Interference — Question 10

Question 10

A far-field fringe estimate can be checked against exact paths

Two in-phase point sources lie at $(0,\pm d/2)$ with $d=1.0\,\mathrm m$. A receiver is at $(R,y)=(10.0,2.0)\,\mathrm m$. Assume equal received amplitudes for the phase calculation and $c=340\,\mathrm{m/s}$.

1. Compute the exact absolute path difference and compare it with the paraxial estimate $dy/R$.

2. Find the lowest positive frequency giving cancellation at that receiver, using exact distances. Quantify the error from using the paraxial path estimate.

Question 10 - Solution

A far-field fringe estimate can be checked against exact paths

(a) Method and calculation

The distances are$$r_{\mathrm{near}}=\sqrt{10^2+1.5^2},\quad r_{\mathrm{far}}=\sqrt{10^2+2.5^2}.$$$$\Delta r=r_{\mathrm{far}}-r_{\mathrm{near}}=0.195890\,\mathrm m.$$The paraxial estimate is $dy/R=0.200\,\mathrm m$, about $2.10\%$ too large.

(b) Interpretation and further calculation

The first cancellation satisfies $f\Delta r/c=1/2$, so$$f=\frac{340}{2\Delta r}=867.83\,\mathrm{Hz}.$$The paraxial value is $340/(2\times0.200)=850\,\mathrm{Hz}$, about $2.05\%$ too low. Inverting an approximate path difference changes the sign of its relative error to first order.

Physical check

The paraxial approximation requires both source spacing and receiver offset small compared with range. Equal received amplitudes are a separate idealization.