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

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The discussion focuses on calculating the focal length of the eye for a person with normal vision and how it changes with age. The initial focal length is calculated as 0.625 cm, which is deemed incorrect since it contradicts the later focal lengths of 2.1 cm to 2.3 cm. The person is likely to suffer from hyperopia due to the increased focal length, indicating difficulty focusing on nearby objects. To correct hyperopia, a convex lens is required, especially if the distance for clear vision is less than 25 cm. The conversation highlights the importance of using appropriate formulas for lenses rather than mirrors when determining focal lengths.
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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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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