A spectrometer slit has a rectangular response of height and width

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Optics & Photonics · Interference and diffraction

Mathematical Methods for Physics, Ch Fourier transforms, spectra and convolution

Problem

A spectrometer slit has a rectangular response of height 11 and width w=0.40 nmw = 0.40\,\mathrm{nm}, so g~(k)=2sin⁡(kw/2)/k\tilde g(k) = 2\sin(kw/2)/k and g~(0)=w\tilde g(0) = w. Two equal, narrow spectral lines sit at x=±d/2x = \pm d/2. Each line's own transform stays within 2 per cent of its value at k=0k = 0 for ∣k∣<16 nm−1|k| < 16\,\mathrm{nm^{-1}}, so treat it as constant there. The transform of the record, f~(k)g~(k)\tilde f(k)\tilde g(k), is measured with an error of fixed size at every kk, at most 1 per cent of the record's value at k=0k = 0. The true transform f~(k)\tilde f(k) is recovered by dividing the record by g~(k)\tilde g(k). ((a)) Show that the error in the recovered f~(k)\tilde f(k), as a fraction of f~(0)\tilde f(0), is at most 0.01/∣sin⁡(kw/2)kw/2∣0.01\big/\left|\dfrac{\sin(kw/2)}{kw/2}\right|. Find the largest kmax⁡k_{\max} such that this stays within 10 per cent for every kk from 00 to kmax⁡k_{\max}. ((b)) Show that the pair of lines has f~(k)=f~(0)cos⁡(kd/2)\tilde f(k) = \tilde f(0)\cos(kd/2) for ∣k∣<16 nm−1|k| < 16\,\mathrm{nm^{-1}}, while a single line would give a constant. For d=0.20 nmd = 0.20\,\mathrm{nm}, evaluate at k=10.0 nm−1k = 10.0\,\mathrm{nm^{-1}} the error bound from part (a) and the fall 1−f~(k)/f~(0)1 - \tilde f(k)/\tilde f(0). Can this record tell the pair from a single line? ((c)) Repeat part (b) for lines 0.05 nm0.05\,\mathrm{nm} apart, and decide whether the record can separate them.

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