A simple model of the human eye ignores its lens entirely. Most of what the eye does to light happens at the outer surface of the transparent cornea. Assume that this surface has a radius of curvature of 6.50 mm and that the eyeball contains just one fluid, with a refractive index of 1.41. Determine the distance from the cornea where a very distant object will be imaged.

the distance from the cornea where a very distant object will be imaged is 23.35 mm

Explanation:

Given the data in the question;

For a spherical refracting surface;

/ + / = ( - )/R

where is the index of refraction of the light of ray in the incident medium

is the object distance

is the index of refraction of light ray in the refracted medium

is the image distance

R is the radius of curvature

Now, let = ∞, such that;

/∞ + / = ( - )/R

0 + / = ( - )/R

we make subject of the formula

R = ( - )

= ( × R ) / ( - )

given that; R = 6.50 mm, = 1.41, we know that = 1.00

so we substitute

= (1.41 × 6.50 mm ) / ( 1.41 - 1.00 )

= 9.165 / 0.41

= 23.35 mm

Therefore, the distance from the cornea where a very distant object will be imaged is 23.35 mm