Polarization by Reflection — Question 8

Section 33-9 · Physics pages 5546–5547

Question 8 Two Brewster reflections select the overlap of their s directions An unpolarized beam of power P0 undergoes two successive Brewster reflections. Their s power reflectances are R1 and R2. In the plane transverse to the intermediate ray, the second s axis is rotated by phi from the first s axis. No other loss occurs between the interfaces. s1 s2 transverse plane of intermediate ray Schematic; labels, not drawing lengths, specify the data. (a) Find the power in the twice-reflected beam. (b) Compare phi=60 degrees and phi=90 degrees with aligned incidence planes.
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Question 8 - Solution Two Brewster reflections select the overlap of their s directions s1 s2 transverse plane of intermediate ray (a) Method and calculation P1 = R1P0/2, E1 ∥ˆs1. P2 = R2P1|ˆs1 · ˆs2|2 = 1 2R1R2P0 cos2 ϕ . (b) Interpretation and further calculation P2(60◦)/P2(0) = 1/4 . P2(90◦) = 0 . Physical check At crossed s axes, the first reflected polarization is p for the second interface and therefore has zero second reflection.

Question 8

Two Brewster reflections select the overlap of their s directions

An unpolarized beam of power P0 undergoes two successive Brewster reflections. Their s power reflectances are R1 and R2. In the plane transverse to the intermediate ray, the second s axis is rotated by phi from the first s axis. No other loss occurs between the interfaces.

Diagram shown in the original SVG worksheet.

  1. Find the power in the twice-reflected beam.

  2. Compare phi=60 degrees and phi=90 degrees with aligned incidence planes.

Question 8 - Solution

Two Brewster reflections select the overlap of their s directions

Diagram shown in the original SVG worksheet.

(a) Method and calculation

P1=R1P0/2,𝑬1∥ŝ1.P_1=R_1P_0/2,\quad\mathbf E_1\parallel\hat s_1.P2=R2P1|ŝ1⋅ŝ2|2=12R1R2P0cos⁡2ϕ.P_2=R_2P_1|\hat s_1\cdot\hat s_2|^2=\boxed{\tfrac12R_1R_2P_0\cos^2\phi}.

(b) Interpretation and further calculation

P2(60∘)/P2(0)=1/4.\boxed{P_2(60^\circ)/P_2(0)=1/4}.P2(90∘)=0.\boxed{P_2(90^\circ)=0}.

Physical check

At crossed s axes, the first reflected polarization is p for the second interface and therefore has zero second reflection.