In a simplified model of the human eye, the aqueous and vitreous humors and the lens all have a refractive index of 1.40, and all the refraction occurs at the cornea, whose vertex is 2.60 cm from the retina. what should be the radius of curvature of the cornea such that the image of an object 40.0 cm from the cornea’s vertex is focused on the retina?
a) The distance from the cornea vertex to the retina is 2.37 cm
b) This distance is shorter than for the normal eye.
a) Let refractive index of air,
n(air) = x = 1
Let refractive index of lens,
n(lens) = y = 1.4
Object distance, s = 36 cm
Radius of curvature, R = 0.65 cm
The distance from the cornea vertex to the retina is the image distance because image is formed in the retina.
Image distance, s' = ?
(x/s) + (y/s') = (y-x)/R
(1/36) + (1.4/s') = (1.4 - 1)/0.65
1.4/s' = 0.62 - 0.028
1.4/s' = 0.592
s' = 1.4/0.592
s' = 2.37 cm
Distance from the cornea vertex to the retina is 2.37 cm
(b) For a normal eye, the distance between the cornea vertex and the retina is 2.60 cm. Since 2.37 < 2.60, this distance is shorter than for normal eye.
where and are refractive indexes of the two surfaces
From the given data and . The distance from the cornea to the retina in this model of the eye is , . Substituting these values:
This means that the object for the cornea must be 3.02cm behind the cornea. Now assume that the glasses are 2.00cm in front of the eye, so then:
From thin lens equation:
This is the focal length in water but to get that of air, we use :
A converging lens is needed.