Question

In an experiment for estimating the value of focal length of converging mirror, image of an object placed at $40cm$ from the pole of the mirror is formed at distance $120cm$ from the pole of the mirror. These distances are measured with a modified scale in which there are $20$ small divisions in $1cm$. The value of error in measurement of focal length of the mirror is $\frac{1}{K}cm$. The value of $K$ is ______

Answer 1

Answer: 32

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\frac{-1}{120}-\frac{1}{40}=\frac{1}{f},f=-30cm$
Now,
$\frac{-1}{v^2}dv-\frac{1}{u^2}du=-\frac{1}{f^2}df$
Also $dv=du=\frac{1}{20}cm$
$\therefore \frac{\frac{1}{20}}{(120{)}^2}+\frac{\frac{1}{20}}{(40{)}^2}=\frac{df}{(30{)}^2}$
On solving
$df=\frac{1}{32}cm$
$\therefore k=32$