## Related questions with answers

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

VerifiedWe 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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