Question

the near point of a hypermetropia eye is 1 m what is power of the lens required to correct this defect​

Answer 1

Answer:

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Explanation:

What is the power of the lens required to correct this defect? A. +1.0D. ... Note- In this particular question we should know that for a hypermetropic eye convex lens is required to correct it.

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Answer 2

Answer:

Given that,

Near point = 1m

Solution :-

Near point = v = -100 cm

Near point for normal eye = -25 cm

u = -25 cm

By Lens Formula,

1/f = 1/v - 1/u

= 1/-100 - 1/-25

= 3/100

f = 100/3  cm

 = 1/3  m

Power = 1/f

        = +3 D

Nature of the lens → Convex Lens