Visual Optics Practice Test

To arrive at the correct reduced axial length of the eye, we start with understanding the relationship between the eye's surface power, the ametropic correction required, and the axial length. The total refractive power of the eye can be expressed in terms of the surface power and the axial length. In this case, the eye has a reduced surface power of +62.5 D, indicating that it is hyperopic (farsighted). To correct this hyperopia, a positive lens with a power of +3.75 D is needed. The total refractive power of the eye system can be determined by the formula: \[ P = \frac{1}{f} \] where \( P \) is the total power and \( f \) is the focal length in meters. The reduced power of the eye, after adding the correction, is thus: \[ 62.5 D + 3.75 D = 66.25 D \] Next, we can use the approximate relationship that links the axial length and the refractive power of the eye. For a typical human eye, we can consider that a power of +60 D corresponds to an axial length of around 22-23 mm. Using the specific corrected total