Question

The pupil is located 3.60 mm from the cornea. Assume that the cornea has a radius of curvature of 7.80 mm and the aqueous humor has an index of refraction of 1.333. How far does the pupil appear to be from the cornea?

Answer 1

$\frac{n_1}d-\frac{n_2}f=\frac{n_1-n_2}R,\implies$

$f=\frac{n_2dR}{n_1R-(n_1-n_2)d}=\frac{1\cdot 3.6\cdot 7.8}{1\cdot 7.8-(1.33-1)\cdot 3.6}=4.25~mm.$