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Q.

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

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a

40 D

b

-4 D

c

-2.5 D

d

0.25 D

answer is C.

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Detailed Solution

For myopic eye.f = - (defected far point)

⇒f=−40cm⇒P=100−40=−2.5D

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