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AIIMS 2014
Exam
AIIMS
Year
2014
Subject
Physics
Chapter
Ray Optics

AIIMS 2014 · Physics · Ray Optics

The near point of distinct vision is at 1 ~m for an elderly person....

The near point of distinct vision is at 1 m1 \mathrm{~m} for an elderly person. What power of lens is to be used to bring the near point to the minimum distance of distinct vision (25 cm)(25 \mathrm{~cm}) ?

Options

  1. A2.5D2.5 \mathrm{D}
  2. B3D3 \mathrm{D}
  3. C3.5D3.5 \mathrm{D}
  4. D4D4 \mathrm{D}

Correct answer

B. 3D3 \mathrm{D}

Step-by-step solution

Here, u=−25 cmv=−1 m=−100 cmu=-25 \mathrm{~cm} v=-1 \mathrm{~m}=-100 \mathrm{~cm} ∴1f=1v−1u=1−100+125=3100\therefore \quad \frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{-100}+\frac{1}{25}=\frac{3}{100} or f=1003 cm=13 mf=\frac{100}{3} \mathrm{~cm}=\frac{1}{3} \mathrm{~m} Therefore, required power, P=1f( m)=11/3P=\frac{1}{f(\mathrm{~m})}=\frac{1}{1 / 3} P=3DP=3 \mathrm{D}
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