= Solution
Let $\beta_0$ be the direction of the <optical camera> <optical axis>. The <grating equation> on that axis is
$$
m\lambda_m=d(\sin\alpha+\sin\beta_0)=K.
$$
The central <wavelengths> in adjacent <diffraction orders> are therefore $\lambda_m=K/m$ and $\lambda_{m+1}=K/(m+1)$, with separation
$$
\boxed{S_{\rm centres}=\lambda_m-\lambda_{m+1}=\frac{K}{m(m+1)},\qquad K=d(\sin\alpha+\sin\beta_0).}
$$
$K$ is the fixed order-times-wavelength product directed onto the <optical camera> axis, or the optical path difference between adjacent grooves in that direction. In a <Littrow configuration>, $K=2d\sin\alpha$. This adjacent-central-wavelength spacing is a conventional <free spectral range of an echelle grating>; at high order it is approximately $\lambda_m/m$.
There is a qualification in the wording of the printed definition. The exact interval for which order $m$ lies closest to the $y$ axis is not, in general, identical to the spacing of two adjacent order centres. For a concrete counterexample take $\beta_0=0$, $0<\alpha<\pi/2$, and a <photodetector> mapping $x=f_{\rm cam}\tan\beta$. Equality of the distances from the axis for orders $m,m+1$ means $\beta_m=-\beta_{m+1}$, so their common boundary is $\lambda=K/(m+1/2)$. The other boundary with order $m-1$ is $K/(m-1/2)$. Consequently the literal closest-order interval has width
$$
\boxed{S_{\rm nearest}=\frac{K}{m-1/2}-\frac{K}{m+1/2}=\frac{K}{m^2-1/4},}
$$
for $m\ge2$ and accessible non-grazing orders. This differs from the printed $K/[m(m+1)]$. For a nonzero $\beta_0$, the exact boundary is determined by $\beta_m+\beta_{m+1}=2\beta_0$ and need not even have the simple half-order form. The two definitions agree to leading order $K/m^2$ when $m$ is large. Thus the printed expression is correct as an adjacent-centre convention or a high-order coverage estimate, rather than an exact consequence of the literal nearest-axis definition. Gap-free coverage should be checked using the actual <photodetector> mapping and neighbouring order endpoints.
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