Question

An object $5\mathrm{c}\mathrm{m}$ high is placed at a distance $60\mathrm{c}\mathrm{m}$ in front of a concave mirror of focal length$10\mathrm{c}\mathrm{m}$, find the size of the image.

Answer 1

Step 1 Given,

1. Height of object, $h_o$ = $5\mathrm{c}\mathrm{m}$
2. Object distance, $u$= - $60\mathrm{c}\mathrm{m}$
3. Focal length, $f$ = - $10\mathrm{c}\mathrm{m}$

Step 2 Formula Used,

1. According to the mirror formula,

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

Step 3 Solution,

$$\begin{gathered} -\frac{1}{10}=-\frac{1}{60}+\frac{1}{v} \\ -\frac{1}{10}+\frac{1}{60}=\frac{1}{v} \\ \frac{-60+10}{10\times 60}=\frac{1}{v} \\ \frac{1}{v}=\frac{-50}{600} \\ v=-\frac{600}{50} \\ v=-12cm \end{gathered}$$

Since$v=-12cm$,

$$\begin{gathered} m=\frac{h_i}{h_o}=\frac{-v}{u} \\ \frac{h_i}{ho}=-\frac{v}{u} \\ {h}_i=-\frac{h_{o}v}{u} \\ {h}_i=-\frac{5\times -12}{-60} \\ {h}_i=\frac{60}{-60} \\ {h}_i=-1cm \end{gathered}$$

The height of the image is 1 cm. The negative sign shows the image is real and inverted