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

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

A

40 D

B

– 4 D

C

– 2.5 D

D

0.25 D

Answer

– 2.5 D

Explanation

Solution

For myopic eye f = – (defected far point)

⇒f=−40 cm\Rightarrow f = - 40 \mathrm {~cm} ⇒ P=100−40=−2.5DP = \frac { 100 } { - 40 } = - 2.5 D