Answer
0.3 cm.
Work Step by Step
For a normal eye, an object at infinity forms an image on the back of the ye. The 2.0 cm length (the distance from the lens to the retina where the image forms) is the image distance for an object at infinity. Therefore, it is the focal length, f, of the lens of the eye.
Calculate the length of the nearsighted eye by finding the image distance (the distance from the lens to the retina where the image forms) for an object located at the far point. Use the lens equation.
$$P=\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$$
$$d_i=\frac{fd_o}{d_o-f}=\frac{(2.0cm)(17cm)}{17cm-2.0cm}=2.27cm$$
Thus the excess length of the nearsighted eye is 2.27 cm-2.0 cm=0.3 cm.
