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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 real object placed at distance $u$ (where $

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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 real object placed at distance $u$ (where $u$ is negative) and a real image formed at distance $v$ (where $v$ is positive),we substitute $u = -|u|$ and $v = |v|$.The formula becomes $\frac{1}{|v|} - \frac{1}{-|u|} = \frac{1}{f}$,which simplifies to $\frac{1}{|v|} + \frac{1}{|u|} = \frac{1}{f}$.Rearranging this gives $\frac{1}{|v|} = \frac{1}{f} - \frac{1}{|u|} = \frac{|u|-f}{f|u|}$,so $|v| = \frac{f|u|}{|u|-f}$.This can be rewritten as $|v| - f = \frac{f|u|}{|u|-f} - f = \frac{f|u| - f(|u|-f)}{|u|-f} = \frac{f^2}{|u|-f}$.Thus,$(|v|-f)(|u|-f) = f^2$. This is the equation of a rectangular hyperbola with asymptotes at $|u|=f$ and $|v|=f$. The graph shown in the solution image represents this relationship.

Let $u$ and $v$ be the distances of the object and the image from a lens of focal length $f$. The correct graphical representation of $u$ and $v$ for a convex lens when $|u|>f$,is

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