Interference of Waves — Question 5

Question 5

Fringe visibility measures an amplitude imbalance under coherence

Two fully coherent scalar waves have weaker-to-stronger amplitude ratio $r\le1$. Let the phase-scanned power visibility be $\mathcal V=(P_{\max}-P_{\min})/(P_{\max}+P_{\min})$.

1. Derive $\mathcal V$ in terms of $r$.

2. For $\mathcal V=.80$, infer $r$ and the individual power ratio. State why the inference needs the coherence assumption.

Question 5 - Solution

Fringe visibility measures an amplitude imbalance under coherence

Method and mathematical working

Write the phase extrema in terms of the amplitude ratio and insert them into the visibility definition. Solve the resulting quadratic and enforce the specified ordering.$$\begin{aligned}\mathcal V&=\frac{(1+r)^2-(1-r)^2}{(1+r)^2+(1-r)^2}=\frac{2r}{1+r^2},\\0.8r^2-2r+0.8&=0.\\[4pt]r&=\frac{2-\sqrt{4-4(0.8)^2}}{1.6}=0.50,\qquad\frac{P_{\mathrm{weak}}}{P_{\mathrm{strong}}}=r^2=0.25.\end{aligned}$$

(a) Calculation and conclusion

Writing powers as $C A^2(1\pm r)^2$ gives $\mathcal V=2r/(1+r^2)$. The common power scale cancels.

(b) Interpretation and limits

$.80(1+r^2)=2r$ has roots $.50$ and $2$. The stated weaker-to-stronger convention selects $r=.50$, with individual power ratio $r^2=.25$. Phase fluctuations can also lower visibility, so without full coherence the observed visibility does not uniquely determine amplitude imbalance.

Physical check

The two algebraic roots are reciprocal because swapping the source labels preserves visibility. The condition $r\le1$ selects $r=0.5$; coherence is essential to this inference.