Interference — Question 9

Question 9

Crossing acoustic beams produce a transverse fringe spacing

Two equal-amplitude plane pressure waves have wave vectors $\mathbf k_\pm=k(\cos\theta\,\hat{\mathbf x}\pm\sin\theta\,\hat{\mathbf y})$. Their source phases and frequencies are equal. Let $\lambda=0.34\,\mathrm m$ and $\theta=30^\circ$.

1. Add the waves and derive the pressure-amplitude dependence on $y$.

2. Find adjacent pressure-node spacing and the first positive node. Explain why the spacing diverges as $\theta\to0$.

Question 9 - Solution

Crossing acoustic beams produce a transverse fringe spacing

(a) Method and calculation

Using the cosine-sum identity,$$p=2P\cos(k\sin\theta\,y)\cos(k\cos\theta\,x-\omega t).$$Thus $A_p(y)=2P|\cos(k\sin\theta\,y)|$, and nodes satisfy$$k\sin\theta\,y=(m+\tfrac12)\pi.$$

(b) Interpretation and further calculation

The node spacing and first positive node are$$\Delta y=\frac\lambda{2\sin\theta}=0.340\,\mathrm m,\qquad y_0=\frac\lambda{4\sin\theta}=0.170\,\mathrm m.$$As $\theta\to0$, the beams become parallel and their wave-vector difference tends to zero. Their phase difference then does not vary across $y$, so finite transverse fringes disappear.

Physical check

The spacing is controlled by the transverse wave-vector difference $2k\sin\theta$, not by $k$ alone.