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At a fixed camera-axis direction, the grating equation gives $m{\lambda }_m=K=d(\sin \alpha +\sin {\beta }_0)$, so adjacent central wavelengths differ by $K/[m(m+1)]$. This differs from the exact wavelength cell where an order is nearest the optical camera axis. For ${\beta }_0=0$ and $x=f\tan \beta$, equal distances of orders $m$ and $m+1$ mean opposite diffraction angles; adding their diffraction grating equations gives $\lambda =K/(m+1/2)$. The full order-$m$ cell is bounded by $K/(m+1/2)$ and $K/(m-1/2)$, hence has width $K/(m^2-1/4)$ for $m\ge 2$ and accessible neighbouring orders. Both widths approach $K/m^2$ at high order.

With fixed spectral resolving power, wavelength interval, diffraction grating geometry, photodetector sampling and collimator focal ratio, a beam diameter scales as the spaxel width $\theta$, but the optical camera focal length stays fixed. A simple on-axis beam-envelope model therefore gives collimator and disperser volumes proportional to ${\theta }^3$ and optical camera volume proportional to ${\theta }^2$. With a fixed number of spaxels per photodetector, replication multiplies these volumes by $M\propto {\theta }^{-2}$. A fixed photodetector field adds transverse clearances and changes the small-$\theta$ limit; these exponents are not universal mechanical-volume laws.

For paraxial rays, the Lagrange optical invariant is $H=n(y_1{u}_2-y_2{u}_1)$. Lossless paraxial ray transfer preserves this quantity: in coordinates $(y,nu)$, free propagation and a thin lens act by matrices with determinant one, which preserve the oriented area of two ray vectors. Consequently a full angular optical slit width $\theta$ at an optical pupil diameter $D$ satisfies $D\theta =A_{\mathrm{cam}}p/f_{\mathrm{cam}}$ for its image width $p$ and optical camera optical pupil width $A_{\mathrm{cam}}$ in the same meridional plane.

An optical camera images incident beams on a photodetector; a spectrograph optical camera maps a diffracted direction into a position in its focal plane.

The optical telescope focuses the sky onto an optical slit; the collimator turns the selected light into a beam incident on the reflection diffraction grating; the optical camera images each diffracted direction onto the photodetector. The drawing also shows the monochromatic optical slit image for uniform optical slit illumination in the geometric, slit-limited approximation.

Near the chosen wavelength $\lambda$, a small wavelength increment moves the image by $q\Delta \lambda$, where $q$ is the magnitude of the linear spectral dispersion. An optical slit image of physical width $p$ therefore has an apparent spectral width $p/q$. With one optical slit width as the adopted separation criterion,Uniform illumination gives a top-hat monochromatic profile of width $p$; an unresolved stellar image need not illuminate the optical slit uniformly, so its profile follows the actual illumination convolved with the instrument response. Negligible optical slit diffraction and optical camera optical aberrations do not remove the finite diffraction grating diffraction width. The formula is the slit-limited result when that width is small compared with $p$, rather than a universal exact line-profile criterion.

Differentiate the grating equation with incidence fixed:The second formula is local to a spectrum centered on the optical camera optical axis, where $dx=f_{\mathrm{cam}}d\beta$ to first order. Since the illuminated surface width $W$ projects to optical camera beam width $A_{\mathrm{cam}}=W\cos \beta$,The expression takes positive diffraction order and width magnitudes. Off-axis, a photodetector mapping $x=f_{\mathrm{cam}}\tan (\beta -{\beta }_0)$ also has a ${\sec }^2(\beta -{\beta }_0)$ factor; the exam's compact formula uses the local paraxial mapping.

The fundamental angular floor is the diffraction limit of a telescope, of order $\lambda /D$. The ideal diffraction grating has $R_g=mW/d$, whereas its slit-limited resolving power has $R\theta D=mW\lambda /d$. Requiring $R\le R_g$ givesThe coefficient depends on the adopted line-resolution and aperture conventions. Below this scale, a narrower spaxel oversamples the same spatial mode rather than creating a new independently resolved element. Equivalently a spatial optical mode occupies etendue of order ${\lambda }^2$.

