The far point of a myopia eye is at $40 c m$ . For removing this defect, the power of lens required will be

40 D

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- 4 D

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- 2.5 D

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2.5 D

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Solution

The correct option is C $- 2.5 D$
For myopic eye $f = -$ ( defect far point )
$\Rightarrow f = - 40 c m \Rightarrow P = \frac{100}{- 40} = - 2.5 D$

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The far point of a myopia eye is at 40 cm. For removing this defect, the power of lens required will be

The correct option is C −2.5 DFor myopic eye f=− ( defect far point )⇒f=−40 cm⇒P=100−40=−2.5 D

The far point of a myopia eye is at $40 c m$ . For removing this defect, the power of lens required will be

The correct option is C $- 2.5 D$
For myopic eye $f = -$ ( defect far point )
$\Rightarrow f = - 40 c m \Rightarrow P = \frac{100}{- 40} = - 2.5 D$