Question

An object 5 cm high is held 25 cm away from a converging lens of focal length 10 cm. Find the position, size at nature of the image formed. Also draw the ray diagram.

Answer 1

Given:
Object distance, u = $-$25 cm
Focal length, f = 10 cm
Height of the object, h = 5 cm
Applying the lens formula, we get:

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$$\begin{gathered} \Rightarrow \frac{1}{v}=\frac{1}{f}+\frac{1}{u} \\ \Rightarrow \frac{1}{v}=\frac{1}{f}+\frac{1}{u} \\ \Rightarrow \frac{1}{v}=\frac{1}{10}+\frac{1}{-25} \\ \Rightarrow \frac{1}{v}=\frac{5-2}{50} \\ \Rightarrow \frac{1}{v}=\frac{3}{50} \\ \Rightarrow v=\frac{50}{3}=+16.6\mathrm{cm} \\ \mathrm{The}\mathrm{image}\mathrm{will}\mathrm{be}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{of}16.6\mathrm{cm}\mathrm{behind}\mathrm{the}\mathrm{lens}. \\ m=\frac{v}{u}=\frac{50}{3} \\ m=\frac{h^{,}}{h}=\frac{h^{,}}{5} \\ \frac{h^{,}}{5}=\frac{-2}{3} \\ {h}^{,}=\frac{-10}{3}=-3.3\mathrm{cm} \\ \mathrm{Thus},\mathrm{the}\mathrm{image}\mathrm{will}\mathrm{be}3.3\mathrm{cm}\mathrm{high}.\mathrm{It}\mathrm{will}\mathrm{be}\mathrm{real}\mathrm{and}\mathrm{inverted}. \end{gathered}$$