Polarization by Reflection — Question 3

Question 3

The surviving s component fixes the reflected power at Brewster incidence

For a lossless nonmagnetic interface use $r_s=(n_1\cos i-n_2\cos t)/(n_1\cos i+n_2\cos t)$. At Brewster incidence $t=90^\circ-i$ and $\tan i=n_2/n_1$. An unpolarized beam carries incident power P0.

1. Simplify the s-polarized power reflectance at this angle.

2. For n2/n1=1.50, find the fraction of total incident power reflected and the reflected polarization.

Question 3 - Solution

The surviving s component fixes the reflected power at Brewster incidence

(a) Method and calculation

$$\cos i=\frac{n_1}{\sqrt{n_1^2+n_2^2}},\quad\cos t=\frac{n_2}{\sqrt{n_1^2+n_2^2}}.$$$$\boxed{R_s=\left(\frac{n_1^2-n_2^2}{n_1^2+n_2^2}\right)^2}.$$

(b) Interpretation and further calculation

$$n_2/n_1=3/2:\quad R_s=\boxed{25/169=0.147929},\quad R_p=0.$$$$\boxed{P_r/P_0=R_s/2=25/338=0.0739645};\quad P_{r,p}=0.$$

Physical check

Only half the unpolarized input power starts in the s component, so the total reflectance is half its s reflectance.