The chapter sums a grating's slits “as a geometric series

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

Optics and Photonics, Ch Fraunhofer diffraction from apertures and gratings

Problem

The chapter sums a grating's NN slits “as a geometric series, as Chapter 11 summed the beams of an etalon”, and simply states the result. Each of the NN slits radiates a wavelet of the same amplitude E0E_0 at angle θ\theta, and successive slits' contributions differ in phase by δ=kdsin⁡θ\delta = kd\sin\theta, so the resultant phasor sum is E(θ)=E0∑j=0N−1eijδE(\theta) = E_0\sum_{j=0}^{N-1}e^{ij\delta}. ((a)) Sum the geometric series in closed form, and hence show that the irradiance is I(θ)=I0(sin⁡(Nδ/2)sin⁡(δ/2))2I(\theta) = I_0\left(\dfrac{\sin(N\delta/2)}{\sin(\delta/2)}\right)^2, where I0=∣E0∣2I_0 = |E_0|^2 is the irradiance one slit alone would give on axis. ((b)) The expression in part (a) is 0/00/0 at a principal maximum, δ=2πm\delta = 2\pi m. Show, by taking the limit δ→2πm\delta\to2\pi m, that the irradiance there is N2I0N^2I_0. ((c)) Set N=2N=2 in the closed form of part (a) and show that it reduces to the two-slit law I=4I0cos⁡2(δ/2)I=4I_0\cos^2(\delta/2) that Chapter 10 used for point slits, confirming that the grating result contains Young's fringes as its simplest case. ((d)) Eight slits of a grating are illuminated, and at some angle off a principal maximum δ=0.30 rad\delta = 0.30\ \text{rad}. Find I/I0I/I_0 at that angle, and compare it with the height N2I0N^2I_0 of the principal maximum itself.

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