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I was able to come until here by myself $$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$$ differentiating both sides wrt time $$0=-\frac{du}{dt* u^2}-\frac{dv}{dt*v^2}$$ so we have from this $$\frac{du}{dt* u^2}=-\frac{dv}{dt*v^2}$$ setting $\frac{du}{dt}$=$v_o$ and $\frac{dv}{dt}=v_i$ we get $$\frac{v_o}{v_i}-m^2$$ where m is the magnification however, my teacher has somehow done everything the same until here and then randomly said $$V_{im}=-m^2v_{om}$$ where $V_{im}$ is the image velocity with respect to the mirror and $V_{om}$ is the velocity of the object wrt the mirror

I'm unable to make sense of this, becasue in the derivation, the mirror is stationary. Could someone help?

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  • $\begingroup$ Rewrite as $1 \over f = 1 \over p + 1 \over q$ to avoid messing up over $v$'s. Rewrite to get $q$ (image, arbitrary) as a function of $p$. Now move $p$. Stay in $p$ and $q$ until the end result, then you may well replace $p\over q$ with $m$... $\endgroup$
    – Jon Custer
    Commented Oct 13, 2022 at 17:26
  • $\begingroup$ I'm unable to understand what you mean $\endgroup$
    – math and physics forever
    Commented Oct 13, 2022 at 17:29

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You incorrectly substituted $m$ for $\dfrac uv$ instead of the correct fraction, $\dfrac vu$.

$\dfrac{du}{dt}\cdot\dfrac {1}{ u^2}=-\dfrac{dv}{dt}\cdot\dfrac {1}{ v^2} \rightarrow v_{\rm o}\cdot\dfrac {1}{ u^2}=-v_{\rm i}\cdot\dfrac {1}{ v^2} \rightarrow -v_{\rm o}\cdot\dfrac {v^2}{ u^2}=v_{\rm i} \rightarrow -v_{\rm o}\cdot m^2=v_{\rm i}$

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