Thin Lens Equation: Nearsighted

[blue_lilly]

Homework Statement

For those fortunate souls who do not need glasses, the lens of the eye adjusts its focal length in order to form a proper image on the retina. This typically means that very distant objects as well as objects as close as 25 cm can be seen clearly.
Many of us need corrective lenses since the lens in our eye cannot adjust sufficiently to produce a clear image over the full range object distances. This may be because the lens itself does not adjust well or because the eye is either longer or shorter than ‘normal’.
In the case of someone who is nearsighted (can see up close) the eye may only be able to see clearly items up to 50cm or 1m away (this would be the far point). In order to see something further away, a lens (either glasses or contacts) is used to produce a virtual image of a distant object at the person’s far point. Their eye can then accommodate the rest of the way and produce a clear image.

1) Suppose a person who has a far point of 71.5 cm is trying to view a distant object. What is the focal length (with correct sign) of a lens that would take a distant object and make an image on the same side of the lens as the object a distance 71.5 cm from the lens?​

2) Lenses are prescribed in terms of their refractive power, which is expressed in terms of diopters (see the text or your favorite search engine for the definition of a diopter). What is the refractive power of this lens in terms of diopters? (do not enter units.)​

Homework Equations

(1/o_d)+(1/i_d)=(1/f)

Power=1/f

The Attempt at a Solution

I know that the object distance is 71.5cm. I think the image distance is -50cm because the object is on the same side as the object which would make it a virtual image and virtual images have a neg number.
(1/o_d)+(1/i_d)=(1/f)
(1/71.5cm)+(-1/50)=(1/f)
-.006013986=(1/f)
(1/f)=-.006013986
f=(1/-.006013986)
f=-166.2790
So the focal length I got was -166.2790cm. However the answer is wrong and I'm not sure what I am doing that is incorrect.

Any help would be greatly appreciated!

 

[ehild]

The Attempt at a Solution

I know that the object distance is 71.5cm. I think the image distance is -50cm because the object is on the same side as the object which would make it a virtual image and virtual images have a neg number.
(1/o_d)+(1/i_d)=(1/f)
(1/71.5cm)+(-1/50)=(1/f)
-.006013986=(1/f)
(1/f)=-.006013986
f=(1/-.006013986)
f=-166.2790
So the focal length I got was -166.2790cm. However the answer is wrong and I'm not sure what I am doing that is incorrect.

Any help would be greatly appreciated!

No, the image distance is not -50 cm. 50-100 cm are the far points of near-sighted persons.

In the problem, the far point is at 71.5 cm from the eye. The image falls on the retina if the object is at 71.5 cm distance. So the image distance is the distance of the retina from the lens.
The near-sighted person needs glasses to bring in infinity to his/her far point which is 71.5 cm in the problem.

ehild