Polarization — Question 3

Section 33-6 · Physics pages 5476–5477

Question 3 Many small polarization rotations transmit more than one large step A beam already polarized along x has intensity I0. It passes through N ideal analyzers with successive axes rotated by equal increments pi/(2N), ending along y. Neglect reflection and all loss except ideal projection. successive transmission axes Schematic; labels, not drawing lengths, specify the data. (a) Derive the transmitted intensity and evaluate it for N=2 and N=3. (b) Find its limiting behavior for large N using log cos(epsilon) approximately equals minus epsilon squared divided by two.
Show solutionHide solution
Question 3 - Solution Many small polarization rotations transmit more than one large step successive transmission axes (a) Method and calculation IN/I0 = [ cos2 ( π 2N )]N . I2/I0 = 1/4, I3/I0 = (3/4)3 = 27/64 . (b) Interpretation and further calculation ln(IN/I0) = 2N ln cos(π/2N) ≃−π2/(4N). IN/I0 ≃e−π2/(4N) −→1 . Physical check The initial beam is already polarized, so no initial one-half factor applies; the loss per small step vanishes quadratically.

Question 3

Many small polarization rotations transmit more than one large step

A beam already polarized along x has intensity I0. It passes through N ideal analyzers with successive axes rotated by equal increments pi/(2N), ending along y. Neglect reflection and all loss except ideal projection.

Diagram shown in the original SVG worksheet.

  1. Derive the transmitted intensity and evaluate it for N=2 and N=3.

  2. Find its limiting behavior for large N using log cos(epsilon) approximately equals minus epsilon squared divided by two.

Question 3 - Solution

Many small polarization rotations transmit more than one large step

Diagram shown in the original SVG worksheet.

(a) Method and calculation

IN/I0=[cos⁡2(π2N)]N.\boxed{I_N/I_0=\left[\cos^2\!\left(\frac{\pi}{2N}\right)\right]^N}.I2/I0=1/4,I3/I0=(3/4)3=27/64.\boxed{I_2/I_0=1/4,\quad I_3/I_0=(3/4)^3=27/64}.

(b) Interpretation and further calculation

ln⁡(IN/I0)=2Nln⁡cos⁡(π/2N)≃−π2/(4N).\ln(I_N/I_0)=2N\ln\cos(\pi/2N)\simeq-\pi^2/(4N).IN/I0≃e−π2/(4N)→1.\boxed{I_N/I_0\simeq e^{-\pi^2/(4N)}\longrightarrow1}.

Physical check

The initial beam is already polarized, so no initial one-half factor applies; the loss per small step vanishes quadratically.