Q142 / 160 · TS EAMCET 2024 (09 May Shift 2) · Physics

If you are using eye glasses of power 2D, your near point is

  • A
    25 cm
  • Correct answer
    B
    50 cm
  • C
    43 cm
  • D
    32 cm

Solution

Given, Power of lens, $P=2 D$ Focal length of lens, $\mathrm{f}=\frac{1}{\mathrm{P}}=\frac{100}{2}=50$ By lens formula $\begin{aligned} & \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{~V}}-\frac{1}{4} \\ & \Rightarrow \frac{1}{50}=\frac{1}{\mathrm{~V}}-\frac{1}{(25)} \Rightarrow \frac{1}{50}=\frac{1}{\mathrm{~V}}+\frac{1}{25} \\ & \Rightarrow \frac{1}{\mathrm{~V}}=\frac{1}{50}-\frac{1}{25}=\frac{-1}{50} \\ & \Rightarrow \mathrm{~V}=-50 \mathrm{~cm} \end{aligned}$

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

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