An object $4 \mathrm{cm}$ high is placed at a distance of 10 from a convex lens of focal length $20 \mathrm{cm}$. Find position, nature and size of the image. Also find power of the lens.

Here, $u=-10 \mathrm{cm}$ (the object assumed to be kept the left of optical center)

f=+20$\mathrm{cm}$ (positive for a convex lens)

$h_1=+4 \mathrm{cm}$ (object kept above the principal axi

Using the lens formula, we get

 $$\begin{gathered} \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{1}{v}-\frac{1}{-10}=\frac{1}{20} \\ \Rightarrow v=-20 \mathrm{cm} \end{gathered}$$

That is, the image is 20cm from the lens, on the sam side as the object. Hence, the image is virtual. Th linear magnification,

$m=\frac{h_2}{h_1}=\frac{v}{u}$

So, size of the image is

$h_2=h_1\left(\frac{v}{u}\right)=4\times \frac{-20}{-10}=8 \mathrm{cm}$

The positive sign indicates that the image is erect (and virtual).

Since, the power of the lens is given by

$P=\frac{1}{f\left(\text{ in }\mathrm{m}\right)}=\frac{1}{+0.2}=+5\mathrm{D}$