Necessary power for reading glasses

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SUMMARY

The discussion focuses on calculating the power of reading glasses needed for a person who struggles to read at a distance of 45 cm. The individual requires glasses to read comfortably at the normal near point of 25 cm, with the glasses positioned 2.0 cm from the eye. The calculations involve using the lens formula, where the object distance (do) is 23 cm and the image distance (di) is 43 cm, resulting in a focal length (f) of 14.98 cm and a power of 6.673 diopters (D). The original poster initially miscalculated but later confirmed the correct approach.

PREREQUISITES
  • Understanding of lens formulas, specifically 1/f = 1/do + 1/di
  • Knowledge of the concept of focal length and its relation to lens power
  • Familiarity with the metric system, particularly converting centimeters to meters
  • Basic principles of optics related to reading glasses
NEXT STEPS
  • Study the lens maker's equation for more complex lens designs
  • Explore the effects of different lens materials on optical power
  • Learn about the impact of distance from the lens on image clarity
  • Research common optical prescriptions and their applications in vision correction
USEFUL FOR

Optometry students, optical technicians, and anyone involved in the design or prescription of corrective eyewear will benefit from this discussion.

grouper
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Homework Statement



A person struggles to read by holding a book at arm's length, a distance of 45 cm away (near point). What power of reading glasses should be prescribed for him, assuming they will be placed 2.0 cm from the eye and he wants to read at the normal near point of 25 cm?

Homework Equations



P=1/f (where f is in meters)

1/f=1/do+1/di where do=distance to object and di=distance to image

The Attempt at a Solution



This seemed simple so I must be missing a detail. We want the object 25 cm away from the eye, which is 23 cm away from the lens because the lens is 2 cm in front of the lens, so do=23 cm. We want the image to be at the near point, which is 45 cm from the eye (43 cm from the lens), so di=43 cm. Using the lens equation above, that makes f=14.98 cm and using the equation for the power of the lens (with f in meters), the power is 6.673 D. This is incorrect, though, and I can't find the error in my process.
 
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