The distance v of the real image formed by a | vedclass

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(D) For a convex lens,the lens formula is given by $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,where $f$ is the focal length,$v$ is the image distance,a

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(D) For a convex lens,the lens formula is given by $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,where $f$ is the focal length,$v$ is the image distance,and $u$ is the object distance.Using the sign convention for a real image,$u$ is negative $(-u)$ and $v$ is positive. The formula becomes $\frac{1}{f} = \frac{1}{v} - \frac{1}{-u}$,which simplifies to $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$.As $u$ increases from $f$ to $\infty$,the value of $\frac{1}{u}$ decreases,which means $\frac{1}{v}$ increases,and therefore $v$ decreases.Specifically,when $u = f$,$v = \infty$,and as $u 	o \infty$,$v 	o f$. This relationship represents a hyperbolic curve where $v$ decreases as $u$ increases,which is correctly depicted in option $D$.

The distance $v$ of the real image formed by a convex lens is measured for various object distances $u$. $A$ graph is plotted between $v$ and $u$. Which one of the following graphs is correct?

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