91Ó°ÊÓ

We have given,

Height of the candle =$2\mathrm{cm}$

Distance from wall = $2\mathrm{m}$

Focal length =$32\mathrm{cm}$

We have to find the position of image and height at that place of the image.

The lens equation is given as,

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

We have to obtain the image at the wall then,

then, $\mathrm{v}=2\mathrm{m}-\mathrm{u}$

put this value in above equation then we will find that,

$$\begin{gathered} \frac{1}{\mathrm{f}}=\frac{1}{2-u}-\frac{1}{u} \\ \frac{1}{0.32m}=\frac{1}{2-u}-\frac{1}{-u} \\ \frac{u+2+u}{u(2+u)}=\frac{1}{0.32} \\ {u}^2-(2m)u+(0.64m^2)=0 \\ u=160cm,40cm \end{gathered}$$

For distance$160\mathrm{cm}$of the candle from the lens , the distance of the image will be ,

$\mathrm{v}=2\mathrm{m}-1.6\mathrm{m}=0.4\mathrm{m}$

Then magnification of the lens will be ,

$$\begin{gathered} \mathrm{m}=-\frac{\mathrm{v}}{\mathrm{u}}=\frac{\mathrm{h}\text{'}}{\mathrm{h}} \\ -\frac{0.4}{1.6}=\frac{\mathrm{h}\text{'}}{0.02} \\ \mathrm{h}\text{'}=-0.5\mathrm{cm} \end{gathered}$$

Here minus sign indicated the inverted image is formed.

For distance $40\mathrm{cm}$of the candle from the lens, the distance of the image will be,

$\mathrm{v}=2\mathrm{m}-0.4\mathrm{m}=1.6\mathrm{m}$

Then magnification of the lens will be ,

$$\begin{gathered} \mathrm{m}=-\frac{\mathrm{v}}{\mathrm{u}}=\frac{\mathrm{h}\text{'}}{\mathrm{h}} \\ -\frac{1.6}{0.4}=\frac{\mathrm{h}\text{'}}{0.02} \\ \mathrm{h}\text{'}=-8\mathrm{cm} \end{gathered}$$

Here minus sign indicated the inverted image is formed.