Chapter 10

NCERT
Class 10
Science
Solutions
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.

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:

Refer Fig. 10.5

Here u = – 25cm, v = –1m = – 100 cm

Lens formula = 1/v - 1/u

                       = 1/ -f

1/(-100) - 1/(-25) = 1/f

⇒ 1/f = -1 + 4 / 100

⇒ f = 100/3

f = 100/3 x 1/100

f = 1/3 cm

Power of lens P = 1/f

P = 1/(1/3)

P = +3D

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