A hypermetropic person having near point at a distance of 0.75 m puts on spectacles of power 2.5 D. The near point now is at ____?
Answer: 0.26 m
Explanation
1/f = 1/v - 1/u
2.5 = 1/0.75 - 1/u
u = -0.26 m
So, the new near point is indeed:
0.26 m
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