Question 9.12: COMPLEX REFRACTIVE INDEX Spectroscopic ellipsometry measurem...
COMPLEX REFRACTIVE INDEX Spectroscopic ellipsometry measurements on a silicon crystal at a wavelength of 826.6 nm show that the real and imaginary parts of the complex relative permittivity are 13.488 and 0.038, respectively. Find the complex refractive index, the reflectance and the absorption coefficient α at this wavelength, and the phase velocity.
We know that \varepsilon_{r}^{\prime}=13.488 \text { and } \varepsilon_{r}^{\prime \prime}=0.038 . Thus, from Equation 9.60, we have
n^{2}-K^{2}=13.488 \quad \text { and } \quad 2 n K=0.038
We can take K from the second equation and substitute for it in the first equation,
n^{2}-\left(\frac{0.038}{2 n}\right)^{2}=13.488
This is a quadratic equation in n^{2} that can be easily solved on a calculator to find n = 3.67. Once we know n, we can find K = 0.038 ∕ 2n = 0.00517. If we simply take the square root of the real part of \epsilon _{r,} we would still find n = 3.67, because the extinction coefficient K is small. The reflectance of the Si crystal is
R=\frac{(n-1)^{2}+K^{2}}{(n+1)^{2}+K^{2}}=\frac{(3.67-1)^{2}+0.00517^{2}}{(3.67+1)^{2}+0.00517^{2}}=0.327
which is the same as simply using (n-1)^{2} /(n+1)^{2}=0.327 , because K is small.
The absorption coefficient α describes the loss in the light intensity I \text { via } I=I_{o} \exp (-\alpha z) By virtue of Equation 9.57,
\frac{d I}{d z}=-2 k^{\prime \prime} I [9.57]
\alpha=2 k^{\prime \prime}=2 k_{o} K=2\left(\frac{2 \pi}{826.6 \times 10^{-9}}\right)(0.00517)=7.9 \times 10^{4} m ^{-1}
Almost all of this absorption is due to band-to-band absorption (photogeneration of electron–hole pairs).
The phase velocity is given by
v=\frac{c}{n}=\frac{3 \times 10^{8} m s ^{-1}}{3.67}=8.17 \times 10^{7} m s ^{-1}