A student measures the focal length of a conv | vedclass

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(C) The lens equation is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Using sign convention for a convex lens,the object distance is negative,so

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(C) The lens equation is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Using sign convention for a convex lens,the object distance is negative,so we take $u = -x$ (where $x > 0$).The equation becomes $\frac{1}{v} - \frac{1}{-x} = \frac{1}{f}$,which simplifies to $\frac{1}{v} + \frac{1}{x} = \frac{1}{f}$.Rearranging for $v$,we get $\frac{1}{v} = \frac{1}{f} - \frac{1}{x} = \frac{x - f}{xf}$,so $v = \frac{xf}{x - f}$.As $x$ increases from $f$ to $\infty$,$v$ decreases from $\infty$ to $f$. This represents a hyperbolic curve in the first quadrant of the $v-u$ plane (considering magnitudes of $u$ and $v$),which corresponds to the shape shown in option $C$.

$A$ student measures the focal length of a convex lens by placing an object pin at a distance '$u$' from the lens and measuring the distance '$v$' of the image pin. The graph between '$u$' and '$v$' plotted by the student should look like:

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