In an optical bench experiment to determine the focal length of the concave mirror, the least count of the bench is 0.1cm. The object needle reading is 10.7cm and the image needle reading is 5.7cm. The mirror is positioned at 15.7cm. Find the maximum possible percent error in the focal length of the mirror with this set of readings. (Round off to nearest integer).

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Detailed Solution

$Δ x_{0} = 0.0 5 cm Δ x_{1} = 0.0 5 cm Δ x_{M} = 0.0 5 cm \begin{array}{l} \Rightarrow Δ u = Δ X_{0} + Δ X_{M} = 0.1 c m \\ Δ v = Δ X_{1} + Δ X_{M} = 0.1 c m \\ u = - [1.57 - 5.7] = - 10 c m \\ v = - [15.7 - 10.7] = - 5 c m \\ F = \frac{u v}{u + v} = \frac{+ 50}{- 15} = - \frac{10}{3} c m \\ \frac{Δ f}{f^{2}} = \frac{Δ u}{u^{2}} + \frac{Δ v}{v^{2}} \\ 9 \times \frac{Δ f}{100} = \frac{0.1}{100} + \frac{0.1}{25} \\ \Rightarrow Δ f = \frac{0.1}{9} + \frac{0.4}{9} = \frac{0.5}{9} \\ \frac{Δ f}{f} \times 100 = \frac{5}{3} = 1.67 \end{array}$