JAMB Physics 2023 Paper

Concept:A converging lens is needed to form a virtual image of a near object exactly at the patient's near point.Explanation:The patient's near point is $50.0\text{ cm}$, so the corrective lens must make an object at $25.0\text{ cm}$ appear to be at $50.0\text{ cm}$.Take the object distance as $u = 25\text{ cm}$.The image must be virtual and on the same side as the object, so $v = -50\text{ cm}$.Using the lens formula:$$\frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{1}{25} - \frac{1}{50}$$$$\frac{1}{f} = \frac{2 - 1}{50} = \frac{1}{50}\text{ cm}^{-1}$$Therefore, $f = 50\text{ cm} = 0.50\text{ m}$.Power of the lens is:$$P = \frac{1}{f} = \frac{1}{0.50} = 2.0\text{ diopters}$$Answer:The corrective lens should have a power of $2$ diopters.