# Reflectivity with Complex Refractive Index

I would like to ask a followup question on a previous post found here.

Starting from the general expression for reflectivity: $$R = \left\lvert\frac{n_1-n_2}{n_1+n_2}\right\rvert^2$$

and substituting in the following expressions:

$$n_1 = \sqrt{{n^2_r}_1 +k_1^2}$$ $$n_2 = \sqrt{{n^2_r}_2 +k_2^2}$$

how does one obtain the provided solution of:

$$R = \frac{({n_r}_1-{n_r}_2)^2+(k_1-k_2)^2}{({n_r}_1+{n_r}_2)^2+(k_1+k_2)^2}$$

If we take the incident medium as air and the second material as a metal, then it is safe to assume that $$n_2 > n_1$$

and the first equation can be rewritten as:

$$R = \left(\frac{n_2-n_1}{n_1+n_2}\right)^2$$

I am unable to simplify this expression into the expression above. Is my initial assumption inappropriate? Or perhaps I am lacking an assumption?

Additionally, this solution is only applicable when radiation is normal to the interface (theta = 0). What would the expression for reflectance be when radiation is non-normal?

I'm not sure I understand how you use complex indices and modules. For a metal, the index is $${\,n}=N(\omega )\left( 1-j \right)$$ with $$N(\omega )=\sqrt{\frac{\gamma }{2{{\varepsilon }_{0}}\omega }}$$
The coefficient of reflection in amplitude is $$\underline{r}=\frac{\underline{{{E}_{0r}}}}{\underline{{{E}_{0i}}}}$$ or $$\underline{r}=\frac{1-\underline{n}}{1+\underline{n}}=\frac{1-\left( 1-j \right)N(\omega )}{1+\left( 1-j \right)N(\omega )}$$.
In energy $$R=\underline{r}{{\underline{r}}^{*}}=\frac{{{(1-N)}^{2}}+{{N}^{2}}}{{{(1+N)}^{2}}+{{N}^{2}}}$$ Usually, since $$N\gg 1$$, one does a limited development that leads to $$R\simeq 1-\frac{2}{N(\omega )}$$