 To calculate the focal length (f) of a lens, we use the formula f = 1/P, where P is the power of the lens in dioptres.
For distant vision correction with a lens of power -5.5 dioptres, the focal length is f = 1/−5.5 which equals approximately -0.182 meters or -18.2 centimetres.
For near vision correction with a lens of power +1.5 dioptres, the focal length is f = 1/1.5 which equals approximately 0.667 meters or 66.7 centimetres.

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## Calculating Focal Length for Vision Correction

Understanding Lens Power and Focal Length: The power of a lens, measured in dioptres, is inversely related to its focal length. The focal length, denoted as f, is calculated using the formula f = 1/P, where P is the power of the lens.

### Correcting Distant Vision

For correcting distant vision, a person requires a lens with a power of -5.5 dioptres. This negative value indicates a concave lens, typically used to correct myopia (near-sightedness).

#### Focal Length for Distant Vision

To find the focal length for the distant vision lens, we apply the formula: f =1/−5.5. This calculation yields a focal length of approximately -0.182 meters, or -18.2 centimetres.

##### Correcting Near Vision

For near vision correction, the person needs a lens with a power of +1.5 dioptres. This positive value suggests a convex lens, commonly used for correcting hypermetropia (far-sightedness).

###### Focal Length for Near Vision

Applying the formula f = 1/1.5 for the near vision lens, we get a focal length of approximately 0.667 meters, or 66.7 centimetres.

These calculations demonstrate how lens power directly influences the focal length, which is crucial for designing glasses that correct specific vision impairments. The distinct focal lengths for distant and near vision correction highlight the tailored approach needed for different visual deficiencies.

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?