A hyperopic person can see distant objects clearly but cannot read a newspaper held at 25 cm without glasses. The far point of his eye is at infinity. Calculate the minimum power (in dioptres) of a thin converging lens that should be placed close to the eye to enable him to read at 25 cm.

Step-by-step solution

1. For a hyperopic eye, the eye can focus parallel rays (far point at infinity) without aid.
2. To read at 25 cm, the corrective lens must make the object appear to be at the far point (infinity).
3. A lens that produces an image at infinity requires the object to be placed at its focal point.
4. Set the focal length f equal to the object distance: f = 25 cm = 0.25 m.
5. Power P = 1/f (in metres) = 1/0.25 = 4 dioptres.
6. Since the lens must converge light, the power is positive: +4 D.

Concept tested: Correction of hyperopia (numerical)

Published 22 August 2026 · How this solution was prepared and checked

Related questions

• 1 A myopic person sees distant objects clearly only up to a far point of 50 cm. What is the minimum power of a lens (in dioptres) required to correct his vision so that he can see objects at infinity clearly? Assume a thin lens placed close to the eye.
• 2 A person with a myopic eye can see clearly up to a distance of 1.2 m. What is the nature and power of the lens required to correct this defect of vision?
• 3 Which part of the eye contributes the greatest refractive power?
• 4 A monochromatic red light (λ = 600 nm) and violet light (λ = 400 nm) pass through a prism whose apex angle is 30°. The refractive index of the prism for red light is 1.50 and for violet light is 1.55. Assuming small angles, calculate the angular dispersion between the red and violet rays after emerging from the prism.
• 5 Which of the following statements correctly describes the changes that occur in the eye during accommodation for a near object?
• 6 Describe what happens to the ciliary muscle, suspensory ligaments and the shape of the lens when the eye focuses on a near object.