Interference of Waves — Question 3

Question 3

Source phase can compensate a propagation-path difference

At a common observation point, two equal-amplitude contributions are $A\cos(\omega t-kr_1+\phi_1)$ and $A\cos(\omega t-kr_2+\phi_2)$. Their path difference is $r_2-r_1=\lambda/4$.

1. Find the source phase difference required for maximum reinforcement.

2. If the sources instead have equal phase, find the resultant amplitude. Explain the sign in the compensation condition.

Question 3 - Solution

Source phase can compensate a propagation-path difference

Method and mathematical working

Compute phase at the observation point, combining source phase and propagation delay with consistent signs. Interference depends on this arrival-phase difference.$$\begin{aligned}\Delta\Phi_P&=(\phi_2-\phi_1)-k(r_2-r_1),\\k\Delta r&=\frac{2\pi}{\lambda}\frac{\lambda}{4}=\frac\pi2.\\[4pt]\phi_2-\phi_1=\pi/2&\Rightarrow\Delta\Phi_P=0,\\\phi_2-\phi_1=0&\Rightarrow\Delta\Phi_P=-\pi/2,\quad A_{\mathrm{res}}=\sqrt2A.\end{aligned}$$

(a) Calculation and conclusion

The arrival phase difference is $\Delta=\phi_2-\phi_1-k(r_2-r_1)$. Maximum reinforcement needs $\Delta=0$ modulo $2\pi$, so $\phi_2-\phi_1=\pi/2$ modulo $2\pi$.

(b) Interpretation and limits

Equal source phases give $\Delta=-\pi/2$, hence amplitude $\sqrt2A$. The longer second path delays that contribution by $\pi/2$; advancing its source phase by the same amount compensates the delay. Source phase and propagation phase must both be retained.

Physical check

A longer path produces a phase lag in the specified cosine convention. A positive source-phase advance of the same size cancels that lag, so path difference alone does not determine the result.