In an experiment for determination of the focal length of a thin...

Given:

Object distance $= 10 \pm 0 . 1cm$

Image distance $= 20 \pm 0 . 2cm$

As per lens formula

$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} . . . \left(i\right)$

$⇒ \dfrac{1}{20} - \dfrac{1}{\left(- 10\right)} = \dfrac{1}{f}$

$⇒ f = \dfrac{20}{3}cm$

Differentiate equation \left(i\right)$on both sides, we get</p><p>\left(- \dfrac{1}{{v}^{2}}\right)dv - \left(\dfrac{- 1}{{u}^{2}}\right)du = \left(\dfrac{- 1}{{f}^{2}}\right)df$

For calculating maximum error, we will add all values

$⇒ \dfrac{1}{{f}^{2}}df = + \dfrac{1}{{v}^{2}}dv + \dfrac{1}{{u}^{2}}du$

$⇒ \left(\dfrac{df}{f}\right) \times 100 = \left[\left(\dfrac{0 . 2}{{20}^{2}} + \dfrac{0 . 1}{{10}^{2}}\right)\dfrac{20}{3}\right] \times 100$

$= \left(\dfrac{0 . 2}{4} + \dfrac{0 . 1}{1}\right)\dfrac{20}{3} = 1$

Therefore,

$\dfrac{df}{f} \times 100 = 1$