Why Do Microscopes Use Magnification Labels and Vision Correction Uses Diopters?

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SUMMARY

This discussion centers on the utility of marking microscope objective lenses by magnification rather than focal length, emphasizing that knowing magnification allows for easier calculations of object size. Additionally, refractive power, measured in diopters, is highlighted as a more convenient metric for vision correction compared to focal length. The total refractive power of the eye is established as 40 D at far point and 43.5 D at near point, with the formula for total refractive power being the sum of the eye's power and that of corrective lenses.

PREREQUISITES
  • Understanding of the thin lens equation: (1/f) = (1/s) + (1/s')
  • Familiarity with magnification calculations: m = (-s'/s)
  • Knowledge of refractive power and its calculation: R = (1/f)
  • Basic concepts of optics related to microscopes and corrective lenses
NEXT STEPS
  • Research the application of the thin lens equation in optical systems
  • Explore the relationship between magnification and object size in microscopy
  • Study the principles of refractive power and its significance in vision correction
  • Investigate the design and function of corrective lenses in relation to focal points
USEFUL FOR

Optics students, microscope manufacturers, vision care professionals, and anyone interested in the principles of magnification and vision correction.

amberita
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Homework Statement


- Why is it more useful to mark an objective lens [in a microscope] by its magnification rather than focal length?
- Why might it be more convenient to think in terms of refractive power rather than focal length? [dealing with vision here]

Homework Equations


- thin lens equation: (1/f) = (1/s)+(1/s')
- magnification (m) = (-s'/s)
- Refractive power (R) = (1/f)
- Total R of eye = R(cornea) + R(lens) = 40 D at far point, 43.5 D at near point.
- Total R = R(eye) + R(corrective lens)

The Attempt at a Solution


- For the first question, I said that if you know m, then you could solve for the size of an object under a microscope pretty easily. Are there any other reasons that anyone could think of?
- For the second, I honestly have no idea. To me, it seems just as reasonable to think about correcting vision in terms of the focal point of the eye/lens. I initially thought "well, glasses and contacts are in diopters, so it makes sense to think about correcting vision in this way," but couldn't we think of corrective lenses in terms of the focal point?
 
The second one is pretty clear from the formulas you provided. You just add the powers if the lenses are close to each other!
 

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