Q133 / 180 · AIIMS 2015 · Physics

A person uses $+1.5 \mathrm{D}$ glasses to have normal vision from $25 \mathrm{~cm}$ onwards. He uses a $+20 \mathrm{D}$ lens as a simple microscope to see an object. What is the maximum magnifying power if he uses the microscope without glasses.

  • A
    8
  • Correct answer
    B
    9
  • C
    10
  • D
    11

Solution

Let $d$ be the distance of near point of the person without glasses. Focal length of glasses, $$ f_1=\frac{100}{P_1}=\frac{100}{1.5}=\frac{200}{3} \mathrm{~cm} $$ Now $u=-25 \mathrm{~cm} ; v=$ ? From lens formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f_1}$ or $\frac{1}{v}=\frac{1}{f_1}+\frac{1}{u}=\frac{3}{200}-\frac{1}{25}=-\frac{5}{200}$ $$ \therefore v=-\frac{200}{5}=-40 \mathrm{~cm} $$ Therefore, the near point of the person is $40 \mathrm{~cm}$. Focal length of microscope, $$ \begin{gathered} f_2=\frac{100}{P_2}=\frac{100}{20}=5 \mathrm{~cm} \\ \therefore M=1+\left|\frac{d}{f_2}\right|=1+\frac{40}{5}=9 \end{gathered} $$

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

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