Focal Length of Eye Homework: C, F Calcs & Myopia/Hyperopia

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Homework Help Overview

The discussion revolves around the focal length of the human eye, particularly in the context of myopia and hyperopia. The original poster presents a problem involving calculations related to the focal length of an eye with normal vision and how it changes with age.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculations for determining the focal length and question the validity of the original poster's approach. There is a focus on the distinction between lenses and mirrors regarding focal length equations. Some participants express confusion about the simplicity of the focal length value provided.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some guidance has been offered regarding the use of focal length in relation to lens types, but there is no explicit consensus on the calculations or the implications for myopia and hyperopia.

Contextual Notes

Participants are navigating assumptions about the eye's focal length and the implications of changes in focal length over time. There is mention of specific distances related to lens types and visual clarity, which may influence the discussion further.

ahrog
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Homework Statement


The diagram shows the eye of a person with normal vision looking at an object that is very far away.
a) Determine the focal length of this persons eye.
b) Later in this person's life, the focal length will change to between 2.1 cm and 2.3 cm. Will this person suffer from myopia or hyperopia? Explain your answer.
c) Will this person require a convex or concave lens to correct this problem?

Homework Equations


Center of curvature= 1/2 diameter
Focal length= 1/2center of curvature


The Attempt at a Solution


a) C=1/2D
=1/2(2.5cm)
=1.25cm

F=1/2C
=1/2 (1.25)
=0.625

It doesn't make sense though, that focal length seems too small/odd. It says in the question that it changes to between 2.1 and 2.3, so my answer seems wayyyyy off...http://img23.imageshack.us/img23/9751/eyecopyw.jpg
 
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Focal length= 1/2center of curvature
This is true for mirrors, not for lens.
In the above problem f = 2.5 cm.
 
Focal Length of 2.5? Wow, that seems...too simple?
 
When a parallel beam of light falls on a lens , it forms an image at focal point.
 
Thanks. Another question though. How would you find the distance in which an object can be seen clearly from by just using the number 2.5 (and possible 2.1 and 2.3)?
 
Using f as 2.5 cm and di as 2.1 and 2.3, find do. If do is lass than 25 cm,convex lens to be used. And if it is more than 25 cm concave lens to be used. For a healthy eye the object should be at 25 cm to see distinctly.
 

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