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Q.

A person wears glasses of power – 2.5 D. The defect of the eye and the far point of the person without the glasses are respectively

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a

Farsightedness, 40 cm

b

Nearsightedness, 40 cm

c

Astigmatism, 40 cm

d

Nearsightedness, 250 cm

answer is B.

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Detailed Solution

Negative power is given, so defect of eye is nearsigntedness 
 Also defected far point =−f=−1p=−100(−2.5)=40cm

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