Final answer:
The relaxed power of a myopic administrator with a far point of 50.0 cm is 2 diopters. If they have the normal 8.00% ability to accommodate, the closest object they can see clearly is approximately 46.3 cm away.
Step-by-step explanation:
The far point of a myopic individual refers to the maximum distance at which they can see objects clearly without accommodation. For a myopic administrator with a far point of 50.0 cm, we can calculate the relaxed power of the eyes using the formula:
Power (P) = 1 / Focal length (f),
where the focal length (f) is in meters. To find the power in diopters (D), we have:
P = 1 / (0.50 m)
= 2 D.
For part (b), if the administrator has the normal 8.00% ability to accommodate, we can calculate the nearest point they can see clearly using the formula:
Total Power (Pt) = P + (P × accommodation),
Pt = 2 D + (2 D × 0.08)
= 2 D + 0.16 D
= 2.16 D.
The closest object he can see clearly, also known as the near point, can be found by taking the reciprocal of the total power in meters:
Near point = 1 / Pt
= 1 / 2.16 D
≈ 0.463 m or 46.3 cm.