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Glasses to fix near points and far points

  1. Apr 13, 2007 #1
    1. The problem statement, all variables and given/known data

    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?



    2. Relevant equations

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

    3. 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.
     
    Last edited: Apr 13, 2007
  2. jcsd
  3. Apr 13, 2007 #2

    andrevdh

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    Homework Helper

    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).
     
  4. Apr 13, 2007 #3
    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.
     
  5. Apr 16, 2007 #4
    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?
     
  6. Apr 16, 2007 #5

    andrevdh

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    Homework Helper

    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.
     
    Last edited: Apr 16, 2007
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