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Homework Help: Optics - near sighted or far sighted

  1. Mar 29, 2005 #1
    A pair of relaxed eyes of a patient have a refractive power of 48.5 diopters.

    a) is this person near or far sighted?

    I think its far sighted because of the short focal length.

    b) Find the near point of this person. (The retina is 2.4cm from the lens).

    Ok, so the f=.02 and I'm assuming we supposed to use 1/do + 1/di = 1/f.

    I'm just not sure what I'm supposed to find. Which variable represents the length of the retina to the lens? Which one am I finding?
  2. jcsd
  3. Mar 31, 2005 #2

    Andrew Mason

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    Are we to assume that this person has retina placement of 2.4 cm behind the lens? If so this person is nearsighted. A refractive power of 48.5 dioptres means that the eye has a focal length of 2.1 cm (f = 1/P in metres). The images from distant locations will focus at the focal length which is in front of the retina. Since the retina is located 2.4 cm behind the lens, the focal length is too short, which means the person is near sighted. For a normal person, the retina is about 1.7 mm from the lens of the eye, so if we assume a normal retina placement, the focus would be behind the retina, in which case the person would be farsighted.

    The near point is the distance of the closest object that is in focus (with no accommodation by the eye, so the focal length does not change). So:

    [tex]\frac{1}{f} = \frac{1}{o} + \frac{1}{i}[/tex] where f = 2.1 cm and i= 2.4 cm

    So [tex]\frac{1}{o} = 48.5 - 41.7 = 6.8[/tex]

    Object distance = 15 cm.

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