Question 9.16: RESTSTRAHLEN ABSORPTION Figure 9.22 shows the infrared extin...
RESTSTRAHLEN ABSORPTION Figure 9.22 shows the infrared extinction coefficient K of GaAs and CdTe. Consider CdTe. Calculate the absorption coefficient α and the reflectance R of CdTe at the Reststrahlen peak, and also at 50 μm and at 100 μm. What is your conclusion?
At the resonant peak, λ ≈ 72 μm, K ≈ 6, and n ≈ 5, so the corresponding free-space wavevector is
k_{o}=\frac{2 \pi}{\lambda}=\frac{2 \pi}{72 \times 10^{-6} m }=8.7 \times 10^{4} m ^{-1}
The absorption coefficient α, by definition, is 2 k^{\prime \prime} in Equation 9.57, so
\frac{d I}{d z}=-2 k^{\prime \prime} I [9.57]
\alpha=2 k^{\prime \prime}=2 k_{o} K=2\left(8.7 \times 10^{4} m ^{-1}\right)(6)=1.0 \times 10^{6} m ^{-1}
which corresponds to an absorption depth 1 ∕ α of about 1 μm. The reflectance is
R=\frac{(n-1)^{2}+K^{2}}{(n+1)^{2}+K^{2}}=\frac{(5-1)^{2}+6^{2}}{(5+1)^{2}+6^{2}}=0.72 \quad \text { or } \quad 72 \%
Repeating the above calculations at \lambda = 50 \mu m, we get \alpha = 8.3 \times 10^{2} m^{−1}, and R = 0.11 or 11 percent. There is a sharp increase in the reflectance from 11 to 72 percent as we approach the resonant peak. At λ = 100 μm, α = 6.3 × 10³ m^{−1} and R = 0.31 or 31 percent, which is again smaller than the peak reflectance. R is maximum around the Reststrahlen peak.