For a hypermetropic person, the near point has shifted to 1.5 m. Using $f = \dfrac{dD}{d-D}$ with $D = 0.25$ m, what focal length convex lens makes the eye normal?

1. Use $f = \dfrac{dD}{d-D}$ with $d = 1.5$ m and $D = 0.25$ m. 2. Numerator: $dD = 1.5 \times 0.25 = 0.375$. 3. Denominator: $d - D = 1.5 - 0.25 = 1.25$. 4. So $f = \dfrac{0.375}{1.25} = +0.3$ m. 5. The positive value confirms a convex lens corrects hypermetropia. 6. A negative result would mean a concave lens, the wrong defect, so $-0.3$ m is a trap. _Source: Samacheer Kalvi (TN SCERT) Class 10 Science, Unit 2 Optics "Solved Problems", p.35_