Why is reflectivity the square of the magnitude of the Fresnel reflection coefficient? Is there a simple way to derive the formula $R=(\frac{n_1-n_2}{n_1+n_2})^2$ by using high school physics and mathematics?
 A: This is only for a special case where light shines at normal incidence.
The general formula is derived from Snell's law and the law of reflection.
Let $\theta_i$ be the angle of incidence, $\theta_r$ angle of reflection, and $\theta_t$ angle of refraction.
Reflectance for s-polarised light: $R_s=(\frac{Z_2cos\theta_i-Z_1cos\theta_t}{Z_2cos\theta_i+Z_1cos\theta_t})^2$
Similarly, reflectance for p-polarised light: $R_p=(\frac{Z_2cos\theta_t-Z_1cos\theta_i}{Z_2cos\theta_t+Z_1cos\theta_i})^2$
Substituting wave impedance $Z_i=\frac{Z_0}{n_i}$, we get:
$R_s=(\frac{n_1cos\theta_i-n_2cos\theta_t}{n_1cos\theta_i+n_2cos\theta_t})^2$
$R_p=(\frac{n_1cos\theta_t-n_2cos\theta_i}{n_1cos\theta_t+n_2cos\theta_i})^2$
Effective reflectance $R_{eff}=\frac{1}{2}(R_s+R_p)$.  In the case of normal incidence, shown in this image, $\theta_i=\theta_t=0$.
 
As such, $R=(\frac{n_1-n_2}{n_1+n_2})^2$ since $cos(0)=1$.
A: The answer is that the Fresnel coefficient refers to the reflection of the electric field $E$, whereas the “reflectivity,” as you’re calling it, refers to the power $P$. Since $P \propto E^2$, reflectivity is the square of the Fresnel coefficient. 
