Hypermetropia, or far-sightedness, is corrected using convex lenses, which converge light rays to focus on the retina. For a hypermetropic eye with a near point of 1 meter, the required lens power is calculated using the formula P =1/f, where f is the focal length in meters. The desired near point is typically 25 cm (0.25 m), so f = 1m−0.25m = 0.75m. Thus, P = 1/0.75 ≈ +1.33 dioptres.

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Correcting Hypermetropia: An Overview

Understanding Hypermetropia: Hypermetropia, commonly known as far-sightedness, is a vision condition where distant objects are seen more clearly than close ones. This occurs because the light entering the eye focuses behind the retina, usually due to a shorter-than-normal eyeball or a lens that is too flat.

The Role of Convex Lenses

To correct hypermetropia, convex lenses are used. These lenses converge light rays, ensuring they focus directly on the retina. This correction allows for clear vision of near objects, which is typically problematic for hypermetropic individuals.

Determining the Near Point

In this case, the near point of the hypermetropic eye is 1 meter. The near point is the closest distance at which the eye can focus on an object. For normal vision, this point is typically about 25 cm (0.25 meters).

Calculating the Required Focal Length

The lens should shift the near point from 1 meter to the normal near point of 25 cm. Therefore, the required focal length (f) is the difference between these distances:
f = 1m−0.25m = 0.75m.

Computing the Lens Power

The power of the lens (P) is calculated using the formula P =1/f. For a focal length of 0.75 meters, the power is P = 1/0.75, which equals approximately +1.33 dioptres.

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Conclusion: Lens Specification for Hypermetropia
To correct the hypermetropic condition of an eye with a near point of 1 meter, a convex lens with a power of about +1.33 dioptres is required. This lens will enable the person to see near objects clearly, effectively correcting the far-sightedness.

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Questions of 10th Science Chapter 10 in Detail

What is meant by power of accommodation of the eye?
A person with a myopic eye cannot see objects beyond 1.2 m distinctly. What should be the type of the corrective lens used to restore proper vision?
What is the far point and near point of the human eye with normal vision?
A student has difficulty reading the blackboard while sitting in the last row. What could be the defect the child is suffering from? How can it be corrected?
A person needs a lens of power –5.5 dioptres for correcting his distant vision. For correcting his near vision he needs a lens of power +1.5 dioptre. What is the focal length of the lens required for correcting (i) distant vision, and (ii) near vision?
The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?
Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect?
Why is a normal eye not able to see clearly the objects placed closer than 25 cm?
What happens to the image distance in the eye when we increase the distance of an object from the eye?
Why do stars twinkle?
Explain why the planets do not twinkle?
Why does the sky appear dark instead of blue to an astronaut?