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

Answer:

0.64814 cm

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

$\dfrac{n_2-n_1}{R}=\dfrac{n_1}{u}+\dfrac{n_2}{v}\\\Rightarrow \dfrac{1.35-1}{R}=\dfrac{1}{\infty}+\dfrac{1.35}{2.5}\\\Rightarrow \dfrac{1.35-1}{R}=\dfrac{1.35}{2.5}\\\Rightarrow \dfrac{0.35}{R}=\dfrac{1.35}{2.5}\\\Rightarrow R=\dfrac{0.35\times 2.5}{1.35}\\\Rightarrow R=0.64814\ cm$

Radius of curvature of the cornea is 0.64814 cm