Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

The problems of this section use the slab relations in normalised form,

NA = n_{1}^{2} - n_{2}^{2}, V = \frac{π d}{λ _{0}} NA, u^{2} + w^{2} = V^{2}, f_{c, m} = \frac{m c}{2 d NA}, TE_{m} : w = u tan (u - \frac{mπ}{2}), TM_{m} : w = \frac{n _{2}^{2}}{n _{1}^{2}} u tan (u - \frac{mπ}{2}), h = \frac{2 u}{d}, α = \frac{2 w}{d}, n_{eff} = \frac{τ}{k _{0}}, τ = k_{0}^{2} n_{1}^{2} - h^{2}, Γ = \frac{1 + \frac{s i n 2 u}{2 u}}{1 + \frac{s i n 2 u}{2 u} + \frac{c o s ^{2} u}{w}},

the coated conductor (the modes of a slab of thickness $2 t$ with $E_{tan} = 0$ at the metal, $V = k_{0} t NA$ , $Z_{s} = j α / (ω ε_{0})$ for TM), and the fibre relations $sin θ_{a} = NA$ , $sin θ_{c} = n_{2} / n_{1}$ , $V = k_{0} a NA$ , single-mode for $V < 2.405$ , $Δ T = (L n_{1} / c) (n_{1} - n_{2}) / n_{2}$ (step index) and $L n_{1} Δ^{2} / (8 c)$ (optimum graded index). The constants are $c = 3 \times 1 0^{8} m/s$ , $μ_{0} = 4 π \times 1 0^{- 7} H/m$ and $ε_{0} = 1 0^{- 9} / (36 π) F/m$ , so that $η_{0} = 120 π Ω$ .

7.6.14 Problems and solutions

In this part

Problems · 4

Home exercises · 2

Further exercises · 5

Worked examples · 2