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If f is the focal length of a convex lens and an object is placed at a distance p from the lens, then its image will be at a distance q from the lens, where f, p, and q are related by the lens equation:

$$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$$

What is the rate of change of p with respect to q if $q=2$ and $f=8$?

I see that I am looking for $\frac{dp}{dq}$ but I can't quite seem to find it. I came up with (which is probably wrong, even if it is not, it surely doesn't seem to help) $-f^{-2}\frac{df}{dq}=-p^{-2}\frac{dp}{dq}-q^{-2}$.

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  • $\begingroup$ You are right you want $\left. \frac {dp}{dq}\right|_{q=2}$. $f$ is fixed, so replace it with $8$. What did you get? It is hard to identify mistakes if we don't see your work (as well as your final answer). $\endgroup$
    – Ross Millikan
    May 5, 2012 at 3:30

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As I said in the comment, $f$ is a constant. So $\frac {df}{dq}=0$

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