Q93 / 125 · BITSAT 2016 · Physics

A biconvex lens has a radius of curvature of magnitude $20 \mathrm{~cm}$. Which one of the following options best describe the image formed of an object of height $2 \mathrm{~cm}$ placed $30 \mathrm{~cm}$ from the lens?

  • A
    Virtual, upright, height $=1 \mathrm{~cm}$
  • B
    Virtual, upright, height $=0.5 \mathrm{~cm}$
  • Correct answer
    C
    Real, inverted, height $=4 \mathrm{~cm}$
  • D
    Real, inverted, height $=1 \mathrm{~cm}$

Solution

In the branch containing \(L\) and \(R_{2}\) \(\begin{aligned} &\mathrm{i}=\frac{\mathrm{E}}{\mathrm{R}_{2}}\left(1-\mathrm{e}^{-\frac{\mathrm{R}_{2} \mathrm{t}}{\mathrm{L}}}\right) \\ &\frac{\mathrm{di}}{\mathrm{dt}}=\frac{\mathrm{E}}{\mathrm{R}_{2}} \mathrm{e}^{-\frac{\mathrm{R}_{2} \mathrm{t}}{\mathrm{L}}} \cdot \frac{\mathrm{R}_{2}}{\mathrm{~L}}=\frac{\mathrm{E}}{\mathrm{L}} \mathrm{e}^{-\frac{\mathrm{R}_{2} \mathrm{t}}{\mathrm{L}}} \\ &\therefore \mathrm{V}_{\mathrm{L}}=\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}=\mathrm{Ee}^{-\frac{\mathrm{R}_{2} \mathrm{t}}{\mathrm{L}}} \\ &=12 \mathrm{e}^{-\frac{2 \mathrm{t}}{400 \times 10^{-3}}}=12 \mathrm{e}^{-5 \mathrm{t}} \mathrm{V} \end{aligned}\)

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

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