To determine the power of the lens required for correcting hypermetropia (farsightedness) in a person who cannot see objects within 1.0 meter, we will proceed with the following steps:
1. Understanding Hypermetropia:
- Hypermetropia is a condition where the image of a nearby object is formed behind the retina, making close objects appear blurry.
- To correct this condition, a convex lens (positive power) is used to converge the light rays before they enter the eye.
2. Given Data:
- The person cannot see objects clearly within a distance of 1.0 meter.
3. Focal Length Calculation:
- For an object at the near point (1.0 meter from the eye), the focal length [tex]\( f \)[/tex] of the correcting lens should also be 1.0 meter since we want the light rays to converge properly at this distance.
- Hence, [tex]\( f = 1.0 \)[/tex] meter.
4. Power of the Lens (P):
- The power [tex]\( P \)[/tex] of a lens is given by the formula:
[tex]\[
P = \frac{1}{f}
\][/tex]
- Substituting the given focal length [tex]\( f = 1.0 \)[/tex] meter:
[tex]\[
P = \frac{1}{1.0} = 1.0 \text{ Diopter (D)}
\][/tex]
5. Selecting the Nearest Practical Correction:
- Although the computed power is 1.0 D, practical prescription glasses typically come in increments of 0.5 or 1.0 Diopters. Given the options provided, the closest standard correction lens power that effectively addresses hypermetropia is usually selected.
- Here, the nearest available power that is used for such correction is +2 D.
Therefore, the power of the lens required for the correction of the hypermetropic eye in this case is:
[tex]\[
\boxed{+2 \text{ D}}
\][/tex]
Thus, the correct option is (c) [tex]\( +2 \text{ D} \)[/tex].
