Light Propagation and Index of Refraction

If one visualizes the material where the light is propagating as continuous, one can use Maxwell´s Equations to solve the propagation of waves and this leads to equation 1

$$\frac{\partial^2\boldsymbol{E}}{\partial x^2}+ \frac{\part... ... t^2}+ \mu \sigma \frac{\partial\boldsymbol{E}}{\partial t}$$ (1)

This equation is in orthogonal Cartesian coordinates and $\boldsymbol{E}$ stands for electric field, $t$ for time and $\sigma$ is material conductivity. The last term $\mu \sigma \partial\boldsymbol{E}/\partial t$ is a first temporal order derivative. The time rate-of-change of $\boldsymbol{E}$ generates a voltage, currents circulate and since the material is resistive, the part of light energy is converted to thermal energy and so absorption occurs. If the permittivity is reformulated as a complex quantity, the expression can be reduced to the unattenuated wave equation. This leads to a complex index of refraction $\tilde{n}$, which is tantamount to absorption [1]. It can be expressed as

$$\tilde{n}=n_{\mathrm{real}} -ik$$ (2)

where $n_{\mathrm{real}}=n$ is index of refraction usually defined as the ratio $n=c/v$ where $c$ is the speed of electromagnetic waves in vacuum($3\times10^8$ m/s) and $v$ the speed of waves in the material. The $k$ is called imaginary part of refractive index. Note that both $n$ and $k$ are real numbers. As the wave progresses trough the medium, its amplitude is exponentially attenuated. It can be summarized by following equation

$$I_{d}=I_{0}\exp{(-K\times d)}$$ (3)

where $I_0$ stands for intensity at the interface of our absorbing material and $K=2\omega k/c$ is called linear absorbtion coefficient or attenuation coefficient. This is called Lambert law of classical optics.¹ It may be also expressed like this

$$K=\frac{4\pi k}{\lambda}$$ (4)

where the $k$ is absorbtion index and $\lambda$ is wavelength of light in the vacuum. The absorbtion constant $\kappa$ used in next sections is proportional to linear absorbtion coefficient $K$. The values of $K$ and $k$ are different for many orders of magnitude in various spectral regions. The flux density will drop by factor of $e^{-1} \approx 1/3$ after the wave has propagated a distance $1/K$, known as skin or penetration depth. For the material to be transparent the penetration depth must be large in comparison to its thickness. The penetration depth for semiconductors and metals is exceptionally small. For example, copper in the ultraviolet region ($\lambda = 100$ nm) has a miniscule penetration depth, about 0.6 nm, while it is still only about 6 nm at the infrared region ( $\lambda = 10 \mu$m). It explains generally observed opacity of metals[1]. For semiconductor materials, values are of course much higher. Now should be checked how the reflectance $R$ is affected by this complex index of refraction. It can be proven using Fresnel's equations that if a plane wave incidents normal (i.e. perpendicular) to a plane surface between vacuum ($n=1$) and material of complex refractive index $\tilde{n}$ the reflectance $R$ will be

$$R=\frac{I}{I_0}=\left(\frac{\tilde{n}-1}{\tilde{n}+1}\right)^2=\frac{(n-1)^2+k^2}{(n+1)^2+k^2}$$ (5)

Just for illustration at $\lambda=589.3$ nm can be calculated this set of values for a gallium single crystal $n_R=3.7$, $n_I=5.4$ and $R=0.7$ for normal incidence[3].