= Solution
Let $x$ be <mirror> <displacement> from equal optical paths in a <Michelson interferometer>, so the <optical path difference> is $2x$. Different spectral components add detected <light intensities>; each component interferes with its own delayed copy. For equal line strengths and ideally balanced arms, write
$$
I(x)=I_0[1+\cos(4\pi x/\lambda_1)]
+I_0[1+\cos(4\pi x/\lambda_2)].
$$
The <trigonometric identity> for the sum of cosines gives
$$
I(x)=2I_0\left[1+
\cos\!\left(2\pi x\left(\frac1{\lambda_1}-\frac1{\lambda_2}\right)\right)
\cos\!\left(2\pi x\left(\frac1{\lambda_1}+\frac1{\lambda_2}\right)\right)
\right].
$$
Thus the sketch consists of rapid fringes within a slowly varying contrast envelope. With $\bar\lambda$ the mean <wavelength> and $|\Delta\lambda|\ll\bar\lambda$, the mirror-displacement fringe spacing is \b[$\bar\lambda/2$]. The <interference visibility> is
$$
\mathcal V(x)=\left|\cos(2\pi x\Delta s)\right|,
\qquad \Delta s=\left|\lambda_1^{-1}-\lambda_2^{-1}\right|.
$$
The first contrast zero is at $x=1/(4\Delta s)$; successive visibility maxima are separated by
$$
\boxed{\Delta x_{\rm revival}=\frac1{2\Delta s}
\simeq\frac{\bar\lambda^2}{2|\Delta\lambda|}.}
$$
Between successive contrast zeros the signed envelope changes sign, shifting the rapid fringes by half a cycle. Unequal line strengths leave a nonzero minimum contrast. Finite linewidth adds a longer-range coherence envelope, while arm imbalance and polarization mismatch can reduce contrast without indicating a different spectral spacing.
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