Question

The cornea of the eye has a radius of curvature of approximately 0.40 cm , and the aqueous humor behind it has an index of refraction of 1.35. The thickness of the cornea itself is small enough that we shall neglect it. The depth of a typical human eye is around 25.0 mm.What would have to be the radius of curvature of the cornea so that it alone would focus the image of a distant mountain on the retina, which is at the back of the eye opposite the cornea?

Answer 1

0.64814 cm

Step-by-step explanation:

`n_1` = Refractive index of air = 1

`n_2` = Refractive index of aqueous humor = 1.35

u = Object distance =
`\infty`

v = Image distance = 25 mm

R = Radius of curvature of the cornea

Lens equation

`(n_2-n_1)/(R)=(n_1)/(u)+(n_2)/(v)\\\Rightarrow (1.35-1)/(R)=(1)/(\infty)+(1.35)/(2.5)\\\Rightarrow (1.35-1)/(R)=(1.35)/(2.5)\\\Rightarrow (0.35)/(R)=(1.35)/(2.5)\\\Rightarrow R=(0.35* 2.5)/(1.35)\\\Rightarrow R=0.64814\ cm`

Radius of curvature of the cornea is 0.64814 cm