Final answer:
The relaxed power of a myopic administrator with a far point of 50.0 cm is 2.00 diopters. With an 8.00% ability to accommodate, the closest object they can see clearly is at 46.0 cm.
Step-by-step explanation:
Eye Power Calculation and Accommodation
The far point of a myopic person is the farthest distance at which they can see objects clearly without correction. For a myopic administrator with a far point of 50.0 cm, we can calculate the relaxed power of the eyes.
(a) To find the relaxed power (P) of the eyes, we use the formula:
P = 1/f
where f is the focal length in meters. Since we have the focal length in centimeters (50.0 cm), we need to convert it to meters by dividing by 100.
f = 50.0 cm / 100 = 0.50 m
Therefore, the relaxed power of the eyes is:
P = 1/0.50 m = 2.00 diopters (D)
(b) To determine the closest object the administrator can see clearly with normal 8.00% ability to accommodate, we use the formula:
Accommodated power (Pa) = P (1 + Accommodation ability)
Accommodated power (Pa) = 2.00 D (1 + 0.08) = 2.00 D (1.08)
Pa = 2.16 D
Now we find the closest distance (near point) that can be seen clearly by using the accommodated power:
Near point (N) = 1/Pa
N = 1 / 2.16 D = 0.46 m or 46.0 cm
The closest object the administrator can see clearly is at a distance of 46.0 cm.