A +3.00 D hyperope corrected with a +1.50 D contact lens viewing an object at 22.0 cm requires how much accommodation to achieve a clear retinal image?

To determine how much accommodation is required for a +3.00D hyperope using a +1.50D contact lens when viewing an object at 22.0 cm, the first step is to calculate the total power of the optical system including both the contact lens and the hyperopia.

Hyperopes require plus lens correction to bring images into focus on the retina. The individual in this scenario has a hyperopic refractive error of +3.00D. When corrected with a +1.50D contact lens, the effective power that the eye experiences becomes:

Effective power = Contact lens power - Hyperopic correction

Effective power = +1.50D - (+3.00D) = -1.50D

This means that, in essence, the eye is acting as a -1.50D system after the contact lens correction.

Next, we can calculate the vergence of light rays coming from the object at 22.0 cm. The distance in meters for the object is:

Distance = 22.0 cm = 0.22 m

Using the formula for vergence (Power = 1/Distance in meters), we find:

V = 1 / 0.22 m ≈ +4