JEE Advanced 2023 Paper 1 · Q10 · Error Propagation

Question

In an experiment for determination of the focal length of a thin convex lens, the distance of the object from the lens is $10 \pm 0.1$ cm and the distance of its real image from the lens is $20 \pm 0.2$ cm. The error in the determination of focal length of the lens is $n$ %. The value of $n$ is ___.

SolveKar AI · Accepted Answer

Lens formula: $\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$. With sign convention, $\dfrac{1}{f} = \dfrac{1}{20} + \dfrac{1}{10}$, so $f = 20/3$ cm. Error propagation: $\dfrac{df}{f^2} = \dfrac{dv}{v^2} + \dfrac{du}{u^2}$. Multiplying by $f^2$: $\dfrac{df}{f} = f\left(\dfrac{dv}{v^2}+\dfrac{du}{u^2}\right) = \dfrac{20}{3}\left(\dfrac{0.2}{400} + \dfrac{0.1}{100}\right) = \dfrac{20}{3}(0.0005 + 0.001) = \dfrac{20}{3}\cdot 0.0015 = 0.01$. So $df/f = 1\%$, giving $n = 1$.

JEE Advanced 2023 Paper 1 · Q10 · Error Propagation

Solution