Correcting Farsightedness · Physics Playground

Strategy

When an object is held 25.0 cm from the person’s eyes, the spectacle lens must produce an image 1.00 m away (the near point). An image 1.00 m from the eye will be 98.5 cm to the left of the spectacle lens because the spectacle lens is 1.50 cm from the eye (see Figure 26.7). Therefore, $d_{i} = - 98.5 cm$ . The image distance is negative, because it is on the same side of the spectacle as the object. The object is 23.5 cm to the left of the spectacle, so that $d_{o} = 23.5 cm$ .

Solution

Since $d_{i}$ and $d_{o}$ are known, the power of the spectacle lens can be found using $P = \frac{1}{d_{o}} + \frac{1}{d_{i}}$ :

\begin{array}{l} P & = & \frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{1}{0.235 m} + \frac{1}{- 0.985 m} \\ = & 4.26 D - 1.02 D = 3.24 D . \end{array}

26.11

Discussion

The positive power indicates a converging (convex) lens, as expected. The convex spectacle produces a case 2 image farther from the eye, where the person can see it. If you examine eyeglasses of farsighted people, you will find the lenses to be thickest in the center. In addition, a prescription of eyeglasses for farsighted people has a prescribed power that is positive.