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Question: Answered & Verified by Expert
In an experiment to measure focal length $(f)$ of convex lens, the least counts of the measuring scales for the position of object $(\mathrm{u})$ and for the position of image $(\mathrm{v})$ are $\Delta \mathrm{u}$ and $\Delta \mathrm{v}$, respectively. The error in the measurement of the focal length of the convex lens will be:
PhysicsMathematics in PhysicsJEE MainJEE Main 2024 (04 Apr Shift 1)
Options:
  • A $2 f\left[\frac{\Delta \mathrm{u}}{\mathrm{u}}+\frac{\Delta \mathrm{v}}{\mathrm{v}}\right]$
  • B $\frac{\Delta \mathrm{u}}{\mathrm{u}}+\frac{\Delta \mathrm{v}}{\mathrm{v}}$
  • C $f^2\left[\frac{\Delta \mathrm{u}}{\mathrm{u}^2}+\frac{\Delta \mathrm{v}}{\mathrm{v}^2}\right]$
  • D $f\left[\frac{\Delta \mathrm{u}}{\mathrm{u}}+\frac{\Delta \mathrm{v}}{\mathrm{v}}\right]$
Solution:
1950 Upvotes Verified Answer
The correct answer is: $f^2\left[\frac{\Delta \mathrm{u}}{\mathrm{u}^2}+\frac{\Delta \mathrm{v}}{\mathrm{v}^2}\right]$
$\begin{aligned} & \mathrm{f}^{-1}=\mathrm{v}^{-1}-\mathrm{u}^{-1} \\ & -\mathrm{f}^{-2} \mathrm{df}=-\mathrm{v}^{-2} \mathrm{dv}-\mathrm{u}^{-2} \mathrm{du} \\ & \frac{\mathrm{df}}{\mathrm{f}^2}=\frac{\mathrm{dv}}{\mathrm{v}^2}+\frac{\mathrm{du}}{\mathrm{u}^2} \\ & \mathrm{df}=\mathrm{f}^2\left[\frac{\mathrm{dv}}{\mathrm{v}^2}+\frac{\mathrm{du}}{\mathrm{u}^2}\right]\end{aligned}$

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