Question

the near point of hypermetropia eye is 100 cm calculate the focal length and power of the lens to correct his defect​

Answer 1

Answer:

• Power of lens for correction = +3 D

Explanation:

Hypermetropia which is also known as far-sightedness is an eye defect. In this eye defect person could see clearly the distant objects, but can't see objects near the eye clearly.

• Hypermetropia is corrected using a converging lens (Convex lens).

So,

We are given that Near point of hypermetropia eye is 100 cm therefore,

→ Position of image, v = -100 cm

and, since near point of view of a healthy eyes is 25 cm therefore,

→ Position of object, u = -25 cm

Now using lens formula

→ 1/f = 1/v - 1/u

[ where f is focal length, v is position of image, u is position of object ]

→ 1/f = 1/(-100) - 1/(-25)

→ 1/f = 1/(-100) + 1/25

→ 1/f = ( -1 + 4 ) / 100

→ 1/f = 3 / 100

→ f = 100 / 3  cm

→ f = 100 / 300  m = 1 / 3  m

Further, Using formula

→ P = 1 / f

[ where P is power in Dioptre and f is focal length in meters unit ]

→ P = 1 / ( 1 / 3 )

→ P = +3 D

Therefore,

• Power of the lens for correction of defect would be +3 Dioptre.