= Solution
Represent each wave by a complex <phasor>. Add <phasors> before squaring: two <amplitudes> give $I\propto|A_1+A_2e^{i\delta}|^2=A_1^2+A_2^2+2A_1A_2\cos\delta$. For $N$ equally illuminated narrow slits, adjacent <phasors> differ by $\delta=2\pi d\sin\theta/\lambda$, so
$$
S_N=\sum_{n=0}^{N-1}e^{in\delta}
=e^{i(N-1)\delta/2}\frac{\sin(N\delta/2)}{\sin(\delta/2)}.
$$
All <phasors> align when $\delta=2\pi m$, giving the <grating equation>
$$
\boxed{d\sin\theta_m=m\lambda,\qquad I_{\rm principal}=N^2I_1.}
$$
The nearest zero has adjacent-phase increment $2\pi/N$: the <phasor> polygon closes once. Thus $d\cos\theta_m\Delta\theta\simeq\lambda/N$, giving
$$
\boxed{\Delta\theta\simeq\frac{\lambda}{Nd\cos\theta_m}=\frac{\lambda}{W\cos\theta_m}.}
$$
The angular linearization is valid away from grazing directions; the underlying exact first-zero condition is $d(\sin\theta_{\rm zero}-\sin\theta_m)=\pm\lambda/N$.
For the <first subsidiary maximum of a large slit array>, use the stipulated $1.5$ <phasor> wraps: $N\delta/2\simeq3\pi/2$ relative to a principal maximum. Its <amplitude> is approximately $2N/(3\pi)$ times one slit's <amplitude>. Therefore
$$
\boxed{\frac{I_{\rm principal}}{I_{\rm subsidiary}}\simeq\frac{9\pi^2}{4}=22.2.}
$$
This uses the question's approximate maximum position, not the exact extremum of the slit-array <intensity>.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-1-ia2014-grating.png]
{title=Original phasor sums for a principal maximum, the adjacent zero and the approximate first subsidiary maximum; calculated large-array intensity shown below}
The <Rayleigh criterion> places one line's principal maximum at the other's first minimum. <Grating dispersion> then gives $|m|N\gtrsim\lambda/\delta\lambda=1.674\times10^6$. Combining with normal incidence, $|m|\lambda=d|\sin\theta|$, gives the <normal-incidence grating width bound>:
$$
\boxed{W\gtrsim\frac{\lambda^2}{\delta\lambda\,|\sin\theta|}
\ge\frac{\lambda^2}{\delta\lambda}=1.289\,\mathrm m.}
$$
The $1.29\,\mathrm m$ value is an ideal lower bound approached near grazing diffraction. A usable finite angle needs greater width; for example $\theta=60^\circ$ requires about $1.49\,\mathrm m$. The exact sine first-zero relation still gives the lower bound when the local angular approximation fails near grazing.
Such a coherently illuminated grating is very large and needs accurate groove placement and suitable collection optics; an unaided far-field screen would require distances much greater than $W^2/\lambda\sim2\times10^6\,\mathrm m$. Focusing optics avoid that screen distance but not the required <resolving power of a diffraction grating>. Real linewidth, slit width and aberrations can require more resolution. The unequal <isotope> abundances also make this ideal Rayleigh estimate less decisive than for equally bright lines.
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