## College Physics (7th Edition)

To correct nearsightedness, a diverging lens is needed to form an image at the far point of an object at infinity. Thus $f=di$, where $di$ is the distance to the far point. When replacing ordinary glasses with contacts, the distance from the lens to the image is now a few centimeters larger, which makes the focal length slightly larger. Since $P=1/f$, an increase of $f$ means a decrease of $P$, thus contacts are weaker than regular lenses.
To correct nearsightedness, a diverging lens is needed to form an image at the far point of an object at infinity. Thus $f=di$, where $di$ is the distance to the far point. When replacing ordinary glasses with contacts, the distance from the lens to the image is now a few centimeters larger, which makes the focal length slightly larger. Since $P=1/f$, an increase of $f$ means a decrease of $P$, thus contacts are weaker than regular lenses.