What is the closest object that can be seen clearly with corrective glasses?

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SUMMARY

The discussion centers on the calculation of the closest object that can be seen clearly with corrective glasses for two farsighted individuals, Bill and Anne. Bill has a near point of 116 cm and Anne has a near point of 74 cm, both corrected to a normal near point of 25 cm with glasses worn 2 cm from their eyes. The power of their glasses was calculated using the formula P = 1/do + 1/di, resulting in 3.47 diopters for Bill and 2.96 diopters for Anne. The initial confusion arose from not properly accounting for the distance between the glasses and the eyes in the calculations.

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Bill is farsighted and has a near point located 116 cm from his eyes. Anne is also farsighted, but her near point is 74.0 cm from her eyes. Both have glasses that correct their vision to a normal near point (25.0 cm from the eyes), and both wear glasses 2.0 cm from the eyes. Relative to the eyes, what is the closet object that can be seen clearly in the following situations.


1/f = 1/do + 1/di
P=1/f


I determined the powers (P) of the people's glasses using P= 1/do + 1/di where do was 0.25-0.02 and di was near point(in meters)+0.02. For Bill: P = 1/(0.23) + 1/-1.16+0.02 = 3.47 diopters. For Anne: P = 1/0.23 + 1/-0.72+0.02 = 2.96 diopters. I then switched the powers around, used the same di as before, and solved for do. I got 48cm for both of them, but this is wrong. Can anybody tell me what I am doing wrong? I am definitely a little confused about when you need to take the distance between the glasses and eye into account and whether you add or subtract it. Thanks!
 
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Nevermind. I figured it out. In addition to just doing the math wrong and forgetting negatives, I wasn't taking the distance between the eye and glasses into consideration for my final answer. Oops.
 

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