In u-v method to find focal length of a concave mirror, if object distance is found to be 10.0 cm and image distance was also found to be 10.0 cm, then find maximum permissible error in f.

Using the mirror formula,

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

We have,

$\frac{1}{-10}+\frac{1}{-10}=\frac{1}{f}$

f=-5 cm or |f|=5 cm

$\frac{-df}{f^2}=-\frac{du}{u^2}-\frac{dv}{v^2}$

This equation can be written as

$|\Delta f{|}_{\max }=\left[\frac{\left|\Delta u\right|}{u^2}+\frac{\left|\Delta v\right|}{v^2}\right]\left(f^2\right)$

Substituting the values we get,

$|\Delta f{|}_{nax}=\left[\frac{0.1}{(10{)}^2}+\frac{0.1}{(10{)}^2}\right](5{)}^2=0.05 \mathrm{cm}$

$\left|f\right|=(5\pm 0.05)\mathrm{cm}$