Question

# A goldfish is swimming at 2.00 cm/s toward the front wall of a rectangular aquarium. What is the apparent speed of the fish measured by an observer looking in from outside the front wall of the tank?

Solution

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We don't see the goldfish directly and its true location - we see its image at a location $q$ (measured from the front wall) which is related to its true location $p$ by Eq. 36.8:

\begin{align*} \frac{n_{\text{water}}}{p} + \frac{n_{\text{air}}}{q} &= \frac{n_{\text{air}}-n_{\text{water}}}{R}\\ \implies \frac{1.333}{p} + \frac{1.000}{q} &= \frac{1.000-1.333}{R}\\ \implies \frac{1.000}{q}&=\frac{-0.333}{R} - \frac{1.333}{p} \end{align*}

The curvature of a flat surface is infinite, i.e. $R=\infty$. Hence, the above equation yields

\begin{align*} \frac{1.000}{q}&=-\frac{1.333}{p}\\ \implies q&=-0.750p \end{align*}

We can differentiate this with respect to time to get

\begin{align*} \dv{q}{t}=0.750\cdot \dv{p}{t} \end{align*}

Note that the two speeds have different signs due to our sign conventions. Here we're not interested in signs of $p$ and $q$ so that it suffices to consider only absolute values:

\begin{align*} \left|\dv{q}{t}\right|=0.750\cdot \left|\dv{p}{t}\right| \end{align*}

Derivative on the right-hand side represents the $\textit{true}$ speed while the one on the left-hand side represents the speed of the image.

Therefore, using $\left|\dv{p}{t}\right|=2.00\mathrm{~\frac{cm}{s}}$ we obtain

\begin{align*} v=\left|\dv{q}{t}\right|=0.750\cdot2.00\mathrm{~\frac{cm}{s}}=\boxed{1.50\mathrm{~\frac{cm}{s}}} \end{align*}

We see that the fish appears to be slower than it really is.

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