Question

The far point of a myopic administrator is 50.0 cm. (a) What is the relaxed power of their eyes? (b) If they have the normal 8.00% ability to accommodate, what is the closest object they can see clearly?

a) -2.00 D; 25.0 cm
b) -2.50 D; 22.5 cm
c) -1.50 D; 20.0 cm
d) -3.00 D; 27.5 cm

Answer 1

The relaxed power of the myopic administrator's eyes is -2.00 D and the closest object they can see clearly is 46.0 cm away. None of the mentioned answer.

Step-by-step explanation:

The relaxed power of the administrator's eyes can be calculated using the formula:

Power = 1/focal length

where the focal length is given by the far point, which is 50.0 cm. So, the relaxed power is -1/0.50 = -2.00 D.

To find the closest object they can see clearly, we can use the formula:

Closest distance = far point - accommodation

where the accommodation is given by 8.00% of the relaxed power. So, the closest distance is 50.0 cm - (8.00% * 50.0 cm) = 50.0 cm - (0.08 * 50.0 cm) = 50.0 cm - 4.0 cm = 46.0 cm.