Polarization by Reflection — Question 2

Question 2

The reverse Brewster angle lies below the critical angle

A boundary separates indices nl and nh with nh greater than nl. Let thetaBL be the Brewster angle for low-to-high incidence, thetaBH for high-to-low incidence, and thetaC the critical angle for high-to-low incidence.

1. Prove the two Brewster angles are complementary.

2. Compare thetaBH with thetaC and evaluate both for nh/nl=1.50.

Question 2 - Solution

The reverse Brewster angle lies below the critical angle

(a) Method and calculation

$$\tan\theta_{BL}=n_h/n_l,\quad\tan\theta_{BH}=n_l/n_h.$$$$\boxed{\theta_{BL}+\theta_{BH}=90^\circ}.$$

(b) Interpretation and further calculation

$$\sin\theta_{BH}=\frac{n_l}{\sqrt{n_l^2+n_h^2}}<\frac{n_l}{n_h}=\sin\theta_C\Rightarrow\boxed{\theta_{BH}<\theta_C}.$$$$\boxed{\theta_{BH}=33.6901^\circ,\quad\theta_C=41.8103^\circ}.$$

Physical check

Brewster transmission remains a propagating ray in the high-to-low case; it occurs before the total-reflection threshold.