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The two sodium lines are distinct optical frequencies, so their rapidly oscillating mutual cross terms average to zero at the detector. Assuming equal line strengths and negligible individual linewidths, add their separate interferograms. Write $\bar{k}=(k_1+k_2)/2$ and $\Delta k=|k_1-k_2|$. With $I_0$ the total mean output intensity,For mirror displacement $d=x/2$ this becomesThis is spectral doublet from Michelson visibility revivals. Since $\Delta \lambda \ll \bar{\lambda }$, $\Delta k\simeq 2\pi \Delta \lambda /{\bar{\lambda }}^2$. Numerically,The first interference visibility zero occurs at $\boxed{d=0.1447\mathrm{mm}}$, with further zeros at odd multiples of that distance. Visibility is restored at integer multiples of $0.2894\mathrm{mm}$. The signed envelope has period $0.5788\mathrm{mm}$: alternate high-visibility packets have opposite fringe phase. Unequal line strengths leave a nonzero minimum visibility, and finite linewidths introduce a further decaying envelope; the wavelength data alone specify neither effect.