For a concave lens of focal length f,the rela | vedclass

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(D) The lens formula is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.For a concave lens,the focal length $f$ is negative,so let $f = -|f|$.Thus,

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(D) The lens formula is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.For a concave lens,the focal length $f$ is negative,so let $f = -|f|$.Thus,$\frac{1}{v} - \frac{1}{u} = -\frac{1}{|f|}$,which simplifies to $\frac{1}{v} = \frac{1}{u} - \frac{1}{|f|} = \frac{|f|-u}{u|f|}$.Therefore,$v = \frac{u|f|}{|f|-u}$.As $u 	o 0$,$v 	o 0$.As $u 	o |f|$,$v 	o \infty$.As $u 	o \infty$,$v 	o -|f|$.Since the question uses magnitudes for $u$ and $v$ in the graphs,we consider the virtual image formed by a concave lens,where $v$ is always between $0$ and $f$. The correct graphical representation for the magnitude of image distance $v$ versus object distance $u$ is shown in option $D$.

For a concave lens of focal length $f$,the relation between object distance $u$ and image distance $v$ from its pole can best be represented by ($u = v$ is the reference line):

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