Question

the power of a physicians eyes is 57.1 D while examinging a patient. how far from her eyes (in m) is the feature being examined. (lens to retina fistance is 2.00 cm)

Answer 1

Answer:

14 cm

Explanation:

Power of eye = 57.1 D

The relation between the focal length and the power is

f = 1 / P = 1 / 57.1 = 0.0175 m = 1.75 cm

The distance between the image and the lens is, v = 2 cm

Let the distance between the object and the eye is u

Use the lens equation

1/ f = 1 / v - 1 / u

1 / 1.75 = 1 / 2 - 1 / u

1 / u = 1 / 2 - 1 / 1.75

1 / u = (1.75 - 2) / 3.5

u = - 14 cm

Thus, the distance between the feature and eye is 14 cm .