A hypermetropic person has a near point of 1.5 m. Calculate the power of the corrective lens needed so that she can read a book placed at 25 cm from her eyes.
Generated by claude-sonnet-4-6 · 2026-06-26 01:08 · grounding rag
Model Answer
Given:
- Near point of hypermetropic person, $v$ = −1.5 m (image must form at her near point)
- Object distance, $u$ = −25 cm = −0.25 m
Using the lens formula:
$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$
$$\frac{1}{f} = \frac{1}{-1.5} - \frac{1}{-0.25} = -\frac{1}{1.5} + \frac{1}{0.25}$$
$$\frac{1}{f} = -0.667 + 4 = 3.33 \text{ m}^{-1}$$
$$\therefore P = +3.33 \text{ D}$$
The corrective lens is a convex lens of power +3.33 D.
Source: Light – Reflection and Refraction, Chapter 9 (Lens Formula); The Human Eye and the Colourful World, Chapter 10 (Hypermetropia)
---
Explanation
- Hypermetropia is corrected by a convex (converging) lens.
- The lens must form a virtual image at the person's own near point (−1.5 m) when the object is at 25 cm — so $v = -1.5$ m and $u = -0.25$ m (both negative by sign convention for real object, virtual image on same side).
- Power = $\frac{1}{f(\text{in metres})}$; a positive power confirms a convex lens.
- Examiners award marks for: correct sign convention, correct substitution in lens formula, and the final answer with unit (D) and nature of lens.