Polarization by Reflection — Question 6

Section 33-9 · Physics pages 5542–5543

Question 6 Analyzer misalignment leaks the dominant polarization of reflected glare An unpolarized beam reflects from a horizontal dielectric surface in a vertical incidence plane. At the chosen angle its power reflectances are Rs=0.16 and Rp=0.01. An ideal analyzer in the reflected beam has its axis at angle phi from the p direction toward s. Normalize all powers to incident P0. s p analyzer view along reflected beam Schematic; labels, not drawing lengths, specify the data. (a) Derive the transmitted analyzer power and find the best suppression orientation. (b) Find the transmitted fraction for phi=10 degrees and compare it with the optimum.
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Question 6 - Solution Analyzer misalignment leaks the dominant polarization of reflected glare s p analyzer view along reflected beam (a) Method and calculation Pr,p = RpP0/2, Pr,s = RsP0/2. PA/P0 = 1 2(Rp cos2 ϕ + Rs sin2 ϕ) . (b) Interpretation and further calculation Rp < Rs ⇒ϕ = 0, PA,min/P0 = Rp/2 = 0.005 . PA(10◦)/P0 = 0.00726153, PA(10◦)/PA,min = 1.45231 . Physical check The s electric field is horizontal and perpendicular to the vertical incidence plane; the analyzer must transmit the transverse p direction to suppress it.

Question 6

Analyzer misalignment leaks the dominant polarization of reflected glare

An unpolarized beam reflects from a horizontal dielectric surface in a vertical incidence plane. At the chosen angle its power reflectances are Rs=0.16 and Rp=0.01. An ideal analyzer in the reflected beam has its axis at angle phi from the p direction toward s. Normalize all powers to incident P0.

Diagram shown in the original SVG worksheet.

  1. Derive the transmitted analyzer power and find the best suppression orientation.

  2. Find the transmitted fraction for phi=10 degrees and compare it with the optimum.

Question 6 - Solution

Analyzer misalignment leaks the dominant polarization of reflected glare

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Pr,p=RpP0/2,Pr,s=RsP0/2.P_{r,p}=R_pP_0/2,\quad P_{r,s}=R_sP_0/2.PA/P0=12(Rpcos⁡2ϕ+Rssin⁡2ϕ).\boxed{P_A/P_0=\tfrac12(R_p\cos^2\phi+R_s\sin^2\phi)}.

(b) Interpretation and further calculation

Rp<Rs⇒ϕ=0,PA,min/P0=Rp/2=0.005.R_p<R_s\Rightarrow\boxed{\phi=0,\quad P_{A,\min}/P_0=R_p/2=0.005}.PA(10∘)/P0=0.00726153,PA(10∘)/PA,min=1.45231.\boxed{P_A(10^\circ)/P_0=0.00726153,\quad P_A(10^\circ)/P_{A,\min}=1.45231}.

Physical check

The s electric field is horizontal and perpendicular to the vertical incidence plane; the analyzer must transmit the transverse p direction to suppress it.