This question concerns the modes of an asymmetric dielectric slab waveguide. A summary of the pertinent theory is rst provided: For an asymmetric slab, the decay constants in the cladding are no longer equal and the field in the core is no longer a pure cosine or sine but a linear superposition of the two which is equivalent to a spatial phase shift. The higher index cladding will have a smaller decay constant than the lower index cladding. Consequently, the mode profile within the core is pulled towards the higher index of the upper or lower cladding to satisfy the boundary conditions. If the two cladding refractive indices approach equality, the spatially phase shifted solutions smoothly evolve into the even parity and odd parity solutions of the symmetric slab-waveguide and may be categorized as such. Following a similar analysis as for the symmetric slab waveguide but accounting for the asymmetry yields the dispersion equations _ _ _ _1 CO1: _ _1 su . Kdm2+2tan (Kd)+2tan (Kid) , TE 2 2 Kd = mE + -1tam_1 HEW Ycovd + -1tan_1 \"gar ysubd ; TM 2 2 ""601: Kd 2 nsub Kd where: 2 2 _ 2 2 \"C(3ka _ [5) _ Two 2 2 _ 2 2 ncorko _ f3 + K 2 2 _ 2 2 \"51!.ka _ l3 _ ysub m is an integer mode index and the subscripts cov, cor, sub refer to cover, core, and substrate regions respectively. All other symbols have their usual meaning. Show that in the case of a symmetric slab waveguide these dispersion equations reduce to those given in (14. Use these equations to explain why the cut-off frequency of the fundamental mode of a symmetric slab waveguide is zero, but not zero when the slab is asymmetric (i.e. the substrate index not equal to cover index). Give a physical explanation for why the dispersion equation for TE and TM differ