In an experiment for determination of the focal length of a thin convex lens, the distance of the object…

Given:

Object distance $=10\pm 0.1 \mathrm{cm}$

Image distance $=20\pm 0.2 \mathrm{cm}$

As per lens formula

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f} ...\left(\mathrm{i}\right)$

$\Rightarrow \frac{1}{20}-\frac{1}{\left(-10\right)}=\frac{1}{f}$

$\Rightarrow f=\frac{20}{3} \mathrm{cm}$

Differentiate equation $\left(\mathrm{i}\right)$ on both sides, we get

$\left(-\frac{1}{v^2}\right)dv-\left(\frac{-1}{u^2}\right)du=\left(\frac{-1}{f^2}\right)df$

For calculating maximum error, we will add all values

$\Rightarrow \frac{1}{f^2}df=+\frac{1}{v^2}dv+\frac{1}{u^2}du$

$\Rightarrow \left(\frac{df}{f}\right)\times 100=\left[\left(\frac{0.2}{{20}^2}+\frac{0.1}{{10}^2}\right)\frac{20}{3}\right]\times 100$

$=\left(\frac{0.2}{4}+\frac{0.1}{1}\right)\frac{20}{3}=1$

Therefore,

$\frac{df}{f}\times 100=1\%=n$