Module 5 d

# Module 5 d - Module 5 Lecture 4 Dielectric...

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Module 5 Lecture 4 Dielectric waveguides (including optical fibres)

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Contents Modes of the infinitely wide slab waveguide Ray optics equivalent solutions Setting up Maxwell’s equations for dielectric structures Channel waveguide structures Optical fibre waveguides Optical devices (“Photonics”)
page 3 The infinite dielectric slab n 0 n 1 n 2 x y End view: waveguide axis into the slide

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Ray optic approach to waveguide resonance imagine a truncated plane wave bouncing inside a high index dielectric through TIR Phase fronts must be continuous while bouncing up and down This imposes an additional condition on the allowable ray angles Side view: waveguide axis horizontal
Matching the phases for the two paths Only those ray angles that satisfy this equation are allowed inside the guide. For a given wg (a, n 1 ,n 0 ) and wavelength ( l = 2 p /k), there is only a fixed number of solutions (one per value of m). NB: here k = k 0 This ensures that the waves are in phase as they propagate along z. The phase shift Φ appearing in the equation is occurring because the wave is totally reflected. It is called a GOOS-HÄNCHEN shift and it depends on the polarization of the light relative to the plane of incidence.

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Goos-Hänchen shift upon TIR For light polarized perpendicular to the plane of incidence (TE), the amplitude reflection coefficient is complex: r = (n 1 sin f + j(n 1 2 cos 2 f n 0 2 ))/ (n 1 sin f - j(n 1 2 cos 2 f n 0 2 )) r = exp(-j Φ ) (i.e. total reflection but with a phase shift) Φ TE = -2 tan -1 (2 D /sin 2 f 1) ½ For TM polarization, the shift is different: Φ TM = -2 tan -1 ( (n 1 /n 0 ) 2 (2 D /sin2 f 1) ½ ) NB GH shifts are usually defined as half those given here
Matching the phases imposes a “dispersion relation” between f and φ (equivalent to b ) Only those ray angles that satisfy this equation are allowed inside the guide. For a given waveguide (a, n 1 ,n 0 ) and wavelength ( l = 2 p /k), there is only a fixed number of solutions (one per value of m). Recall that Φ depends on polarization, hence solutions are different for TE and TM NB: here k = k 0

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Mode derivation from Maxwell’s equations Ray picture has its limitations Maxwell’s equations and problem geometry provide full exact solutions for modes. Derive a wave equation as usual from M.E. Impose uniformity of the geometry along one direction (z) in order to find mode- like solutions of “pure” waveguides. Non uniformities in z will be dealt with later. Are there solutions that have mode-like behaviour, i.e. a field dependence that is periodic along z and no power propagating away from the core (i.e. from the z axis)?
Postulates of the wave analysis Single frequency (monochromatic) Note that even for GHz modulation of light signals, Dw/w is of the order of 10 -5 NB: w 2 e 0 m 0 = w 2 / c 2 = k 0 2 = (2p/l) 2 (and k = nk 0 ) – Higher refractive index means larger k, smaller l and smaller phase velocity

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Module 5 d - Module 5 Lecture 4 Dielectric...

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