 For a myopic person with a far point of 80 cm, a concave lens is needed for correction. The focal length of the lens, f, is equal to the far point distance, which is -80 cm (negative because it’s a concave lens). The power P of the lens is calculated using P = 1/f, where f is in meters. Thus, P = 1/−0.80 = −1.25 dioptres. The required lens is concave with a power of -1.25 dioptres.

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## Correcting Myopia: Lens Requirements for a Specific Case

Understanding Myopia and Far Point: Myopia, or near-sightedness, is a condition where distant objects appear blurry. The far point in myopia is the farthest distance at which objects are seen clearly. For the person in question, this distance is 80 cm, indicating that they cannot see objects clearly beyond this point.

### The Role of Concave Lenses in Myopia

To correct myopia, concave lenses are used. These lenses diverge light rays, effectively extending the point where the image comes into focus onto the retina. This adjustment allows for clear vision of distant objects.

#### Determining the Focal Length

The focal length of the corrective lens is crucial. In this case, the far point of the person is 80 cm from the eye. Therefore, the focal length of the required lens is -80 cm (negative sign indicates a concave lens).

##### Calculating the Power of the Lens

The power of a lens (P) is calculated using the formula P =1/f, where f is the focal length in meters. For a focal length of -80 cm, or -0.80 meters, the power is P = 1/−0.80.

###### Resulting Lens Power

The calculation yields a lens power of -1.25 dioptres. This value indicates the strength of the concave lens needed to correct the myopic condition of the individual.

Conclusion: Lens Specification
To correct the myopic condition of this individual, a concave lens with a power of -1.25 dioptres is required. This lens will enable them to see distant objects clearly, correcting the short-sightedness caused by their myopia.

Discuss this question in detail or visit to Class 10 Science Chapter 10 for all questions.
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?