What Focal Length Lens Corrects a Near Point from 80cm to 30cm?

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SUMMARY

The discussion centers on calculating the focal length of a lens required to correct a person's near point from 80 cm to 30 cm. The correct formula used is 1/(80 cm) - 1/(30 cm) = 1/f, which yields a focal length of 48 cm. This calculation is essential for understanding how corrective lenses function in optics, specifically for individuals with hyperopia. The user successfully applied the formula to arrive at the correct focal length, demonstrating a practical application of lens equations.

PREREQUISITES
  • Understanding of basic optics principles, specifically lens equations.
  • Familiarity with the concept of near point and its significance in vision correction.
  • Knowledge of focal length and its relationship to lens power.
  • Ability to manipulate equations involving fractions and reciprocals.
NEXT STEPS
  • Study the derivation of the lens formula: 1/f = 1/do + 1/di.
  • Learn about different types of corrective lenses and their applications.
  • Explore the concept of lens power and its measurement in diopters.
  • Investigate the effects of varying focal lengths on vision correction.
USEFUL FOR

This discussion is beneficial for students studying optics, optometrists, and anyone interested in understanding the principles of corrective lenses and vision science.

klilly
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Homework Statement



A person had a near point of 80 cm. What focal length lens would be required to correct this persons near point to 30cm, allowing then to read at this distance

Homework Equations



Maybe: 1/di + 1/do = 1/f = P
I found another similar question which used 1/(the given near point aka the 80cm) - 1/(the desired near point aka 30) = 1/f
This got me the right answer, but I still don't understand why.

The Attempt at a Solution



The answer is 48cm
 
Last edited:
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. I got this answer by using the equation I found for similar problems, which is 1/(the given near point aka the 80cm) - 1/(the desired near point aka 30) = 1/f
 

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