HW2 solutions

HW2 solutions - 2.5 Dielectric slab waveguide Consider a...

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Unformatted text preview: 2.5 Dielectric slab waveguide Consider a dielectric slab waveguide which has a thin GaAs layer of thickness 0.2 μ m between two AlGaAs layers. The refractive index of GaAs is 3.66 and that of the AlGaAs layers is 3.40. What is the cut-off wavelength beyond which only a single mode can propagate in the waveguide assuming that the refractive index does not vary greatly with the wavelength? If a radiation of wavelength 870 nm (corresponding to bandgap radiation) is propagating in the GaAs layer, what is the penetration of the evanescent wave into the AlGaAs layers? What is the mode field distance of this radiation? Solution Given n 1 = 3.66 (AlGaAs), n 2 = 3.4 (AlGaAs), 2 a = 2 × 10-7 m or a = 0.1 μ m, for only a single mode we need V = 2 π a λ n 1 2- n 2 2 ( ) 1/ 2 < π 2 ∴ λ > 2 π a n 1 2- n 2 2 ( ) 1/ 2 π 2 = 2 π (0.1 μ m) 3.66 2- 3.40 2 ( ) 1/ 2 π 2 = 0. 542 μ m. The cut-off wavelength is 542 nm. When λ = 870 nm, V = 2 π (1 μ m) 3.66 2- 3.40 2 ( ) 1/ 2 (0.870 μ m) = 0.979 < π /2 Therefore, λ = 870 nm is a single mode operation. For a rectangular waveguide, the fundamental mode has a mode field distance 2 w o = MFD ≈ 2 a V + 1 V = (0.2 μ m) 0.979 + 1 0.979 = 0.404 μ m. The decay constant α of the evanescent wave is given by, α = V a = 0.979 0.1 μ m = 9.79 ( μ m)-1 or 9.79 × 10 6 m-1 . The penetration depth δ = 1/ α = 1/ [9.79 ( μ m)-1 ] = 0.102 μ m. The penetration depth is half the core thickness. 2.8 A multimode fiber Consider a multimode fiber with a core diameter of 100 μ m, core refractive index of 1.475 and a cladding refractive index of 1.455 both at 850 nm. Consider operating this fiber at λ = 850 nm. a Calculate the V-number for the fiber and estimate the number of modes. b Calculate the wavelength beyond which the fiber becomes single mode. c Calculate the numerical aperture. d Calculate the maximum acceptance angle. e Calculate the modal dispersion Δ τ and hence the bit rate × distance product given that rms dispersion σ ≈ 0.29 Δ τ where Δ τ is the full spread. Solution Given n 1 = 1.475, n 2 = 1.455, 2 a = 100 × 10-6 m or a = 50 μ m and λ = 0.850 μ m. The V-number is, V = 2 π a λ n 1 2- n 2 2 ( ) 1/ 2 = 2 π (50 μ m) 1.475 2- 1.455 2 ( ) 1 / 2 (0.850 μ m) = 89.47 Number of modes M , M = V 2 2 = 89.47 2 2 ≈ 4002 The fiber becomes monomode when, V = 2 π a λ n 1 2- n 2 2 ( ) 1/ 2 < 2.405 or λ > 2 π a n 1 2- n 2 2 ( ) 1/ 2 2.405 = 2 π (50 μ m) 1.475 2- 1.455 2 ( ) 1/ 2 2.405 = 31.6 μ m For wavelengths longer than 31.6 μ m, the fiber is a single mode waveguide. m, the fiber is a single mode waveguide....
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This note was uploaded on 12/02/2009 for the course ECE 000 taught by Professor O during the Spring '09 term at UCSB.

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HW2 solutions - 2.5 Dielectric slab waveguide Consider a...

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