thanks for the help.
 
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I will have a go at this soon.
A 'relaxed eye' means that the image being viewed is at infinity...most comfortable for the eye.
Near point means what you have already stated, the image is viewed at 25cm for the average eye, this means greater magnification but there is strain on the eye muscles.
 

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