# Longitudinal magnification confusion in concave mirror

If you move a rod lying on the principal axis of any concave mirror, when the near end of the rod is at a distance equal to half it's own length from the mirror, then the instantaneous velocity of image is four times the object.

Can this be proved by calculation?

My try at doing this was by taking assumption that the rod has velocity say $V$ and the image then will have velocity $4V$. Let the rod length be $L$. Now i tried to prove that for the above relation of velocities the center of the rod will be at distance L from the concave mirror.

I differentiated the mirror equation$$\frac{1}{v}+\frac{1}{u}~=~\frac{1}{f}$$for two ends of the rod and replaced $u'$ with $V$ and $v'$ with $4V.$

Perhaps I'm not clear on what you mean, but my answer would be "no". You make no statement about the focal length of the mirror, which means this should apply to one with any focal length; if we choose an infinite focal length, i.e. a flat mirror, then if the rod is moving away from the mirror with velocity $V$, from the rod's viewpoint its reflection will be moving away from the mirror with velocity $2v$.
You have the right approach: indeed differentiating in that way the Gaussian mirror equation tells you how far axially the image of an on-axis point will move in response to that point's motion. I get, from $v^{-1} + u^{-1} = f^{-1}$:
$$-\frac{\dot{v}}{v^2}-\frac{\dot{u}}{u^2}=0$$
which gives you $\frac{\dot{v}}{\dot{u}} = -\frac{v^2}{u^2}$, which, given that the transverse magnification is $\frac{v}{u}$, is a statement of the well known result that the longitudinal magnification is the transverse magnification squared, since $\frac{\dot{v}}{\dot{u}}$ is the ratio of the relative longitudinal motions for small motions.