# Nearsightedness, Refractive Power, Optics

1. Oct 28, 2008

### yayirunin2car

1. The problem statement, all variables and given/known data
The relaxed eyes of a patient have a refractive power of 50.1 diopters. (For the purposes of this problem, treat the eye as a single-lens system, with the retina 2.40 cm from the lens.)
(a) Is the patient nearsighted or farsighted? (b) If the patient is nearsighted, find the far point. If this person is farsighted, find the near point.

2. Relevant equations

1/f = 1/d0 + 1/di

3. The attempt at a solution

I know that the person is nearsighted.

I guess where I'm going wrong is by thinking that the lens is a pair of glasses 2.40 cm in front of the eye, which is all the problems I have done so far. I have been using 1/50.1 to get f and converting it to centimeters where f = 1.9960 cm.

What I use to get my answer:
F is the far point.

1/f = 1/infinity + 1/(F-2.40)

The answer I get, 4.3960 cm is wrong. Also, when I make di negative since it should be in front of the eye, I get .4039 which is also wrong.

2. Oct 28, 2008

### Redbelly98

Staff Emeritus
d0 and di refer to the distance of the object or image (respectively to the lens)

1. Since the image is to be focused on the retina, that means di=___?
2. d0 is not infinity for an object located at the far point.
3. And as you said, f=1.996 cm.

3. Oct 29, 2008

### yayirunin2car

If I take di as 2.4, should I make it negative since it is in front of the lense?

4. Oct 29, 2008

### Redbelly98

Staff Emeritus
5. Oct 29, 2008

### Perillux

so is 'di' positive because it is behind the lens?
Or were you just correcting him, and 'di' is negative because it is behind the lens?
thanks.

6. Oct 30, 2008

### Redbelly98

Staff Emeritus
It's like a standard lens problem, with do and di on opposite sides of the lens, so both are positive. The problem is in translating from the description in the problem statement into values for do, di, and f.