Acoustic interference

The wave length of the sound results $\lambda = c / f = 0.4 \,\text{m}$ . For destructive interference, the paths of the two waves must differ by an odd multiple of half the wavelength $\frac{(2 n + 1)}{2} \lambda = r_2 - r_1 = \sqrt{x^2 + y_0^2} - x \quad \text{mit} \quad n = 0,1,2,\dots$ with radial distances $r_1$ and $r_2$ between source and receiver. Solving for $x$ results $\begin{aligned} \Rightarrow & & \frac{(2 n + 1)}{2} \lambda + x &= \sqrt{x^2 + y_0^2} \\ \Rightarrow & & \frac{(2 n + 1)^2}{4} \lambda^2 +(2 n+1) \lambda x + x^2 &= x^2 + y_0^2 \\ \Rightarrow & & x &= \frac{y_0^2}{(2 n + 1) \lambda} - \frac{2 n +1}{4} \lambda \end{aligned}$ For sufficiently large $n$ , the position $x$ becomes negative, so that there is no longer a valid solution. The maximum value of $n$ is given by zeroing $x$ $\begin{aligned} & & 0 &= \frac{y_0^2}{(2 n^\ast +1) \lambda} - \frac{2 n^\ast+1}{4} \lambda \\ \Rightarrow & & (2 n^\ast +1)^2 &= \frac{4 y_0^2}{\lambda^2} \\ \Rightarrow & & n^\ast &= \frac{y_0}{\lambda} - 1 = 16.5 \end{aligned}$ Therefore, valid numbers are in the range $n \in \{0,1,\dots,16\}$ , so that there are 17 interference minima.