Review question

# If we know the object's speed, can we find the speed of the image? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8625

## Solution

The distances $u$ and $v$ of an object and its image respectively from a lens of focal length $f$ are connected by the formula $\begin{equation*} \frac{1}{u} + \frac{1}{v} = \frac{1}{f}. \end{equation*}$

Determine the speed of the image of an object when the object is $\quantity{3}{m}$ from a lens of focal length $\quantity{\dfrac{1}{2}}{m}$ and is moving away from the lens at $\quantity{5}{m\,s^{-1}}$.

We want to find the speed of the image, $\dfrac{dv}{dt}$, at the moment when $u=3$ and $\dfrac{du}{dt}=5$. So let’s try implicitly differentiating the formula with respect to $t$. \begin{align*} u^{-1} + v^{-1} &= 2 \\ \implies\quad -u^{-2}\frac{du}{dt} -v^{-2}\frac{dv}{dt} &= 0 \\ \iff\quad \frac{1}{u^2}\frac{du}{dt} &= -\frac{1}{v^2}\frac{dv}{dt} \end{align*} Substituting in the given values, we have $\begin{equation} \frac{dv}{dt} = -\frac{5}{9}v^2. \label{eq:dvdt} \end{equation}$ Substituting values into the original formula, we have \begin{align*} \frac{1}{3} + \frac{1}{v} &= 2 \\ \implies\quad v^2 &= \frac{9}{25} \end{align*} and combining with $\eqref{eq:dvdt}$ we have $\begin{equation*} \frac{dv}{dt} = -\frac{5}{9}\frac{9}{25} = -\frac{1}{5}. \end{equation*}$

So the speed of the image is $\quantity{0.2}{m\,s^{-1}}$.

Alternatively, we could have rearranged the given formula as $v=\frac{u}{2u-1}$ and differentiated using the quotient rule to achieve the same result.