The graph shows relationship between object distance and image distance for an equiconvex lens. Then, focal length of the lens is [2006]

Question

Smart Achievers · Accepted Answer

(3)

From the graph, $u = - 10 cm, v = 10 cm, Δ u = Δ v = 0.1$

From the lens formula, $\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{10} - \frac{1}{- 10}$

$∴$ $f = 5 cm$

Differentiating the lens formula,

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ $\Rightarrow \frac{Δ f}{f^{2}} = \frac{Δ v}{v^{2}} + \frac{Δ u}{u^{2}}$ $(for maximum error in f)$

$\Rightarrow \frac{Δ f}{25} = \frac{0.1}{{(10)}^{2}} + \frac{0.1}{{(10)}^{2}}$

$\Rightarrow Δ f = 25 \times 0.1 \times 2 \times 0.01 = 0.05$

$∴$ $Focal length, f \pm Δ f = (5.00 \pm 0.05) cm .$

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