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?


You should use the complex valued refractive indices, not the absute values, in the Fresnel formulas. Data on metal RIs can be found on. Refractiveindex.info. You can not rely on simple formulas as suggested in another answer.

  • $\begingroup$ I agree with this position. What I developed is a very simple model of reflection on a metal. The goal was simply to show how one could use a complex index. $\endgroup$ – Vincent Fraticelli Jan 9 at 18:59

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 $R\simeq 1-\frac{2}{N(\omega )}$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.