Question

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

Answer 1

$ \frac{1}{ v }+\frac{1}{ u }=\frac{1}{ f } $
$ \frac{-1}{120}-\frac{1}{40}=\frac{1}{ f }, \,\,\, f =-30\, cm$
Now,
$\frac{-1}{v^2} d v-\frac{1}{u^2} d u=-\frac{1}{f^2} d f$
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$