Eyes: A nearsighted person has her distant vision corrected using a -2.0-diopter eyeglasses. Her uncorrected near point is 15 cm. What is her near point using this lens if the lens is 2.0 cm from the eye?

Question

Quizplus · Accepted Answer

- The uncorrected near point is 15 cm, so without any correction, the person can only see objects clearly at a distance of 15 cm or less from the eye.
- The distance between the eye and the lens is 2.0 cm, so the effective power of the lens is -1/2.0 = -0.5 diopters.
- To find the new near point, we can use the formula: 
   1/f = 1/do + 1/di 
   where f is the focal length of the lens (which is equal to the lens power in diopters), do is the distance between the object and the lens, and di is the distance between the lens and the image (the person's eye).
- Solving for di, we get: 
   di = 1/f - 1/do 
         = 1/-2.0 - 1/15 
         = -0.5 + (-0.067) 
         = -0.567 
- The negative sign means that the image is formed on the same side of the lens as the object (i.e. it is a virtual image).
- The absolute value of di is 1.77 cm, so the near point is 2.0 + 1.77 = 3.77 cm.
- However, we need to convert this to the equivalent distance if the person had normal vision, which is 25 cm. This can be done using the formula: 
   1/near point = 1/far point + 1/di 
- Solving for the far point, we get:
   far point = 1/(1/near point - 1/di) 
                   = 1/(1/25 - 1/1.77) 
                   = 1.98 m 
- Therefore, the person's corrected near point is 20 cm (i.e. she can now see objects clearly at a distance of 20 cm or less from the eye).

Eyes: a Nearsighted Person Has Her Distant Vision Corrected Using