Final answer:
To correct hypermetropia with a 1.00 m near point to a normal near point of 25.0 cm, the required spectacle lens power is 4 D. The distance between the lens and the eye (1.50 cm in frames) is typically neglected in the calculation.
Step-by-step explanation:
The question concerns the correction of hypermetropia (farsightedness), which is a common vision defect where a person has difficulty focusing on nearby objects. The condition arises when the eye is too short, causing light to focus behind the retina. To correct this defect, a converging lens is used to bring the image into focus on the retina. For a hypermetropic person whose near point is 1.00 m, they require a spectacle lens that can adjust the focal point for closer objects.
To calculate the required power of a corrective lens, we use the formula P = 1/f, where P is the power in diopters (D) and f is the focal length in meters (m). Considering the near point of the hypermetropic eye is at 1.00 m, and it needs to be brought to the normal near point of 25.0 cm, the corrective lens should have a focal length of 25.0 cm or 0.25 m. Hence the power P needed would be P = 1/0.25 m, which equals 4 D.
However, when eyeglass frames hold the spectacle lens at a certain distance from the eye, in this case 1.50 cm, this distance is typically neglected in the calculation as it is very small compared to the focal lengths involved. Therefore, the power of the lens calculated without considering this distance is a suitable approximation.