 Solution

First intensity minimum. In Fraunhofer diffraction, the path difference between adjacent slits is $d\sin \theta$, so the relative wave phase is $\delta =2\pi d\sin \theta /{\lambda }_1$. Summing the $N$ equal complex amplitudes as a geometric series gives the diffraction grating interference factorThe central maximum has $\delta =0$. Its nearest zero occurs at $N\delta /2=\pi$, where the denominator is nonzero. ThusThere is a symmetric first zero on each side of the incident direction. The narrow slits are treated as having a sufficiently broad single-slit diffraction envelope.

Wavelength resolution and the missing order specification. Adopt the Rayleigh criterion: the principal maximum of one wavelength lies at the adjacent minimum of the other in the same nonzero diffraction order $m$. For ${\lambda }_2>{\lambda }_1$ and $m>0$, those directions obeyConsequently the ideal resolving power of a diffraction grating is ${\lambda }_1/\Delta \lambda =mN$, givingThe usual first-order reading therefore gives $64\mathrm{nm}$. Zero diffraction order cannot separate the two wavelengths, because both central maxima coincide.

The paper does not specify which diffraction order is observed, so $64\mathrm{nm}$ is not an order-independent minimum. In the ideal infinitely narrow-slit model, the largest usable order for this positive wavelength separation must satisfy $(m+1/N){\lambda }_1\le d$. Since $d/{\lambda }_1=125$, this permits $m\le 124$, giving the formal all-order resultThat pair meets the criterion at $\theta \simeq 83.1^{\circ }$. Order $125$ of ${\lambda }_1$ is at grazing incidence and the corresponding longer-wavelength peak cannot propagate. In a real device, finite slit width, weak high-order intensity and screen acceptance may prevent such an extreme order from being used; those data are absent. Thus the specified-order expression, and $64\mathrm{nm}$ when first order is intended, are the defensible resolution answers under the stated criterion.