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An inverse-square field spreads through area proportional to squared distance. More generally the flux contribution of a point source is proportional to the solid angle subtended by a surface element; the oriented solid angles sum to $4\pi$ for an enclosed source and zero for an exterior one. Superposition extends this to a mass distribution.

Newton's law of universal gravitation gives $\mathbf{g}=-GM\hat{\mathbf{r}}/r^2$ for a point mass. A surrounding sphere therefore has flux $-4\pi GM$, soFor the tunnel, approximate its water as an infinite uniform cylinder with $R=3.5\mathrm{m}$ and density $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$. The bob is at radial distance $r=R+1.5\mathrm{m}=5.0\mathrm{m}$. A coaxial Gaussian surface givesFor the plumb-line deflection from cylindrical gravity, horizontal equilibrium gives $d/L\simeq |g_r|/g$. Draining removes the attraction toward the tunnel, so the bob shifts away byThe unchanged rock and ordinary gravitational field cancel in the before/after comparison; end effects and mechanical disturbances are neglected.

For the tidal water loading of a spring balance, approximate the extra water at high tide as an infinite plane slab of thickness $h=8\mathrm{m}$, with the suspended mass above it. The gravitational field of an infinite plane sheet, derived with a pillbox Gaussian surface, gives added downward field $\Delta g=2\pi G\rho h$. With initial spring extension $x_0=mg/k=1\mathrm{m}$,The extension increases at high tide. These estimates use $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and the stated water-density approximation; salinity, finite estuary geometry and terrestrial tidal deformation alter a real measurement.

Represent each wave by a complex phasor. Add phasors before squaring: two amplitudes give $I\propto |A_1+A_2{e}^{i\delta }{|}^2=A_1^2+A_2^2+2A_1{A}_2\cos \delta$. For $N$ equally illuminated narrow slits, adjacent phasors differ by $\delta =2\pi d\sin \theta /\lambda$, soAll phasors align when $\delta =2\pi m$, giving the grating equationThe nearest zero has adjacent-phase increment $2\pi /N$: the phasor polygon closes once. Thus $d\cos {\theta }_m\Delta \theta \simeq \lambda /N$, givingThe angular linearization is valid away from grazing directions; the underlying exact first-zero condition is $d(\sin {\theta }_{\mathrm{zero}}-\sin {\theta }_m)=\pm \lambda /N$.

For the first subsidiary maximum of a large slit array, use the stipulated $1.5$ phasor wraps: $N\delta /2\simeq 3\pi /2$ relative to a principal maximum. Its amplitude is approximately $2N/(3\pi )$ times one slit's amplitude. ThereforeThis uses the question's approximate maximum position, not the exact extremum of the slit-array intensity.

The Rayleigh criterion places one line's principal maximum at the other's first minimum. Grating dispersion then gives $|m|N\gtrsim \lambda /\delta \lambda =1.674\times 10^6$. Combining with normal incidence, $|m|\lambda =d|\sin \theta |$, gives the normal-incidence grating width bound: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 $W^2/\lambda \sim 2\times 10^6\mathrm{m}$. 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.