A 2 meter long scale with least count of 0 . 2cm is used to measure...

The formula for a lens is given by $\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} . . . ( i )$

$u = \left(100 \pm 0 . 2\right)cm - \left(80 \pm 0 . 2\right)cm = \left(20 \pm 0 . 4\right)cm$

$v = \left(180 \pm 0 . 2\right)cm - \left(100 \pm 0 . 2\right)cm = \left(80 \pm 0 . 4\right)cm$

Substituting the values in the equation of focal length,

$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{80} - \left(- \dfrac{1}{20}\right) = \dfrac{5}{80}$

$⇒ f = 16cm$

Differentiating the equation (i)

$\dfrac{df}{{f}^{2}} = \dfrac{dv}{{v}^{2}} + \dfrac{du}{{u}^{2}}$(Only positive sign is considered for error calculation)

$= \dfrac{0 . 4}{6400} + \dfrac{0 . 4}{400} = 0 . 4 \times \dfrac{17}{6400}$

$⇒ df = 0 . 4 \times \dfrac{17}{6400} \times {16}^{2} = 0 . 272$

The percentage error is

$\dfrac{df}{f} \times 100 = \dfrac{0 . 272}{16} \times 100 = 1 . 70$