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)?

  • 2
    $\begingroup$ Well, what's the definition of "near/far point" in your text? $\endgroup$ – Carl Witthoft Apr 3 '14 at 17:29
  • $\begingroup$ Professor doesnt follow a text $\endgroup$ – yankeefan11 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.

  • $\begingroup$ This will be the same if I use $\frac 1o {\bf +}\frac 1i$ right $\endgroup$ – yankeefan11 Apr 3 '14 at 18:03
  • $\begingroup$ I'm not familiar with that notation. What are 'o' and 'i' in that expression? $\endgroup$ – wgrenard Apr 3 '14 at 18:07
  • $\begingroup$ Nevermind I just looked it up. Yes, in my equation above s=object distance, and o equals the image distance. So essentially s=o and s'=i. $\endgroup$ – wgrenard Apr 3 '14 at 18:10
  • $\begingroup$ @yankeefan11 please check my edit. I apologize, I had a typo. My sign was backward in the thin lens equation. $\endgroup$ – wgrenard Apr 3 '14 at 23:07

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