Question 9.2: GaAs DISPERSION RELATION For GaAs, from λ = 0.89 to 4.1 μm, ...
GaAs DISPERSION RELATION For GaAs, from λ = 0.89 to 4.1 μm, the refractive index is given by the following dispersion relation,
n^{2}=7.10+\frac{3.78 \lambda^{2}}{\lambda^{2}-0.2767} [9.18]
where λ is in microns (μm). What is the refractive index of GaAs for light with a photon energy of 1 eV?
At hf = 1 eV,
\lambda=\frac{h c}{h f}=\frac{\left(6.62 \times 10^{-34} J s \right)\left(3 \times 10^{8} m s ^{-1}\right)}{\left(1 eV \times 1.6 \times 10^{-19} J eV ^{-1}\right)}=1.24 \mu m
Thus,
n^{2}=7.10+\frac{3.78 \lambda^{2}}{\lambda^{2}-0.2767}=7.10+\frac{3.78(1.24)^{2}}{(1.24)^{2}-0.2767}=11.71
so that n = 3.42
Note that the n versus λ expression for GaAs is actually a Sellmeier-type formula because when \lambda^{2} \gg \lambda_{1}^{2} , then A_{1} can be simply lumped with 1 to give 1 + A_{1} = 7.10.