JEE MAIN - Physics (2024 - 4th April Morning Shift - No. 10)
Explanation
First, let's understand the relationship between the object distance ($$u$$), the image distance ($$v$$), and the focal length ($$f$$) of a convex lens, which is given by the lens formula:
$$ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} $$Now, to find the error in the focal length ($$\Delta f$$) due to the errors in the measurements of $$u$$ and $$v$$ ($$\Delta u$$ and $$\Delta v$$, respectively), we have to differentiate the lens formula with respect to $$u$$ and $$v$$, keeping in mind the propagation of error.
By differentiating both sides of the lens formula with respect to $$v$$ and $$u$$, and also considering the negative reciprocal relation (given $$f$$ is a constant for a specific lens), we have:
$$ \frac{\Delta f}{f^2} = \frac{\Delta v}{v^2} + \frac{\Delta u}{u^2} $$Rearranging this equation to find $$\Delta f$$, we get:
$$ \Delta f = f^2 \left( \frac{\Delta v}{v^2} + \frac{\Delta u}{u^2} \right) $$Therefore, the correct option showing the error in the measurement of the focal length of the convex lens, taking into account the least counts ($$\Delta u$$ and $$\Delta v$$) of the measuring scales for the position of the object ($$u$$) and for the position of the image ($$v$$), is:
Option C: $$f^2\left[\frac{\Delta \mathrm{u}}{\mathrm{u}^2}+\frac{\Delta \mathrm{v}}{\mathrm{v}^2}\right]$$
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