(a) A bat uses ultrasound to find its way among trees. If this bat can detect echoes 1.00ms apart, what minimum distance between objects can it detect? (b) Could this distance explain the difficulty that bats have to find an open door when they accidentally get into a house?

A simple speed formula can be given as,

$\mathit{v}\mathbf{=}\frac{\mathbf{\Delta }\mathbf{d}}{\mathbf{\Delta }\mathbf{t}}$ ….. (1)

Here, is the speed of sound, is the distance between the transmission and reflecting surface which is the distance between two objects (tree and bat.

And represents the time when ultrasound waves just fall on the reflecting surface of the tissue of the tree.

Consider the given data as below.

The speed of sound in air (between two tress),$v=330m/s$

Time delay,$\Delta t=1ms=1\times {10}^{-3}s$

This is the time of round trip of sound wave.

Here the first trip is completed when the wave starts from the first tree and ends on the second tree

And the second trip is completed when this wave again comes back to the first tree

Then time to reach the reflection surface (second tree) will be $\frac{\Delta t}{2}$.

Then actual formula from depth will be

$v=\frac{2\Delta d}{\Delta t}$

The depth$\Delta d$ will be as follows.

$$\begin{gathered} 2\Delta d=v\Delta \\ t\Delta d=\frac{v\Delta t}{2} \end{gathered}$$

Now put the values

$\begin{array}{rcl}\Delta d& =& \frac{330\times 1\times {10}^{-3}}{2}\\ & =& 0.165m\\ & =& 16.5cm\end{array}$

If the bat can detect echoes at 1.00msthen the minimum distance between objects should be 16.5cm.

When a bat accidentally gets into the house they will find it difficult to find their way to come again outside because the walls of the house have a significant distance (much greater than ) and for walls detection range is not sufficient so the bat can probably be stuck there in the house for some time.