A child’s near point is 50 cm, but normal near point for a human eye is 25 cm. What is the required power of the lens to correct the child's near point?
+2.0 D
A near point of 50 cm (instead of 25 cm) means the child cannot see nearby objects clearly – this is hypermetropia (long-sightedness), corrected by a convex lens.
The lens must form a virtual image at the child's near point for an object kept at 25 cm. So \(u = -25\) cm and \(v = -50\) cm.
\(\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = -\frac{1}{50} + \frac{1}{25} = \frac{1}{50}\), so \(f = +50\) cm = +0.5 m.
Power \(P = \frac{1}{f(\text{m})} = \frac{1}{0.5} = +2.0\) D (positive, a convex lens).
Hence, the required power of the lens is +2.0 D.
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