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;
[tex]P = \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}[/tex]
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.
[tex]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[/tex]
Therefore, the power of the eye is 50.29 D
