What Lens Power Does a 77-Year-Old Need to Correct Near-Sightedness?

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Homework Help Overview

The discussion revolves around determining the lens power needed for a 77-year-old man with a near point of 115 cm to see objects clearly at the normal near point of 25 cm. The context involves optics and the use of contact lenses to correct near-sightedness.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the lens formula and the need to convert distances to meters for calculations. There are questions about the definitions and values used, particularly regarding the near point and its implications for lens power.

Discussion Status

Participants are actively engaging with the problem, questioning the definitions and values involved. Some have provided clarifications about the near point, while others are exploring the implications of the distances on the lens calculations. There is a focus on ensuring the correct interpretation of the problem setup.

Contextual Notes

There are uncertainties regarding the values of the near point and the distances involved, with some participants suggesting alternative interpretations. The discussion reflects a mix of assumptions about the human eye's near point and the specific requirements for lens correction.

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A 77 year old man has a near point of 115 cm. What power lens (in diopters) does the man need in order to be able to see an object clearly at the normal near point of 25 cm? Assume that the man will get contact lenses, that is neglect the two centimeters that glasses would be away from the eye.


ok so the goal here is to move the objects image from 115 cm to 25 cm.

i am going to use the equation 1/s+1/s'=1/f and 1/f=P

so this looks really simple, but i am not getting the right answer. for 1/f or P i am getting 4.87 d. and i converted to meters before i did the calculation. please help me.
 
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I'm not familiar with the term near point (sorry), but are the numbers really 25cm and 115cm? Those seem around 10x too big. If the near point is the focul distance of the human eye, that's more like a 2.5cm number...
 
i tired 2.5 cm and 11.5 cm and that didn't work. any other suggestions
 
What's the definition of near point?
 
near point is the nearest point of an object that a relaxed eye can see. usually for a normal eye it is about 25 cm.
 
The average near point for a human is indeed 25cm.

You want the man to be able to read something 25cm away from his face like a normal person right? That way he won't have to hold a newspaper 115cm away from his face to read it!

We know the closest he can see is 115cm, and we want him to be able to see an object at 25cm. Now, the object is at 25cm, so its obvious what the object distance is. What would the image distance be if we want the image to appear at his near point?
 
The object ditance will be 25 centimeters. That is he wants to hold the object at a distance of 25 centimeters from his eyes when he looks at it. The image needs to form at a distance of 115 centimeters from his eye. This means that the image distance will be - 115 centimeters (the contact lens forms a virtual image at this position which enables the senior citizen to see it with his unaided eyes). The function of the contact lens is therefore to shift the object from 25 to 115 centimeters for his eyes to be able see it.

You just need to convert these distances to meters before inserting them into the equation in order to get the answer in diopters.
 
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