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Chapter 2: Problem 79

The far point of a myopic administrator is \(50.0 \mathrm{cm}\). (a) What is the relaxed power of his eyes? (b) If he has the normal \(8.00 \%\) ability to accommodate, what is the closest object he can see clearly?

Short Answer

Expert verified
The relaxed power of the myopic administrator's eyes is 2.00 diopters. The closest object he can see clearly when fully accommodated is approximately 46.3 cm away.

Step by step solution

01

Convert the far point distance to meters.

The given far point distance is 50 cm. To convert it to meters, simply divide by 100: \(50 \ \mathrm{cm} \cdot \frac{1 \ \mathrm{m}}{100 \ \mathrm{cm}} = 0.5 \ \mathrm{m}\).
02

Calculate the relaxed power of the eye.

Using the thin-lens formula, we can find the relaxed power of the eye since the far point is the distance where the person can see clearly without any accommodation. Therefore, \(P = P_0\) at the far point: \[ P_0 = \frac{1}{f} = \frac{1}{0.5} = 2.00 \ \mathrm{D} \] So, the relaxed power of the myopic administrator's eyes is 2.00 diopters. #b) Find the closest object the person can see clearly.#
03

Calculate the total power of the eye when fully accommodated.

We're given that the person has an 8% ability to accommodate, so we can calculate the accommodative power as follows: \[ P_a = P_0 \cdot 0.08 = 2.00 \ \mathrm{D} \cdot 0.08 = 0.16 \ \mathrm{D} \] The total power of the eye when fully accommodated is the sum of the relaxed power and the accommodative power: \[ P = P_0 + P_a = 2.00 \ \mathrm{D} + 0.16 \ \mathrm{D} = 2.16 \ \mathrm{D} \]
04

Calculate the nearest distance the person can see clearly.

Now, we can use the thin-lens formula to find the nearest distance the person can see clearly by plugging in the total power of the eye when fully accommodated: \[ f = \frac{1}{P} = \frac{1}{2.16} = 0.46296 \ \mathrm{m} \] Therefore, the closest object the myopic administrator can see clearly is approximately 0.463 meters, or 46.3 cm, away.

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