# make a diagram to show how hypermetropia is corrected the near point of hypermetropic eye is 1m

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## 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.

Click hereπto get an answer to your question βοΈ 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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## 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 : 2022-09-05

Solution Verified by Toppr

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

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

Given,

Object distance,u=β25 cm

Image distance, v=β100 cm

From lens formula, v 1 β β u 1 β = f 1 β β100 1 β β β25 1 β = f 1 β

Focal length,f=100/3 cm=1/3 m

Power, P= f 1 β = 1/3 1 β =3 D

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## 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? Assume that the near point of the normal eye is 25 cm.

Q.7.Β Β Β Β 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? Assume that the near point of the normal eye is 25 cm.

# NCERT

**Q.7. **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? Assume that the near point of the normal eye is 25 cm.

**Master Physics with Foundation Course for Class 10**

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**Answers (1)**

D Divya Prakash Singh

Given the near point for the hypermetropic eye is 1m.

Assume the near point of the normal eye is 25 cm.

So,

The object distance will be, (Normal near point).

The image distance, (Near point of this defective eye) or .

Then the focal length can be found from the lens formula:

Substituting the values in the equation, we obtain

Hence the power of the lens will be:

**Thus, the power of the convex lens required will be**

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## NCERT Q7

NCERT Question 7 The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cmGiven,Near point of the person is 1 m.Since the hypermetropia.The type of lens used to correct hypermetropia is

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## NCERT Question 7 - Chapter 11 Class 10 - Human Eye and Colourful World (Term 1)

Last updated at Sept. 10, 2019 by Teachoo

## 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? Assume that the near point of the normal eye is 25 cm.

### Transcript

NCERT Question 7 The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm Given, Near point of the person is 1 m. Since the hypermetropia. The type of lens used to correct hypermetropia is a convex lens. Near point = 1 m means that person can see object placed at 25 cm clearly if the image is formed at 1 m. To find the power of the lens, we need to find its focal length. So taking, Since the object is in front of the eye, object distance will be negative. Object distance = u = β 25 cm Since the image is made in front of the lens, image distance will be negative. Image distance = v = β 1 m = β100 cm We have to find focal length. Using lens formula, 1/π = 1/π£ β 1/π’ 1/π = 1/((β100)) β 1/((β25)) 1/π = (β1)/100 + 1/25 1/π = (β1 + 4)/100 1/π = 3/100 f = 100/3 cm f = 100/3 Γ 1/100 m f = 1/3 m Now, Power of the lens = 1/(πΉππππ πππππ‘β (ππ π)) = 1/((1/3) ) = 3 D Therefore, Power of the lens is 3 D and since focal length was positive, it is a convex lens.

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