Polarization by Reflection — Question 9

Question 9

Small angular error produces quadratic p leakage near Brewster incidence

For air-to-index-n incidence, use $r_p=(n\cos i-\cos t)/(n\cos i+\cos t)$ and $\sin i=n\sin t$, with n greater than one. Set $i=\theta_B+\delta$, where $\theta_B=\tan^{-1}n$ and delta is small in radians.

1. Differentiate at Brewster incidence to find the leading reflected amplitude.

2. Find the leading power reflectance and estimate it for n=1.50 and delta=1.0 degree.

Question 9 - Solution

Small angular error produces quadratic p leakage near Brewster incidence

(a) Method and calculation

$$\left.\frac{dt}{di}\right|_B=\frac{\cos\theta_B}{n\cos t_B}=\frac1{n^2}.$$$$\left.\frac{dr_p}{di}\right|_B=\frac{-n\sin\theta_B+\sin t_B/n^2}{2n\cos\theta_B}=\boxed{-\frac{n^4-1}{2n^3}}.$$

(b) Interpretation and further calculation

$$\boxed{r_p\simeq-\frac{n^4-1}{2n^3}\delta,\quad R_p\simeq\left(\frac{n^4-1}{2n^3}\right)^2\delta^2}.$$$$\delta=\pi/180:\quad\boxed{R_p\simeq1.10340\times10^{-4}};\quad R_p(2\delta)\simeq4R_p(\delta).$$

Physical check

The amplitude crosses zero linearly while its squared magnitude has a quadratic minimum. The estimate neglects higher orders in angular error.