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 :- 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. 

 

Answer:-

Chapter-11-The-Human-Eye-and-the-Colourful-World-CBSE-Class-10th-Science

As we know that hypermetropia is corrected by using a convex lens.
Here, The near point of a hypermetropic eye is 1 m and the near point of the normal eye is 25 cm.
Therefor, v= − 1 metre = − 100 cm and u= −25 cm
As per lens formula :

1 1 1



v u f
1 1 1



− 100 − 25 f
− 1 + 4 1


100 f
100
f =
cm
3
1
f =
m
3
1
P =
f
1 × 3
P =
Dioptre
1
P = + 3.0 D