There is also an engineering floor before arbitrary shrinkage: the optical camera focal length, photodetector size and support clearances do not shrink, while the number of photodetectors increases as ${\theta }^{-2}$. Thus the shrinking-beam model's constant combined optical camera volume is eventually overtaken by fixed per-camera overheads. For fixed brightness, smaller spaxels receive fewer photons, so read noise and photon shot noise can constrain useful sampling before the formal optical floor. Diffraction limits independent spatial information; fixed photodetector and mechanical dimensions limit practical volume reduction.

The number of spaxels is $M=(\beta /\theta {)}^2$. To obtain definite scaling exponents, keep the spectral resolving power, wavelength interval, diffraction grating angles and groove spacing, input focal ratio, and photodetector sampling of a spectral resolution element fixed. A fixed photodetector then accommodates a fixed number $M_0$ of spectra, so $N\simeq M/M_0$ (rounded up in an actual instrument). The corresponding etendue per spaxel scales as ${\theta }^2$. Angular ratios are independent of whether both angles are in arcseconds; optical invariant equations use radians.

Write $b$ for the collimated beam diameter. The diffraction grating result $R\theta D=W(\sin \alpha +\sin \beta )$ gives $b\propto W\propto \theta$. A fixed collimator focal ratio gives $f_{\mathrm{coll}}\propto b\propto \theta$. With a fixed image width $p$ in photodetector detector pixels, the invariant $\theta D=p{A}_{\mathrm{cam}}/f_{\mathrm{cam}}$ and $A_{\mathrm{cam}}\propto \theta$ instead imply $f_{\mathrm{cam}}$ is constant: the optical camera focal ratio increases as $1/\theta$.

In a simple on-axis beam-envelope model, each collimator and collimated disperser space has area proportional to $b^2$ and length proportional to $b$, whereas the optical camera cone has area proportional to $b^2$ and fixed length. HenceThe constants include $1/M_0$ and fixed design parameters. At fixed field size, the combined collimator and diffraction grating volumes scale as $M{\theta }^3\propto \theta \propto M^{-1/2}$, while the combined optical camera beam-cone volume scales as $M{\theta }^2\propto 1$.

These are integral-field spectrograph volume scaling laws for the shrinking beam, not a claim that the whole apparatus can shrink without a floor. A photodetector of fixed transverse size $d_0$ needs space for its field: an optical camera envelope interpolating between optical pupil width $b$ and photodetector width $d_0$ has volume proportional to $f_{\mathrm{cam}}(b^2+b{d}_0+d_0^2)$, rather than just $f_{\mathrm{cam}}{b}^2$. Replication then adds terms scaling as $M\theta$ and $M$, which can grow as the spaxels shrink. Other field clearances and mechanical margins also change the asymptote. Without the fixed spectral/design assumptions above, the information in the question does not determine unique physical-volume exponents. The role of fixed photodetector size and beam spread is discussed in Allington-Smith's instrument scaling model.

Let ${\beta }_0$ be the direction of the optical camera optical axis. The grating equation on that axis isThe central wavelengths in adjacent diffraction orders are therefore ${\lambda }_m=K/m$ and ${\lambda }_{m+1}=K/(m+1)$, with separation$K$ 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 $x=f_{\mathrm{cam}}\tan \beta$. 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 widthfor $m\ge 2$ 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.

Choose photodetector $x$ along increasing main grating dispersion, and $y$ along increasing wavelength for the cross-disperser. The echellogram shows two order traces and their intersections with the $y$ axis, which passes through the optical camera optical axis.

For fixed incidence, $d\beta /d\lambda =m/(d\cos \beta )>0$, so increasing wavelength goes to the right along each order. The cross-disperser also sends increasing wavelength upward in the chosen orientation; the traces therefore slope upward. At $x=0$, ${\lambda }_m=K/m>{\lambda }_{m+1}=K/(m+1)$, so the trace labeled $m$ lies above the trace labeled $m+1$. Reflecting either plotting axis would reverse its arrow but leave the optical content unchanged. The cross-disperser separates orders that would overlap in the main dispersion direction.