Glasses to fix near points and far points

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To determine the power of glasses needed for a person with a near point of 0.6m and a far point of 2m to see infinitely far, calculations show that the glasses should create an image at -1.98m, resulting in a power of approximately -0.505 diopters. For the new near point, the calculations indicate that the image should form at 0.58m from the glasses, leading to a new near point of about 0.82m, although the correct answer is stated as 0.82m. In part b, to bring the near point to 0.25m, the required power is calculated to be approximately -2.62 diopters, but the new far point calculation yields an incorrect result of 0.32m instead of the expected 0.34m. The discussion highlights the importance of accurately determining object distances and the effects of lens placement on vision correction.
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Homework Statement



a)A person has a near point of .6m and far point of 2m . What power of glasses would be needed for the person to see infinitely far away. Assume that glasses are 2cm from the person's eyes. If the person used those glasses, what would his new near point be?

b) What power of glasses would be needed to bring the person's near point to .25m. If the person used those glasses, what would their new far point be?



Homework Equations



1/f = 1/o + 1/i
P=1/f

The Attempt at a Solution



This is what I have so far.

I know that
1/f = 1/o + 1/i

I want an image at infinity so 1/f = 1/o
The object is 2m from the eyes so it is 1.98 m from the glasses. The direction of the light is toward the glasses while the image is in the opposite direction. so o is negative.

1/-1.98 = -.505


And this is the correct answer according to the answer key.

I have problem solving for the new near point though.

1/f = 1/o + 1/i
-.505 = 1/o + 1/i

I want to create a new image for a distance of .6 m from the eyes or .58 from the glasses

-.505 = 1/.58 + 1/-i
-1/1.98 - 1/.58 = -1/i
I get i = .44 meters

but the correct answer is .82 meters. What did i do wrong?

I am having a similar problem for part b so I think I can get it if I find out what I did wrong here in part a.
 
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a) The object is at infinity (what the person wants to look at). The image needs to form at -1.98 meters (his far point, in front of the glasses).
 
Thanks for your help.

I just realized that I need help for part b as well.

b)

1/f = 1/o + 1/i

The person's far point is .6 or .58 from the glasses. I want him to see at .25 m or .23 m from the glasses.

1/-f = 1/.58 + 1/-.23
1/-f = -2.62
1/f = 2.62

that is the power. I am getting the wrong answer when I have to calculate the new near point though.

-2.62 = 1/1.98 + 1/-i

I get i = .32 m but the correct answer is .34m. Is there anything wrong with the 1/f and 1/1.98? I know that i/f =2.62 according to the answer key so I think I either rounded something wrong or have 1/1.98 wrong. However, I don't know what the object distance would be if it is not 1.98.

Thanks in advance for your help.
 
If -2.62 is 1/f and the answer is indeed .34 for as in the answer key then o must be greater than 3 meters away. How is that possible? Do I somehow combine that 1.98 m with the other part a, which is .82 meters?
 
a) For the new near point you have that 1/f = - 0.51 m
The image needs to form at a distance of 0.58 m in front of the lens in order for him to see it. That is i = - 0.58 m. The object distance that you get will then be his new near point. That is it is the new closest point (+ 2 cm) that he can hold something in front of him and still see it in focus with his glasses on.
 
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