Interference of Waves — Question 2

Question 2

Unequal amplitudes prevent a perfect destructive null

Two coherent waves have amplitudes $A_1\ge A_2>0$. Varying their relative phase gives maximum resultant amplitude $9\,\mathrm{mm}$ and minimum $3\,\mathrm{mm}$.

1. Find both component amplitudes.

2. Find the minimum-to-maximum power ratio and the relative phase at each extreme.

Question 2 - Solution

Unequal amplitudes prevent a perfect destructive null

Method and mathematical working

Use the sum and absolute difference of the component amplitudes at the two phase extrema. Order the amplitudes first to remove the absolute-value ambiguity.$$\begin{aligned}A_1+A_2&=9\,\mathrm{mm},\qquad A_1-A_2=3\,\mathrm{mm},\\A_1&=\frac{9+3}{2},\qquad A_2=\frac{9-3}{2}.\\[4pt]A_1&=6\,\mathrm{mm},\quad A_2=3\,\mathrm{mm},\qquad\frac{P_{\min}}{P_{\max}}=\left(\frac39\right)^2=\frac19.\end{aligned}$$

(a) Calculation and conclusion

The extremes are $A_1+A_2=9$ and $A_1-A_2=3\,\mathrm{mm}$. Hence $A_1=6$ and $A_2=3\,\mathrm{mm}$.

(b) Interpretation and limits

The power ratio is $(3/9)^2=1/9$. The maximum occurs at phase $0$ modulo $2\pi$ and minimum at $\pi$ modulo $2\pi$. Opposite phase cancels only equal amplitudes; the excess amplitude remains.

Physical check

Both recovered amplitudes are positive and reproduce both extrema. The power contrast uses the square of $3/9$, rather than the amplitude ratio itself.