Past exam of the physics A course of the University of Cambridge / 2014 / Paper 2 / Section B / B8

For uniform illumination, the aperture is $h(x)=1$ for $|x|<a/2$ and zero otherwise. With $q=2\pi \sin \theta /\lambda$,Zeros occur at $\sin \theta =j\lambda /a$, $j\ne 0$; the central maximum extends between $\pm \lambda /a$. A monochromatic laser expanded into a collimated beam can illuminate the slit. A lens with the screen in its back focal plane produces the Fraunhofer pattern; alternatively use a sufficiently distant screen, $z\gg a^2/\lambda$. Align the slit and beam and avoid detector saturation.

For $N$ equally spaced slits, write the aperture as a convolutionThe convolution theorem converts this into a single-slit transform times the finite-array transform. ThereforePrincipal maxima are at $b\sin \theta =m\lambda$ for normal incidence, inside the single-slit envelope. Their height scales as $N^2$ times the envelope, and their first zeros are separated from a peak by $\delta (\sin \theta )=\lambda /(Nb)$.

The envelope-to-order-spacing ratio is $b/a$: increasing it admits more orders within the central envelope and changes the relative order strengths. In particular, an order is missing when $ma/b$ is a nonzero integer. Statements about absolute angular spacing need a fixed dimension: at fixed $a$, increasing $b$ reduces order spacing; at fixed $b$, decreasing $a$ broadens the envelope.

For fixed incident angle, differentiating the grating equation gives an order displacement $\delta (\sin \theta )=|m|\delta \lambda /b$. Equating this to the peak-to-first-zero width givesHere $W\simeq Nb$ is the illuminated grating width; this is the ideal Rayleigh limit, before slit width or aberrations broaden the spectral image.

With $b=(1600\mathrm{m}{\mathrm{m}}^{-1}{)}^{-1}=0.000625\mathrm{mm}$, the required count is $N\ge 372.7/(2\times 0.2)=931.75$. Thus 932 illuminated slits, or $W_{\min }\simeq 0.5825\mathrm{mm}$, suffice ideally. Checking diffraction-order feasibility, normal incidence cannot produce this second order because $2\lambda /b=1.193>1$; an oblique spectrograph geometry is required. A feasible reflection-Littrow example has $\sin \theta =m\lambda /(2b)=0.5963$, or $\theta \simeq 36.6^{\circ }$.