To correct the vision of a person who is nearR sighted, we use a diverging lens to cause the rays from the distant object to diverge slightly, as if they were coming from a closer object, allowing the eye to focus the image on the retina. In this example, a nearRsighted eye has near and far points of 12 cm and 17 cm, respectively. Assume that the prescribed lens is 2.0 cm from the eye. a) What lens power is needed for this person to see a distant object clearly? For a distant object, the lens must put the image at the far point of the eye so that the person can focus on it. i) What is the object distance? ii) What is the image distance? iii) Now you can use the thin lens equation to find the focal length of the lens that must be prescribed: iv) What is the power of the lens? Solution i) Object distance is u= infinity ii) Now the image from the lens should be at the far point. Since the image is in front of the lens, it is negative Image distance is v= -15cm iii) Thin lens equation => 1/f=1/u+1/v =>1/f=1/inf+1/(-15) =>f=-15cm iv) Power =1/f(in m)=-1/0.15=-6.67D.