A diffraction grating of period has illuminated grooves

PHYSICS · Subjective · Fermi University Physics · Optics & Photonics · Interference and diffraction · Interference and diffraction

What an Instrument Can See, Ch Scalar diffraction theory, derived

Problem

A diffraction grating of period Λ\Lambda has NN illuminated grooves, so its illuminated width is L=NΛL=N\Lambda. Its amplitude transmittance can be written as an infinite comb of period Λ\Lambda multiplied by a window of width LL. (a) Using the convolution theorem, argue that the diffracted amplitude near the mmth order (at sin⁡θm=mλ/Λ\sin\theta_m=m\lambda/\Lambda) is a narrow peak whose width in sin⁡θ\sin\theta, from its centre to its first zero, is λ/L\lambda/L. (b) Differentiate the grating equation Λsin⁡θ=mλ\Lambda\sin\theta=m\lambda to relate a small wavelength change dλd\lambda to a small change d(sin⁡θ)d(\sin\theta) in the position of order mm. (c) Apply the Rayleigh criterion (two wavelengths just resolved when the peak of one order falls on the first zero of the other) to show that the resolving power is R=λ/Δλ=mNR=\lambda/\Delta\lambda=mN, independent of Λ\Lambda. (d) A grating ruled at 800800 lines per millimetre is illuminated over a width of 75 mm75\ \mathrm{mm} and used in second order (m=2m=2) to examine the Hβ\beta line at λ=486.1 nm\lambda=486.1\ \mathrm{nm}. Find NN, RR, and the smallest wavelength difference Δλ\Delta\lambda the grating can resolve at this order.

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