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BITSAT 2016
Exam
BITSAT
Year
2016
Subject
Physics
Chapter
Ray Optics

BITSAT 2016 · Physics · Ray Optics

A biconvex lens has a radius of curvature of magnitude 20 cm ....

A biconvex lens has a radius of curvature of magnitude 20cm20 \mathrm{ cm}. Which one of the following options best describe the image formed of an object of height 2cm2 \mathrm{ cm} placed 30cm30 \mathrm{ cm} from the lens?

Options

  1. AVirtual, upright, height =1cm=1 \mathrm{ cm}
  2. BVirtual, upright, height =0.5cm=0.5 \mathrm{ cm}
  3. CReal, inverted, height =4cm=4 \mathrm{ cm}
  4. DReal, inverted, height =1cm=1 \mathrm{ cm}

Correct answer

C. Real, inverted, height =4cm=4 \mathrm{ cm}

Step-by-step solution

In the branch containing LL and R2R_{2} i=ER2(1−e−R2tL)didt=ER2e−R2tL⋅R2L=ELe−R2tL∴VL=Ldidt=Ee−R2tL=12e−2t400×10−3=12e−5tV\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}
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