Find the power of a convex lens if the image formed is at a distance of 16 cm from the lens when the object is placed on the other side of the lens at 20 cm from the optical centre?

According to the Cartesian sign convention for a convex lens:Object distance ($u$) = $-20\text{cm}=-0.2\text{m}$ (placed on the left side)Image distance ($v$) = $+16\text{cm}=+0.16\text{m}$ (formed on the other side, right side)Using the lens formula:
$$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$$
The power ($P$) of a lens is defined as $P=\frac{1}{f}$ (where $f$ is in meters). Thus, we can compute $P$ directly in diopters ($\text{D}$) by converting distances to meters, or work in centimeters first:
$$\frac{1}{f}=\frac{1}{16}-\left(\frac{1}{-20}\right)=\frac{1}{16}+\frac{1}{20}$$
$$\frac{1}{f}=\frac{5+4}{80}=\frac{9}{80}{\text{cm}}^{-1}$$
Thus, focal length $f=\frac{80}{9}\text{cm}=\frac{80}{9\times 100}\text{m}=\frac{8}{90}\text{m}$.Power ($P$) is:
$$P=\frac{1}{f(\text{in meters})}=\frac{90}{8}=11.25\text{D}$$
Since it is a convex lens, the power is positive, so $P=11.25$ diopters.