Question: You find that you have to hold a book 35 from your eyes for the print to be in sharp focus. What power lens is needed to correct your farsightedness?
1. **State the problem:** You need to find the power of the lens required to correct farsightedness given that the near point is at $35$ cm instead of the normal $25$ cm.
2. **Formula used:** The power $P$ of a lens is given by
$$P = \frac{100}{f}$$
where $f$ is the focal length in centimeters.
3. **Important rule:** For correcting farsightedness, the lens should form a virtual image at the person's near point when the object is at the normal near point ($25$ cm).
4. **Calculate focal length:** Using the lens formula for virtual image,
$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$
where $u = -25$ cm (object distance, negative because object is on the same side as incoming light), and $v = -35$ cm (image distance, negative for virtual image).
5. Substitute values:
$$\frac{1}{f} = \frac{1}{-35} - \frac{1}{-25} = -\frac{1}{35} + \frac{1}{25} = \frac{1}{25} - \frac{1}{35}$$
6. Find common denominator and simplify:
$$\frac{1}{f} = \frac{7}{175} - \frac{5}{175} = \frac{2}{175}$$
7. Calculate focal length:
$$f = \frac{175}{2} = 87.5 \text{ cm}$$
8. Calculate power:
$$P = \frac{100}{87.5} = 1.14 \text{ diopters}$$
**Final answer:** The power of the lens needed is approximately $+1.14$ diopters.