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