A student measures the focal length of a conv | vedclass

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(B) The lens formula is given by: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Using the sign convention for a convex lens,the object distance $u$ is neg

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(B) The lens formula is given by: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Using the sign convention for a convex lens,the object distance $u$ is negative $(-u)$ and the image distance $v$ is positive $(+v)$.Substituting these into the lens formula: $\frac{1}{v} - \frac{1}{-u} = \frac{1}{f} \implies \frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.Rearranging for $v$: $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{u-f}{uf} \implies v = \frac{uf}{u-f}$.When $u$ is between $f$ and $2f$,$v$ is positive and greater than $2f$. As $u$ increases from $f$ to $\infty$,$v$ decreases from $\infty$ to $f$. This represents a hyperbolic curve in the first quadrant where $u$ and $v$ are both positive values (taking magnitudes for the graph).

$A$ student measures the focal length of a convex lens by obtaining an image of an object pin at a distance '$v$' from the lens,where the object pin is at a distance '$u$' from the lens. What will the graph between '$u$' and '$v$' look like?

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