Calculating Lens Power: Q&A for Diopters

[nophun6]

I have a question dealing with diopters.
Q: Assume that the distance from the lens of your eye to the retina is 1 inch (in effect, the image distance). Calculate the power of the lens in your eye (in diopters) for both your far point and near point. The difference between these two lens powers is called the “power of accommodation”, and it decreases with age, as
shown in the table below:
Age (years) 10 15 20 25 30 35 40 45 50 55 60 65
Accommodation (D) 14 12 10 8.5 7 5.5 4.5 3.5 2.5 1.7 1.0 0.5
Does your power of accommodation agree with the table?

Ok, so q= -2.54cm and p= my near point (which is 20 cm)
So I use the thin lens equation: 1/p + 1/q = 1/f and get f= -.029 m
Then to calculate lens power: 1/f = -32.4 Diopters

Did I do this correctly?

 

[dimensionless]

I got 0.023m for the focal length at your near point.

-->1/(1/0.20+1/0.0254)
 ans  =

    0.0225377

You should use a positive value for q.