What Focal Length Eyepiece Provides a -125 Magnification?

AI Thread Summary
To achieve an overall magnification of -125 with a microscope that has a medium-power objective lens focal length of 3.75 mm, the focal length of the eyepiece lens must be calculated using the magnification formula. The discussion highlights the challenge of solving for two unknowns, di and feyepiece, while emphasizing the importance of using the working distance obtained from Part A to find di. Participants confirm that values from previous parts of the problem are essential for solving subsequent parts. The relationship between the parts suggests that understanding the working distance is crucial for determining the eyepiece focal length. This approach ensures a comprehensive solution to the magnification problem.
Anna M
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Homework Statement


The medium-power objective lens in a laboratory microscope has a focal length fobjective = 3.75 mm.

What is the focal length of an eyepiece lens that will provide an overall magnification of -125? Assume student's near-point distance is N=25cm.

Homework Equations


M=(-di/fobjective)(N/feyepiece)

The Attempt at a Solution


I have set up my equation and plugged in all the known values, but I do not understand how to solve when you have 2 unknown values (di and feyepiece).

Part A of this question says: If this lens produces a lateral magnification of -46.0, what is its "working distance"; that is, what is the distance from the object to the objective lens?
I have worked Part A out, but I am not sure if they correlate and I am supposed to use values from that portion or if they are completely separate.
 
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Yes, you need to use the working distance found in part A to solve part B. Otherwise you don't know ##di##.
Generally, if a problem consists of more than 1 part, and you're missing a value for a later part, the missing value can be found using given information or your answers in previous parts. That's been my experience at least.
 
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