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Question: A person wears glasses of power – 2.5 D. The defect of the eye and the far point of the person witho...

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

A

Farsightedness, 40 cm

B

Nearsightedness, 40 cm

C

Astigmatism, 40 cm

D

Nearsightedness, 250 cm

Answer

Nearsightedness, 40 cm

Explanation

Solution

Negative power is given, so defect of eye is

Nearsigntedness

Also defected far point =−f=−1p=−100(−2.5)=40 cm= - f = - \frac { 1 } { p } = - \frac { 100 } { ( - 2.5 ) } = 40 \mathrm {~cm}