- A hyperopic eye has a near point of 75 cm. If a lens of power .75 d is placed in front of the eye, what is the near point of the eye-plus spectacle?
- P= 1/p + 1/q
I found the power of the eye itself by doing P= 1/.25 m + 1/ -.75 m. I used negative .75 m because the image would be on the same side of the object. I got a power of 2.67 d. The I added that to the .75 d of the lens. Then I used the same equation and did 3.42= 1/.25 m +1/q to find q, and got -1.72 m. However, the answer is 48 cm.