In an experiment for estimating the value of focal length of converging mirror, image of an object placed at $40\ \text{cm}$ from the pole of the mirror is formed at distance $120\ \text{cm}$ from the pole of the mirror. These distances are measured with a modified scale in which there are $20$ small divisions in $1\ \text{cm}$. The value of error in measurement of focal length of the mirror is $\dfrac1K\ \text{cm}$. The value of $K$ is ______.
Numerical value type. Enter your answer.
Answer: 32
$\dfrac1f=\dfrac1u+\dfrac1v=\dfrac1{40}+\dfrac1{120}\Rightarrow f=30\ \text{cm}$. The least count is $\Delta u=\Delta v=\dfrac1{20}=0.05\ \text{cm}$.
Differentiating: $\dfrac{\Delta f}{f^2}=\dfrac{\Delta u}{u^2}+\dfrac{\Delta v}{v^2}$
$$\Delta f=900\times0.05\left(\frac1{1600}+\frac1{14400}\right)=45\times\frac{10}{14400}=\frac1{32}\ \text{cm}$$
So $K=32$.
Solution by Sreeraj P, M.Sc Physics