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Make a diagram to show how hypermetropia is corrected. The nea

Science

Class 10

NCERT

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The far point of a myopic person is

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in front of the eye. What is the nature and power of lens required to correct the problem?

Make a diagram to show how hypermetropia is corrected. The near point of hypermetropic eye is 1 m. What is the power of lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.

Why is a normal eye not able to see clearly the objects placed closer than 25 cm?

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Question

Hard

Solving time: 4 mins

Make a diagram to show how hypermetropia is corrected. The near point of hypermetropic eye is 1 m. What is the power of lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.

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Updated on: Feb 10, 2024

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Text solutionVerified

Hypermetropia can be corrected by using a convex lens. A convex lens converges the incog light such that the image is formed on the retina.
Solution Image

An object at 25 cm forms an image at the near point of the hypermetropic eye. Here, near point is 1 m.

Given,

u = - 25 c m

v = - 100 c m

From lens formula, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

Focal length, $f = 100/3 c m = 1/3 m$

\frac{1}{- 100} - \frac{1}{- 25} = \frac{1}{f}

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Question Text	Make a diagram to show how hypermetropia is corrected. The near point of hypermetropic eye is 1 m. What is the power of lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.
Updated On	Feb 10, 2024
Topic	The Human Eye and Colourful World
Subject	Science
Class	Class 10
Answer Type	Text solution:1 Video solution: 15
Upvotes	1771
Avg. Video Duration	4 min

Make a diagram to show how hypermetropia is corrected. The near point of hypermetropic eye is 1 m. What is the power of lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.

Text solutionVerified

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