The patient cannot see clearly an object closer than 50 cm Therefore, the patient needs a lens that would enable him see clearly, objects placed 25 cm from the lens So, we take the object to a distance of 25 cm from the lens so that the image forms at 50 cm in front of the lens u=25cm ;v=-50cm (virtual image); p= 1f=1u + 1v ⇒1f =150 ⇒1f = 125-150 ⇒1f = f = 50cm =0.5m ⇒1f = 2−150 p = 1f p = 10.5 p = 2 diopters ; The patient needs a converging lens with a power of 2 diopters
