In an experiment for estimating the value of focal length of...

Given here, $u = - 40cm, v = - 120cm$

Using mirror formula, $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$

$\dfrac{- 1}{120} - \dfrac{1}{40} = \dfrac{1}{f} ⇒ f = - 30cm$

Now, differentiating mirror equation, we get

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

Also, the least count of scale, $dv = du = \dfrac{1}{20}cm$

Putting in above equation,

$∴ \dfrac{\dfrac{1}{20}}{( 120{)}^{2}} + \dfrac{\dfrac{1}{20}}{( 40{)}^{2}} = \dfrac{df}{( 30{)}^{2}}$

On solving, we get

$df = \dfrac{1}{32}cm$

$∴ K = 32$.