Question

An optical bench has 1.5 m long scale having four equal divisions in each cm. While measuring the focal length of a convex lens, the lens is kept at 75 cm mark of the scale and the object pin is kept at 45 cm mark. The image of the object pin on the other side of the lens overlaps with image pin that is kept at 135 cm mark. In this experiment, the percentage error in the measurement of the focal length of the lens is______.

Answer 1

For the given lens
$u = -30\, cm$
$v = 60\, cm$
$\& \frac{1}{f} = \frac{1}{v}-\frac{1}{u}$ on solving : $f = 20\, cm$
also $\frac{1}{f} = \frac{1}{v}-\frac{1}{u}$
on differentiation
$\frac{df}{f^{2}} = \frac{dv}{v^{2}}+\frac{du}{u^{2}}$
$\frac{df}{f} = f\left[\frac{dv}{v^{2}}+\frac{du}{u^{2}}\right]$
$\& \frac{df}{f} \times 100 = f\left[\frac{dv}{v^{2}}+\frac{du}{u^{2}}\right]\times100\%$
$f = 20cm, du = dv = \frac{1}{4} cm$
Since there are 4 divisions in 1 cm on scale
$\therefore \frac{df}{f}\times100 = 20\left[\frac{1/ 4}{\left(60\right)^{2}}+\frac{1/ 4}{\left(30\right)^{2}}\right]\times100\%$
$= 5\left[\frac{1}{3600}+\frac{1}{900}\right]\times100\%$
$= 5\left[\frac{5}{36}\right]\% = \frac{25}{36}\% \approx 0.69\%$