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

$|v|=60cm$
$v=+60cm$
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\frac{1}{f}=\frac{1}{60}-\frac{1}{-30}$
$\Rightarrow f=20cm$
$\Delta v=0.5cm$
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
Differentiating,
$-\frac{1}{v^2}\Delta v+\frac{1}{u^2}\Delta u=-\frac{1}{f^2}\Delta f$
${\left(\frac{\Delta f}{f^2}\right)}_{max}=\frac{\Delta v}{v^2}+\frac{\Delta u}{u^2}$
$\Delta u=\Delta v=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$ [4 equal divisions in 1 cm]
$\frac{\Delta f}{(20{)}^2}=\frac{1}{2\left(60\right)\left(60\right)}+\frac{1}{2\left(30\right)\left(30\right)}$

$\frac{\Delta f}{f}=\frac{5}{36}\times \frac{1}{10}$

$\frac{\Delta f}{f}\times 100\%=\frac{5}{36}\times \frac{1}{10}\times 100\%=\frac{50}{36}\%=1.39\%$

Answer $1.39$%