**1. The problem statement, all variables and given/known data**

**Prescribing Eyeglasses – Hyperopia**

A)Suppose you have hyperopia (you are farsighted). You can clearly focus on very distant objects, but any object which is closer than 56.3 cm from your eyes will appear blurry unless you use glasses. The purpose of glasses would be to make objects that are close to you appear to be at your near point. What would be the refractive power (in diopters – do not enter units) for your prescription if you want to be able to clearly read by holding the paper

B)If your glasses make close objects appear farther away, it will make objects that are farther appear to be even more distant. If the object gets too far away, the image will become real and be located such that you can no longer focus on the image. What is the farthest distance that an object can be and still be able to see it clearly with your reading glasses on?

**2. Relevant equations**

(1/do)+(1/di)=(1/FL) [do=object distance; di=image distance; FL= Focal Length]

Reactive Power=1/FL [when the FL is in meters]

**3. The attempt at a solution**

A) From the problem, I am given near-point=56.3cm, the far-point=infinity. I also know that becuase this is an corrective lens, the image distance(di) will always be negative.

(1/do)+(1/di)=(1/FL)

di=-56.3cm FL=?

(1/-56.3)=(1/FL)

-.017761=(1/FL)

FL= 1/-.017761

FL= -56.3cm

Then I need to take my Focal Length and convert it into meters because reactive power uses meter and not cm.

-56.3 cm = -0.563 m

Reactive Power=1/FL

=1/-0.563

P=-1.77619This answer is incorrect.

B) The point where a virtual image becomes a real image on a converging lens is at the Focal Point. I found the Focal Length in the previous answer

FL= -56.3cmThis answer is incorrect.

Any help would be greatly appreciated!!!

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