Answer
2.77 cm.
Work Step by Step
The image and object distances are related to the indices of refraction and to the radius of curvature.
$$\frac{n_a}{s}+\frac{n_b}{s’}=\frac{n_b-n_a}{R}$$
$$\frac{1.00}{36.0cm}+\frac{1.40}{s’}=\frac{1.40-1.00}{0.75cm}$$
$$s’=2.77cm$$
This distance in the nearsighted eye between the cornea’s vertex and the retina is greater than in a normal eye, where it’s about 2.6 cm.
