Question

Assume the lens-to-retina distance of the human eye is 2.00 cm. Calculate the power, P, in diopters, of the eye when viewing an object 3.5 m away.

Answer 1

Answer:

The power of the eye is 50.29 D

Explanation:

Given;

distance between the lens and retina, d₁ = 2 cm = 0.02 m

distance between the object and lens, d₀ = 3.5 m

The power of the eye is given by;

$P = \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$

where;

f is the focal length of the eye's lens

For a clear view, the distance between the lens and retina must be equal to distance between the lens and the image.

$P = \frac{1}{d_o} + \frac{1}{d_i}\\\\P = \frac{1}{3.5} + \frac{1}{0.02}\\\\P = 50.29 \ m^{-1}\\\\P = 50.29 \ diopters$

Therefore, the power of the eye is 50.29 D