if you want to remove an article from website contact us from top.

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

Category :

### Mohammed

Guys, does anyone know the answer?

get make a diagram to show how hypermetropia is corrected the near point of hypermetropic eye is 1m from screen.

## 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 hereπto get an answer to your question βοΈ 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.

Question

## 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.

Medium Open in App

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=β25 cm

Image distance, v=β100 cm

From lens formula, v 1 β β u 1 β = f 1 β β100 1 β β β25 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

Video Explanation

225 40

ΰ€Έΰ₯ΰ€°ΰ₯ΰ€€ : www.toppr.com

## 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

View Details

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

ΰ€Έΰ₯ΰ€°ΰ₯ΰ€€ : learn.careers360.com

## 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

Check sibling questions

## 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 = β100 cm We have to find focal length. Using lens formula, 1/π = 1/π£ β 1/π’ 1/π = 1/((β100)) β 1/((β25)) 1/π = (β1)/100 + 1/25 1/π = (β1 + 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/(πΉππππ πππππ‘β (ππ 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.

Next: NCERT Question 8 β

### CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 12 years and a teacher from the past 16 years. He teaches Science, Economics, Accounting and English at Teachoo

Get Professional Certification in Accounts and Taxation

Register me

ΰ€Έΰ₯ΰ€°ΰ₯ΰ€€ : www.teachoo.com

Do you want to see answer or more ?
Mohammed 15 day ago

Guys, does anyone know the answer?