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= \(\frac{1}{f}\)=\(\frac{1}{u}\) + \(\frac{1}{v}\) ⇒\(\frac{1}{f}\) =\(\frac{1}{50}\) ⇒\(\frac{1}{f}\) = \(\frac{1}{25}\)-\(\frac{1}{50}\) ⇒\(\frac{1}{f}\) = f = 50cm =0.5m ⇒\(\frac{1}{f}\) = \(\frac{2-1}{50}\) p = \(\frac{1}{f}\) p = \(\frac{1}{0.5}\) p = 2 diopters ; The patient needs a converging lens with a power of 2 diopters