A $2$ metre long scale with least count of $0.2\ \text{cm}$ is used to measure the locations of objects on an optical bench. While measuring the focal length of a convex lens, the object pin and the convex lens are placed at $80\ \text{cm}$ mark and $1\ \text{m}$ mark, respectively. The image of the object pin on the other side of lens coincides with image pin that is kept at $180\ \text{cm}$ mark. The % error in the estimation of focal length is
Answer: (B) 1.70
$u=20\ \text{cm}$, $v=80\ \text{cm}$, $f=\dfrac{uv}{u+v}=16\ \text{cm}$.
Each distance is the difference of two scale readings, so its error is $0.2+0.2=0.4\ \text{cm}$. From $\dfrac1f=\dfrac1u+\dfrac1v$:
$$\Delta f=f^2\left(\frac{\Delta u}{u^2}+\frac{\Delta v}{v^2}\right)=256\left(\frac{0.4}{400}+\frac{0.4}{6400}\right)=0.272\ \text{cm}$$
$\dfrac{0.272}{16}\times100=1.70\%$.
Solution by Sreeraj P, M.Sc Physics