Answer
a. Nearsighted.
b. 20.2 cm.
Work Step by Step
a. The lens power is negative so it is a diverging lens. It produces virtual images that are closer to the person than the actual object. The person is nearsighted.
b. For an object at infinity, find the image distance.
$$P=\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$$
$$-5.50D=\frac{1}{\infty}+\frac{1}{d_i}$$
$$d_i=-\frac{1}{5.50D}=-0.182m$$
This is the distance of the image from the lens, and the negative sign tells us it is a virtual image, on the same side of the lens as the object. The lens is 0.020 m away from the eye, so the distance of the image from the eye is 0.182m + 0.020m = 0.202 m = 20.2 cm.
That is the eye’s far point without glasses, because the glasses are designed to take an object at infinity and make an image 20.2 cm away from the eye that the person can focus upon.
