Q180 / 180 · KCET 2013 · Physics

The distance between an object and its image produced by a converging lens is $0.72 \mathrm{~m}$. The magnification is 2 . What will be the magnification when the object is moved by $0.04 \mathrm{~m}$ towards the lens?

  • A
    2
  • Correct answer
    B
    4
  • C
    3
  • D
    6

Solution

Given, $m=2$ and the image formed is real $\therefore$ For a real image $\frac{1}{f}=\frac{1}{v}+\frac{1}{u} \quad(\because u$ is negative $)$ $\begin{array}{ll} \text { Or } & 1+\frac{1}{m}=\frac{u}{f} \\ \Rightarrow & 1+\frac{1}{2}=\frac{u}{f} \\ \Rightarrow & \frac{3}{2}=\frac{0.72}{f} \\ \text { or } \quad & f=\frac{0.72 \times 2}{3}=0.16 \mathrm{~m} \end{array}$ As the object is moved by $0.04 \mathrm{~m}$ towards the lens, the new $u_{1}=0.72-0.04=0.68 \mathrm{~m}$ Again real image is formed So, $1+\frac{1}{m}=\frac{u_{1}}{f}$ $\Rightarrow \quad \frac{1}{m}=\frac{0.04}{0.16}$ $\Rightarrow \quad m=4$

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

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