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The far point of a myopic administrator is 50.0 cm. (a) What is the relaxed power of his eyes?

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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.

User RonakThakkar
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