This result was published
in: C. Summonte, Mat. Res. Soc. Symp. Proc. Vol. 297, 1993,
395
The absorption coefficient
$\alpha$ of a thin film material on a finite, transparent substrate can be
directly retrieved with very good approximation from reflectance and
transmittance ($R\&T$) using the formulation below.
Basing on the observation
that the quantity $T/(1-R)$ does not oscillate, the original paper shows how
the oscillating terms actually cancel out, thus yielding an invertible
equation.
In the calculations, the
following data are needed:
- $R,T =$ spectral reflectance and
trasmittance of a sample on a transparent substrate
- $n_0 =$ refractive index of the ambient
(Typically air, $n_0 = 1$)
- $n_2 =$ refractive index of the substrate
(suggested: 1.45 for quartz, 1.55 for borosilicate glass)
- $n =$ refractive index of the film. Due to
Kramers Kronig correlation, if an $\alpha \neq 0$ exists, that is why we
want to determine it, $n$ is typically wavelenght dependent in the range
of interest. However, a fixed $n$ is a good approximation. In the original
paper, a very simple procedure is suggested to evaluate accurately such
fixed $n$ from $R\&T$. The error on $\alpha$ deriving from an error on
$n$ is also quantified.
- $d =$ thickness of the film. If it is
known only approximately, the same error bar simply tranfers to $\alpha$.
|
$$T_i=\frac{T}{1-R}$$ |
$$C_c=2(n_0^2-n^2)(n^2-n_2^2)$$ |
|
$$T_s=\frac{4n_2n_0}{(n_2+n_0)^2}$$
|
$$C_A=(n_0-n)^2(n+n_2)^2$$ |
|
$$T_{NUM}=16n_0n_2n^2$$ |
$$C_B=(n_0^2+n)^2(n-n_2)^2$$ |
|
$$R_s=\frac{(n_2-n_0)^2}{(n_2+n_0)^2}$$
|
$$C_9=(n_0+n)^2(n+n_2)^2$$ |
|
$$Z=(C_A+C_B)^2-T_{NUM}^2+C_C^2$$
|
$$C_{10}=(n_0-n)^2(n-n_2)^2$$ |
|
$$Y=2C_C(C_A+C_B)$$ |
$$D=\frac{T_ST_{NUM}}{T_i}-R_S\frac{ZC_9-YC_C}{C_9^2-C_C^2}$$
|
$$\alpha
d=\ln\left[{\frac{D+\sqrt{D^2-4(C_9-C_A-R_SC_B)(C_{10}-C_B-R_SC_A)}}{2(C_9-C_A-R_SC_B)}}\right]$$