Percentage Error in Convex Lens Focal Length Measurement

To find the percentage error in the determination of the focal length of the thin convex lens, we use the lens formula:

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

According to the Cartesian sign convention for a convex lens forming a real image:

• Object distance, $u = -10\text{ cm}$ with an uncertainty $\Delta u = 0.1\text{ cm}$
• Image distance, $v = +20\text{ cm}$ with an uncertainty $\Delta v = 0.2\text{ cm}$

Substituting the values into the lens formula: $\frac{1}{f} = \frac{1}{20} - \left(-\frac{1}{10}\right) = \frac{1}{20} + \frac{1}{10} = \frac{3}{20}\text{ cm}^{-1}$ $f = \frac{20}{3}\text{ cm}$

Differentiating the relation $\frac{1}{f} = \frac{1}{v} + \frac{1}{|u|}$ to determine the maximum fractional error: $\left| \frac{\Delta f}{f^2} \right| = \frac{\Delta v}{v^2} + \frac{\Delta |u|}{|u|^2}$

Multiplying both sides by $f$: $\frac{\Delta f}{f} = f \left( \frac{\Delta v}{v^2} + \frac{\Delta u}{u^2} \right)$

Substitute the given values: $\frac{\Delta f}{f} = \frac{20}{3} \left( \frac{0.2}{20^2} + \frac{0.1}{10^2} \right)$ $\frac{\Delta f}{f} = \frac{20}{3} \left( \frac{0.2}{400} + \frac{0.4}{400} \right)$ $\frac{\Delta f}{f} = \frac{20}{3} \times \frac{0.6}{400} = \frac{12}{1200} = \frac{1}{100} = 0.01$

The percentage error in the focal length is: $\text{Percentage Error} = \frac{\Delta f}{f} \times 100\% = 0.01 \times 100\% = 1\%$

Thus, the value of $n$ is $1$.