Question

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

• A. Virtual, upright, height = 1 cm
• B. Virtual, upright, height = 0.5 cm
• C. Real, inverted, height = 4 cm
• D. Real, inverted, height = 1 cm

Answer 1

Answer: (C)

Given,
Radius of curvature, $R=20cm$
Height of object, $h_0=2$
Object distance, $u=30cm$
We have, $\frac{1}{f}=(\mu -1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
$\begin{array}{l}=\left(\frac{3}{2}-1\right)\left[\frac{1}{20}-\left(-\frac{1}{20}\right)\right]\\ \Rightarrow \frac{1}{f}=\left(\frac{3}{2}-1\right)\times \frac{2}{20}\end{array}$
$\therefore f=20cm$
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\Rightarrow \frac{1}{20}=\frac{1}{v}+\frac{1}{30}$
$\frac{1}{v}=\frac{1}{20}-\frac{1}{30}$
$=\frac{10}{600}$
$v=60cm$
$m=\frac{h_i}{h_0}=\frac{v}{u}$
$\Rightarrow h_i=\frac{v}{u}\times h_0$
$=\frac{60}{30}\times 2=-4cm$
so, image is inverted.