Elementary Corrective Lens Problem

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SUMMARY

The discussion centers on a physics problem involving a farsighted woman using eyeglasses with a refractive power of 1.655 diopters. The woman must hold a newspaper 39.4 cm from her eyes, and the distance from her eyes to the eyeglasses is 1.90 cm. The correct calculation for her near point is determined to be 1.15 meters, derived from the formula 1/f = 1/(do) + 1/(di), where do is the object distance measured from the lens. The confusion arises from misinterpreting the object distance as the distance from the eyes instead of from the lens.

PREREQUISITES
  • Understanding of optical physics concepts, specifically lens equations.
  • Familiarity with the concept of refractive power in diopters.
  • Knowledge of virtual images and their characteristics.
  • Ability to perform unit conversions, particularly between centimeters and meters.
NEXT STEPS
  • Study the lens maker's equation to understand how lens curvature affects refractive power.
  • Learn about the significance of object distance (do) and image distance (di) in optics.
  • Explore the concept of virtual images in more detail, including their formation and properties.
  • Practice solving similar problems involving corrective lenses and near point calculations.
USEFUL FOR

Students studying optics, particularly those focusing on corrective lenses and vision problems, as well as educators teaching physics concepts related to lenses and refraction.

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Homework Statement



A farsighted woman breaks her current eyeglasses and is using an old pair whose refractive power is 1.655 diopters. Since these eyeglasses do not completely correct her vision, she must hold a newspaper 39.4 cm from her eyes in order to read it. She wears the eyeglasses 1.90 cm from her eyes. How far is her near point from her eyes?



Homework Equations



1/f = 1/(do) + 1/(di)

di < 0 for virtual images (what this is)

The Attempt at a Solution



do is .394m and 1/f=1.655 diopters

so 1/(1/.394m - 1.655) = di

di = 1.13m

Near Point = 1.13m + .019m = 1.15m

Where did I go wrong? Thanks!
 
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Hi EsimatedEyes,

EstimatedEyes said:

Homework Statement



A farsighted woman breaks her current eyeglasses and is using an old pair whose refractive power is 1.655 diopters. Since these eyeglasses do not completely correct her vision, she must hold a newspaper 39.4 cm from her eyes in order to read it. She wears the eyeglasses 1.90 cm from her eyes. How far is her near point from her eyes?



Homework Equations



1/f = 1/(do) + 1/(di)

di < 0 for virtual images (what this is)

The Attempt at a Solution



do is .394m

What is the variable do? It is called the object distance, but what is it? In particular, what two points is it measured between? Once you answer that, I think you'll see what do needs to equal here.
 
I just figured out that the distance was from the lens so I kept getting it wrong because I was using the eyes value that they gave.
 

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