# What is the near point for the eye

So I am a teaching assistant for an introductory physics class. One of the problems on this weeks workshops is:

Where is the near point (far point) of an eye for which a contact lens with power of +2.75 (-0.83) diopter is prescribed.

So I know that the power is the inverse of the focal length, so I can find f in the thin lens equation. I'm pretty sure the near point is the image of what someone with the lens would see? So would I plug in 20/25 cm in for object distance (which is roughly near point of the eye) and then solve fir image distance?

And then for far point, I figure I probably just need to take object to infinity and then the image will actually be the power (in m)?

• Well, what's the definition of "near/far point" in your text? Apr 3 '14 at 17:29
• Professor doesnt follow a text Apr 3 '14 at 17:59

The near point is defined as the closest distance on which the eye can focus. “Normal” vision is usually considered to be vision with a near point of $$\newcommand{\cm}{\:\mathrm{cm}}25\cm$$. So, say there is a person who has a near point of $$100\cm$$ rather than the normal $$25\cm$$. To correct this vision, his/her prescription should be designed so that the lenses will take an object at $$25\cm$$ and create a virtual image at $$100\cm$$, so the “non-normal” eye can see it.
If we know the power, and the “normal” near point, we can find the near point of the “non-normal” eye by the thin lens equation: $${1 \over f} = {1 \over s} + {1 \over s'}$$ In your situation this equation becomes: $$2.75\:\mathrm m^{-1}={1 \over 0.25\:\mathrm m} + {1 \over s'}$$This means we are taking an object at $$25\cm$$, refracting the light through the $$2.75$$ diopter lens and we are solving for $$s'$$, the virtual image distance to which the $$25\cm$$ object is focused. This is the “non-normal” eye’s near point. Note, $$s'$$ is going to be negative because this is a virtual image.
• This will be the same if I use $\frac 1o {\bf +}\frac 1i$ right Apr 3 '14 at 18:03