If we know the object's speed, can we find the speed of the image?

If we know the object's speed, can we find the speed of the image?

Question

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 $3\,\text{m}$ from a lens of focal length $\dfrac{1}{2}\,\text{m}$ and is moving away from the lens at $5\,{\mathrm{m\,s^{-1}}}$.

Solution

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

$$\begin{equation} \frac{dv}{dt} = -\frac{5}{9}v^2. \end{equation}$$

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

$(dvdt)$

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 $0.2\,{\mathrm{m\,s^{-1}}}$.

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.