= Solution
A physical delay is a time, whereas the stated $\lambda/(4\sqrt2)$ is a length. First interpret it literally as an additional optical path length $\ell=\lambda/(4\sqrt2)$ 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 gives
$$
\boxed{I_{\rm new}(y)=I_0+\operatorname{Re}\Gamma\left(\frac{\sqrt2y}{c}+\frac{\lambda}{4\sqrt2c}\right).}
$$
Delaying 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 $y+\ell/\sqrt2=y+\lambda/8$.
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:
$$
\Gamma(\tau+\tau_d)\simeq e^{-i\omega\tau_d}\Gamma(\tau),\qquad
\phi_d=\omega\tau_d=\frac{\pi}{2\sqrt2}.
$$
Hence
$$
\boxed{I_{\rm new}(y)\simeq I_0+\cos\!\left(\frac{\pi}{2\sqrt2}\right)\operatorname{Re}\Gamma(\tau(y))
+\sin\!\left(\frac{\pi}{2\sqrt2}\right)\operatorname{Im}\Gamma(\tau(y)).}
$$
This 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, $\Delta y=\lambda/(4\sqrt2)$, the equivalent delay is $\sqrt2\Delta y/c=\lambda/(4c)$ and the <wave phase> shift is $\pi/2$. Under that different interpretation the narrow-band result is $I_0+\operatorname{Im}\Gamma(\tau(y))$, 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.
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