Answer to Question #282121 in Physics for Ella

Question

Directions: Solve the problems below and use the rubric as your guide in answering Use a separate sheet of paper for your solution.

Focus on the Eye. The comea of the eye has a radius of curvature of approximately 0.50 cm, and the aqueous humor behind it has an index of refraction of 1.35. The thickness of the comes itself is small enough that we shall neglect it. The depth of a typical human eye is around 25 mm.

(a) What should be the radius of curvature of the cornea so that it could focus the image of a distant mountain on the retina located at the back of the eye opposite the comea?

(b) If the cornea focused the mountain correctly on the retina as described in part (a), will it also focus the text from a computer screen on the retina if the screen was placed 25 cm in front of the eye? If not, where will it focus the text, in front of or behind the retina?

Answer 1

a) Find the focal length for an object located in infinity:

$$\frac 1f=\frac 1o+\frac 1i=\frac 1{\infty}+\frac 1i=\frac 1i.\\\space\\ f=i=25\text{ mm}.$$

For such a focal length:

$$\frac 1f=\frac{(n-1)}R,\\\space\\ R=f(n-1)=25(1.35-1)=8.75\text{ mm}.$$

b) No, the image will not be in focus. The position of the image can be found with the thin lens equation:

$$\frac 1{f}=\frac 1o+\frac 1i=\frac 1{250}+\frac 1{i}.\\\space\\ f=25\text{ mm},\\\space\\ \frac 1{25}=\frac 1{250}+\frac 1{i},\\\space\\ i=27.8\text{ mm (behind the retina)}.$$