Final+sol - 1. You have available a source of unpolarized...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
1. You have available a source of unpolarized light (intensity I o ) and a number of "perfect" linear polarizers (each of them transmitting without losses all the light polarized parallel to its "transmission axis" and blocking totally all light polarized perpendicular to this axis.) (A) Using a sketch of the light polarization and the polarizer orientation , derive and show the simple relation for the light intensity I( θ ) relative to I o transmitted by two polarizers as a function of the angle θ between their transmission axes. I 0 I 1 I 2 ! I 1 = 1 2 I 0 , and I 2 = 1 2 I 0 cos 2 θ
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
In considering the electric field components, we can obtain the relations between intensities before and after each polarizer. E ! " E E ! E | | I o = 1 2 ε o c {(E ) 2 + (E ) 2 } I 1 = 1 2 ε o c (E ) 2 = 1 2 I o I 2 = 1 2 ε o c (E cos θ ) 2 = 1 2 I o cos 2 θ (B) Imagine two polarizers are set with their transmission axes "crossed" at 0° and 90°. What is the transmitted light intensity? For θ = 90°, I 2 = 0.
Background image of page 2
(C) Five additional polarizers are placed between the two "crossed" polarizers with their transmission axes set at 60° , 45° , 75° , 15° , and 30° in that order, with the 60° polarizer adjacent to the 0° polarizer and so on (all angles are measured from the axis of the first 0° polarizer). What is the intensity relative to the incident one, transmitted by these seven polarizers? The orientation angles of a series of polarizers are θ o = 0°, θ 1 = 60°, θ 2 = 45°, θ 3 = 75°, θ 4 = 15°, θ 5 = 30°, and θ 6 = 90°. I 1 = 1 2 I o I 2 = I 1 cos 2 ( θ 1 θ o ) = 1 2 I o cos 2 ( 60 – 0) I 3 = I 2 cos 2 ( θ 2 θ 1 ) = 1 2 I o cos 2 (60) cos 2 (15) I 4 = I 3 cos 2 ( θ 3 θ 2 ) = 1 2 I o cos 2 (60) cos 2 (15) cos 2 (30) I 5 = I 4 cos 2 ( θ 4 θ 3 ) = 1 2 I o cos 2 (60) cos 2 (15) cos 2 (30) cos 2 (60) I 6 = I 5 cos 2 ( θ 5 θ
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 08/06/2008 for the course PHYS 352 taught by Professor Dierolf during the Fall '04 term at Lehigh University .

Page1 / 11

Final+sol - 1. You have available a source of unpolarized...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online