 Solution

A physical delay is a time, whereas the stated $\lambda /(4\sqrt{2})$ is a length. First interpret it literally as an additional optical path length $\ell =\lambda /(4\sqrt{2})$ in arm 1, with the other arm unchanged. In vacuum this delays its full field by ${\tau }_d=\ell /c$. The exact stationarity argument from part (b) then givesDelaying arm 2 changes the plus sign before ${\tau }_d$ to a minus sign. As a function of $y$, the arm-1 result is the old pattern evaluated at $y+\ell /\sqrt{2}=y+\lambda /8$.

For a quasi-monochromatic source, ${\tau }_d$ is much shorter than the coherence time. The envelope correlation can then be treated as unchanged, while the carrier wave phase cannot:HenceThis literal optical-path delay is not a quarter-cycle wave phase shift. If the intended length instead specifies a displacement of the entire recorded fringe pattern by one quarter of its period, $\Delta y=\lambda /(4\sqrt{2})$, the equivalent delay is $\sqrt{2}\Delta y/c=\lambda /(4c)$ and the wave phase shift is $\pi /2$. Under that different interpretation the narrow-band result is $I_0+\mathrm{Im}\Gamma (\tau (y))$, with its sign reversed for the opposite displacement. The exact expression is still a shifted argument of $\Gamma$, not merely a wave phase multiplier when the envelope varies. The original wording does not specify such a fringe-pattern displacement, so these interpretations must be distinguished.