Q138 / 180 · AIIMS 2017 · Physics

The least distance of vision of a longsighted person is $60 \mathrm{~cm}$. By using a spectacle lens, this distance is reduced to $12 \mathrm{~cm}$. The power of the lens is

  • A
    $+5.0 \mathrm{D}$
  • Correct answer
    B
    $+(20 / 3) \mathrm{D}$
  • C
    $-(10 / 3) \mathrm{D}$
  • D
    $+2.0 \mathrm{D}$

Solution

Using, $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ Here, $v=-60 \mathrm{~cm}, u=-12 \mathrm{~cm}$. $\begin{gathered}\therefore \quad \frac{1}{f}=\frac{1}{(-60)}-\frac{1}{(-12)}=\frac{-1}{60}+\frac{1}{12}=\frac{1}{15} \\ \frac{1}{f}=\frac{1}{15} \Rightarrow f=15 \mathrm{~cm}\end{gathered}$ Power, $P=\frac{100}{15}=\frac{20}{3} \mathrm{D}$

Hence, the correct answer is option: 'B'.

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