Question

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 1

Answer:

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=−25 cm

Image distance, v=−100 cm

From lens formula,

v

1

−

u

1

=

f

1

−100

1

−

−25

1

=

f

1

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

Power, P=

f

1

=

1/3

1

=3 D

Explanation:

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Answer 2

Answer:

Near point of defective eye is 1 m and that of normal eye is 25 cm.

Here, u = -25 cm, v = -1m = 100 cm.

Using lens formula

1/f = 1/v – 1/u

1/f = 1/-100 + 1/25 = 3/100

f = 100/3 cm = 1/3m.

P = 1/f(m) = 1/0.33 = +3.0 D.