Final answer:
The natural near point of a farsighted person prescribed +1.64 diopters corrective lenses is 36.9 cm from the eye, which is approximately the given option of 39 cm when rounded.
Step-by-step explanation:
The subject of this question is Physics, and it pertains to the topic of optics, specifically correcting farsightedness with lenses. The power of a lens in diopters (D) is the reciprocal of its focal length in meters (1/D = f), and the power prescribed for a person's corrective lenses depends on how much the person's natural near point is deviated from the standard near point.
Given that a farsighted person has been prescribed lenses with a power of +1.64 diopters and assuming that the glasses are held 2 cm from the eyes, we calculate the person's natural near point using the formula:
p = 1/do + 1/di
where p is the power of the lenses, do is the object distance from the lens (natural near point to the glasses), and di is the image distance from the lens (25 cm - 2 cm = 23 cm, the adjusted standard near point due to the glasses-eye distance).
Now we need to rearrange this formula to solve for do, the person's natural near point:
do = 1/(p - 1/di)
After substituting the given values, we get:
do = 1/(1.64 D - 1/0.23 m)
do = 1/(1.64 D - 4.35 m-1)
do = 1/(-2.71 m-1)
do = -0.369 m or -36.9 cm
Since the near point is taken to be a positive distance from the eye, the person's natural near point is 36.9 cm away from the eye, which corresponds to one of the choices provided to the student (39 cm when rounded).