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 $=1cm$
• B. Virtual, upright, height $=0.5cm$
• C. Real, invertedm height$=4cm$
• D. Real, invertedm height$=1cm$

Answer 1

The correct option is C Real, invertedm height$=4cm$

$\begin{array}{c}\text{Given,}\\ \text{Radius of curvature}=20cm\\ \text{Height of object,}{h}_0=2cm\\ \text{object distance,}u=-30cm\\ \text{we have,}\\ \begin{array}{cc}\frac{1}{f}& =(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})\\ & =(\frac{3}{2}-1)[\frac{1}{20}-(-\frac{1}{20})]\\ \Rightarrow \frac{1}{f}& =(\frac{3}{2}-1)\times \frac{2}{20}\\ & \therefore f=20cm\\ & \Rightarrow \frac{1}{f}=\frac{1}{v}-\frac{1}{u}\\ \Rightarrow & \frac{1}{20}=\frac{1}{v}-(-\frac{1}{30})\\ \Rightarrow & \frac{1}{v}=\frac{1}{20}-\frac{1}{30}\Rightarrow v=60cm\end{array}\end{array}$

$\begin{array}{cc}m& =\frac{hi}{h_0}=\frac{v}{u}\\ \Rightarrow & hi=\frac{v}{u}\times h_0\\ \Rightarrow hi& =\frac{60}{-30}\times 2=-4cm\\ \text{hence, image is inverted.}& \\ \text{So, option (c) is correct}& \end{array}$