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  • A biconvex lens has radius of curvature of magnitude 20 cm. Which one of the following options describes best the image formed of an object of height 2 cm placed 30 cm from the lens?<div class=

A biconvex lens has radius of curvature of magnitude 20 cm. Which one of the following options describes best the image formed of an object of height 2 cm placed 30 cm from the lens?

Option: 1

Real, inverted, height = 1 cm


Option: 2

Virtual, upright, height = 1 cm


Option: 3

Virtual, upright, height = 0.5 cm


Option: 4

Real, inverted, height = 4 cm


Answers (1)

best_answer

\\ Given, \\ Radius of curvature, R=20 \mathrm{cm}$ \\ Height of object, $h_{0}=2$ \\ Object distance, $u=-30 \mathrm{cm}$ \\ We have, $\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$ $=\left(\frac{3}{2}-1\right)\left[\frac{1}{20}-\left(-\frac{1}{20}\right)\right]$ \\ $\Rightarrow \frac{1}{f}=\left(\frac{3}{2}-1\right) \times \frac{2}{20}$ \\ $\therefore f=20 \mathrm{cm}$

\\ \frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ \\ $\Rightarrow \frac{1}{20}=\frac{1}{v}+\frac{1}{30}$ \\ $\frac{1}{v}=\frac{1}{20}-\frac{1}{30}$ $=\frac{10}{600}$ \\ \\ $v=60 \mathrm{cm}$ \\ \\ $m=\frac{h_{i}}{h_{0}}=\frac{v}{u}$ \\ $\Rightarrow h_{i}=\frac{v}{u} \times h_{0}$ $=\frac{60}{-30} \times 2=-4 \mathrm{cm}$ \\ so, image is inverted.

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