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)?
Best Answer
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.