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

From  the graph  $U=-10cm;v=10cm\Delta u=\Delta v=0.1$ 
From lens formula,    $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{10}-\frac{1}{-10}\therefore f=5cm$
    
Differentiating lens  formula, 
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\Rightarrow \frac{\Delta f}{f^2}=\frac{\Delta v}{v^2}+\frac{\Delta u}{u^2} \Rightarrow \frac{\Delta f}{25}=\frac{0.1}{{\left(10\right)}^2}+\frac{0.1}{{\left(10\right)}^2}$

$\Rightarrow \Delta f=25\times 0.1\times 2\times 0.01=0.05$

$\therefore Focal length f\pm \Delta f=(500\pm 0.05)cm$

$\Rightarrow \frac{\Delta f}{25}=\frac{0.1}{{\left(10\right)}^2}+\frac{0.1}{{\left(10\right)}^2} \Rightarrow \Delta f=25\times 0.1\times 2\times 0.01=0.05$

$\therefore Focal length, f\pm \Delta f=(5.00\pm 0.05)cm$