Polarization by Reflection — Question 7

Section 33-9 · Physics pages 5544–5545

Question 7 A parallel Brewster window transmits p polarization through both faces A lossless plate of index square root of three is surrounded by air. Light enters at its Brewster angle. The faces are parallel. Count only the directly transmitted beam; internally reflected secondary beams are excluded. Use the Brewster s reflectance Rs = [(n2 −1)/(n2 + 1)]2. n = √ 3 Schematic; labels, not drawing lengths, specify the data. (a) Show that the second face is also at its reverse Brewster angle and find direct p and s power transmission fractions. (b) For unpolarized input, find direct total transmission and degree of linear polarization.
Show solutionHide solution
Question 7 - Solution A parallel Brewster window transmits p polarization through both faces n = √ 3 (a) Method and calculation θB,in = 60◦, θinside = 30◦= tan−1(1/n). Rs = 1/4 ⇒Tp,direct = 1, Ts,direct = (1 −Rs)2 = 9/16 . (b) Interpretation and further calculation Pdirect/P0 = (1 + 9/16)/2 = 25/32 . Pdirect = 1 −9/16 1 + 9/16 = 7/25 = 0.28 . Physical check The reverse Brewster angle matches the internal angle at the second parallel face. The s result here deliberately excludes later emerging beams.

Question 7

A parallel Brewster window transmits p polarization through both faces

A lossless plate of index square root of three is surrounded by air. Light enters at its Brewster angle. The faces are parallel. Count only the directly transmitted beam; internally reflected secondary beams are excluded. Use the Brewster s reflectance Rs=[(n2−1)/(n2+1)]2R_s=[(n^2-1)/(n^2+1)]^2.

Diagram shown in the original SVG worksheet.

  1. Show that the second face is also at its reverse Brewster angle and find direct p and s power transmission fractions.

  2. For unpolarized input, find direct total transmission and degree of linear polarization.

Question 7 - Solution

A parallel Brewster window transmits p polarization through both faces

Diagram shown in the original SVG worksheet.

(a) Method and calculation

θB,in=60∘,θinside=30∘=tan⁡−1(1/n).\theta_{B,\mathrm{in}}=60^\circ,\quad\theta_{\mathrm{inside}}=30^\circ=\tan^{-1}(1/n).Rs=1/4⇒Tp,direct=1,Ts,direct=(1−Rs)2=9/16.R_s=1/4\Rightarrow\boxed{T_{p,\mathrm{direct}}=1,\quad T_{s,\mathrm{direct}}=(1-R_s)^2=9/16}.

(b) Interpretation and further calculation

Pdirect/P0=(1+9/16)/2=25/32.\boxed{P_{\mathrm{direct}}/P_0=(1+9/16)/2=25/32}.𝒫direct=1−9/161+9/16=7/25=0.28.\boxed{\mathcal P_{\mathrm{direct}}=\frac{1-9/16}{1+9/16}=7/25=0.28}.

Physical check

The reverse Brewster angle matches the internal angle at the second parallel face. The s result here deliberately excludes later emerging beams.