Figure, shows two points sources${\mathbf{S}}_1$and ${\mathbf{S}}_2$that emit sound of wavelength $\mathbf{\lambda }\mathbf{=}\mathbf{\text{2.00m}}$The emission are isotropic and in phase, and the separation between the sources is $\mathbf{d}\mathbf{=}\mathbf{\text{16.0m}}$. At any point Pon the, the wave from${\mathbf{S}}_1$and the wave from ${\mathbf{S}}_2$
interfere. When Pis very away,$(\mathbf{x}\mathbf{=}\yen )$what are (a) the phase difference between the arriving waves from ${\mathbf{S}}_1\mathbf{}\text{and}\mathbf{}{\mathbf{S}}_2$and (b) the type of interference they produce? Now move pointP along the $\mathbf{x}\mathbf{}\mathbf{-}\mathit{a}\mathit{x}\mathit{i}\mathit{s}$ towards. (c) Does the phase difference between the waves increase or decrease? At what distancedo the waves have a phase difference of (d) 0.50(e)1.00 (f)1.50$\mathbf{\lambda }$?

• Wavelength of the sound is, $\lambda =2\text{m}$.

• Separation between the source,$d=1\text{6m}$.

Sound waves are sinusoidal waves, which consist of wavelength and frequency. Calculate the phase difference between the two waves.

Formula is as follow:

$f=\frac{2\pi }{\lambda }\text{Δ}L$

Here, L is length, $\lambda$ is wavelength and f is frequency.

Point P is infinitely far from the source. Therefore, the small distance d in between the two sources is of no consequence. Thus, there is no effect of distance on the phase difference. Therefore, there is no phase difference between the two waves.

$\begin{array}{c}f=\frac{2\pi }{\lambda }\text{Δ}L\\ =\frac{2\pi }{\lambda }\times 0\\ =0\end{array}$

Hence, the phase difference between $s_1$and $s_2$is zero.

If two sources oscillate inthephase, they produce constructive interference. In this problem, source$s_1$and source$s_2$are inthatphase, so they produce constructive interference.

Hence, type of interference produced by $s_1$and$s_2$is constructive.

Now, point P is moving towards source. Supposethedistance between source$s_1$and point P is x. By using Pythagorean theorem, findthedistance between source$s_2$and point p as,

$q=\sqrt{d^2+x^2}$$x>0,$

So, that path difference between two waves is,

$\text{Δ}L=\sqrt{d^2+x^2}-x$

Where, i.e., the path of the two sources is different. So, the phase difference has non-zero value, and it is clearly increasing.

Hence,the phase difference betweenthewaves increases.

From the above explanation, write the phase difference as,

$\begin{array}{l}f=\frac{\text{Δ}L}{\lambda }\\ f=\frac{\sqrt{d^2+x^2}-x}{\lambda }\end{array}$

In terms of $\lambda ,$the phase difference is identical to the path length difference,

$\begin{array}{l}\text{Δ}L=\frac{\lambda }{2}\\ \frac{\text{Δ}L}{\lambda }=\frac{\sqrt{d^2+x^2}-x}{\lambda }\end{array}$

On solving the above equation,

$$\begin{gathered} \text{Δ}L=\sqrt{d^2+x^2}-x \\ \frac{\lambda }{2}=\sqrt{d^2+x^2}-x \\ x+\frac{\lambda }{2}=\sqrt{d^2+x^2} \end{gathered}$$

Squaring both sides,

$\begin{array}{c}{x}^2+\frac{{\lambda }^2}{4}+x\lambda =d^2+x^2\\ \frac{{\lambda }^2}{4}+x\lambda =d^2\\ \frac{\lambda }{4}+x=\frac{d^2}{\lambda }\\ x=\frac{d^2}{\lambda }-\frac{\lambda }{4}\end{array}$

In general, if $\text{Δ}L=n\lambda$where n is some multiplier,

$\text{Δ}L=\sqrt{d^2+x^2}-x$

Substitute $n\lambda$for $\text{Δ}L$into the above equation,

$\begin{array}{c}n\lambda =\sqrt{d^2+x^2}-x\\ x^2+{\left(n\lambda \right)}^2+xn\lambda =d^2+x^2\\ {\left(n\lambda \right)}^2+2xn\lambda =d^2\\ n\lambda +2x=d^2/n\lambda \end{array}$

Express the above equation in terms of x

$\begin{array}{c}n\lambda +2x=d^2/n\lambda \\ x=\frac{d^2}{2n\lambda }-\frac{1}{2}n\lambda \\ x=\frac{d^2}{2n\lambda }-\frac{1}{2}n\lambda \\ x=\frac{{16}^2}{2n\times 2}-\frac{1}{2}n\times 2\\ x=\frac{256}{4n}-n\end{array}$

$\Rightarrow x=\frac{64}{n}-n$

Now, phase difference $\text{Δ}L=0.50\lambda ,$i.e.$n=0.50$, Find distance x as,

Substitute $0.50$for into the above equation,

$\begin{array}{l}x=\frac{64}{n}-n\\ =\frac{64}{0.5}-0.5\\ =127.\text{5m}\\ =1\text{28m}\end{array}$

Hence, at distance $x=128\text{m,}$the waves have a phase difference of$0.50\lambda$ .

Now, phase difference$\text{Δ}L=1.00\lambda ,$i.e.$n=\mathrm{1.00.}$,Find distance x as,

$\begin{array}{c}x=\frac{64}{n}-n\\ =\frac{64}{1.00}-1\\ =63\text{m}\end{array}$

Hence, at distance $x=63.\text{0m,}$the waves have a phase difference of $1.00\lambda$.

Now, phase difference$\text{Δ}L=1.50\lambda ,$i.e.$n=\mathrm{1.50.}$,Find distance x as,

$\begin{array}{l}x=\frac{64}{n}-n\\ =\frac{64}{1.50}-1.50\\ =41.2\text{m}\end{array}$

Hence, at distance$x=41.\text{2m,}$ the waves have a phase difference of$1.50\lambda$