A person with a myopic eye can see clearly up to a distance of $1.2\text{ m}$. What is the nature and power of the lens required to correct this defect of vision?

Step-by-step solution

1. Identify the object distance $u$ for a normal eye to see distant objects: $u=-\infty$.
2. Identify the image distance $v$, which is the myopic far point: $v=-1.2\text{ m}$.
3. Apply the lens formula: $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$.
4. Substitute the values into the formula: $\frac{1}{f}=\frac{1}{-1.2}-\frac{1}{-\infty }=-\frac{1}{1.2}{\text{ m}}^{-1}$.
5. Calculate the focal length: $f=-1.2\text{ m}$.
6. Calculate the power $P$ of the lens using $P=\frac{1}{f(\text{in metres})}$: $P=\frac{1}{-1.2}\approx -0.83\text{ D}$.
7. Since the focal length and power are negative, the required lens is a concave lens.

Concept tested: Correction of myopia using concave lens

Published 23 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 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.
• 3 Which part of the eye contributes the greatest refractive power?
• 4 Describe what happens to the ciliary muscle, suspensory ligaments and the shape of the lens when the eye focuses on a near object.
• 5 Which of the following statements correctly describes the changes that occur in the eye during accommodation for a near object?
• 6 List three common causes of eye injuries in daily life and suggest one preventive measure for each cause.