I'm trying to manipulate some data to see if my analysis method is reliable: I want to use transmission and reflection measurements within a certain wavelength range to get the index of refraction (real and imaginary parts) of a material of very well established index of refraction, like regular silica glass.
The information I get out of the apparatus is the normalized transmission $T(\lambda)$ (the intensity transmitted through the material divided by the intensity transmitted through only air) and the normalized reflection $R(\lambda)$ (the intensity reflected off the material divided by the intensity reflected off a nearly perfect reflector). I also have the width of the sample, and the light is at normal incidence so I don't need to worry about any angle stuff or polarization.
So from what we learned in class the standard procedure is to write out the fields in each medium and make the boundary conditions ($E$ continuous and $dE/dx$ continuous) match up at the boundaries. My notation is visible in this diagram:
Where $k_1$ and $k_2$ are determined from $n_1$ and $n_2$ through $k = \frac{\omega}{c}n$. Applying the boundary conditions gives us 4 equations for the 5 coefficients and allows us to get $F/A$ and $B/A$ (which are the electric field amplitudes of the transmitted and reflected waves, respectively) in terms of only $k_1$,$k_2$, and $L$.
Now for a given wavelength, $T = \frac{|F|^2}{|A|^2}$ and $R = \frac{|B|^2}{|A|^2}$. So, that gives me two equations for two unknowns (the real and imaginary parts of $n_2$).
So, is there any reason this shouldn't work for solving (numerically, definitely not analytically) for both parts of $n_2$ if I have $k_1$, $L$, $T(\lambda)$ and $R(\lambda)$?
Best Answer
Rather than use the field strengths, you can use the Fresnel equations which give the fraction transmitted and reflected at an interface. For normal incidence, the reflectivity is $\left( \frac {n_1-n_2}{n_1+n_2}\right)^2$, which for air at index $1$ and glass at index $1.5$ gives $4\%$ reflection at each interface. The derivation of this reflectivity is basically what you are alluding to-it has been done for you. So if you can measure the reflectivity accurately, you are there.