The graph shows relationship between object distance and image distance for a equiconvex lens. Then, focal length of the lens is

Solution :
We know that in case of a convex lens when object is placed at $$C',$$ the image is obtained at $$C.$$ This situation is represented in the graph by the point corresponding to $$u = - 10\,cm, v = 10\,cm.$$
Therefore, $$R = 10\,cm$$
$$ \Rightarrow \,\,\frac{R}{2} = 5\,cm = f$$

Lens formula is
$$\eqalign{ & \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \cr & \Rightarrow \,\,\frac{{\Delta f}}{{{f^2}}} = \frac{{\Delta v}}{{{v^2}}} + \frac{{\Delta u}}{{{u^2}}} \cr} $$
(for maximum error in $$f$$)
$$\eqalign{ & \Rightarrow \,\,\frac{{\Delta f}}{{25}} = \frac{{0.1}}{{{{\left( {10} \right)}^2}}} + \frac{{0.1}}{{{{\left( {10} \right)}^2}}}\left[ {\Delta u = \Delta v = 0.1\,{\text{from graph}}} \right) \cr & \Rightarrow \,\,\Delta f = 25 \times 0.1 \times 2 \times 0.01 = 0.05 \cr} $$
Therefore, the focal length $$ = \left( {5.00 \pm 0.05} \right)cm.$$