Polarization — Question 10

Section 33-6 · Physics pages 5490–5491

Question 10 Removing a polarizer can reduce the light reaching a fixed final axis An x-polarized beam of intensity I0 encounters ideal polarizers at 30 degrees, 60 degrees, and 90 degrees, in that order. The initial polarization and final axis remain fixed. 30◦ 60◦ 90◦ Schematic; labels, not drawing lengths, specify the data. (a) Find the final transmission with all three polarizers and with the middle 60-degree element removed. (b) Find the transmission after swapping the 30-degree and 60-degree elements, and quantify the order dependence.
Show solutionHide solution
Question 10 - Solution Removing a polarizer can reduce the light reaching a fixed final axis 30◦ 60◦ 90◦ (a) Method and calculation I30,60,90 = I0(cos2 30◦)3 = 27I0/64 . I30,90 = I0 cos2 30◦cos2 60◦= 3I0/16 . (b) Interpretation and further calculation I60,30,90 = I0 cos2 60◦cos2 30◦cos2 60◦= 3I0/64 . I30,60,90/I60,30,90 = 9 . Physical check Each transmission depends on the immediately preceding polarization, so changing the order changes the product of projections.

Question 10

Removing a polarizer can reduce the light reaching a fixed final axis

An x-polarized beam of intensity I0 encounters ideal polarizers at 30 degrees, 60 degrees, and 90 degrees, in that order. The initial polarization and final axis remain fixed.

Diagram shown in the original SVG worksheet.

  1. Find the final transmission with all three polarizers and with the middle 60-degree element removed.

  2. Find the transmission after swapping the 30-degree and 60-degree elements, and quantify the order dependence.

Question 10 - Solution

Removing a polarizer can reduce the light reaching a fixed final axis

Diagram shown in the original SVG worksheet.

(a) Method and calculation

I30,60,90=I0(cos⁡230∘)3=27I0/64.I_{30,60,90}=I_0(\cos^230^\circ)^3=\boxed{27I_0/64}.I30,90=I0cos⁡230∘cos⁡260∘=3I0/16.I_{30,90}=I_0\cos^230^\circ\cos^260^\circ=\boxed{3I_0/16}.

(b) Interpretation and further calculation

I60,30,90=I0cos⁡260∘cos⁡230∘cos⁡260∘=3I0/64.I_{60,30,90}=I_0\cos^260^\circ\cos^230^\circ\cos^260^\circ=\boxed{3I_0/64}.I30,60,90/I60,30,90=9.\boxed{I_{30,60,90}/I_{60,30,90}=9}.

Physical check

Each transmission depends on the immediately preceding polarization, so changing the order changes the product of projections.