p34_039

p34_039 - Consequently the transmitted intensity is I = fI...

This preview shows page 1. Sign up to view the full content.

39. Let I 0 be the intensity of the incident beam and f be the fraction that is polarized. Thus, the intensity of the polarized portion is fI 0 . After transmission, this portion contributes fI 0 cos 2 θ to the intensity of the transmitted beam. Here θ is the angle between the direction of polarization of the radiation and the polarizing direction of the Flter. The intensity of the unpolarized portion of the incident beam is (1 f ) I 0 and after transmission, this portion contributes (1 f ) I 0 / 2 to the transmitted intensity.
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Consequently, the transmitted intensity is I = fI cos 2 θ + 1 2 (1 − f ) I . As the Flter is rotated, cos 2 θ varies from a minimum of 0 to a maximum of 1, so the transmitted intensity varies from a minimum of I min = 1 2 (1 − f ) I to a maximum of I max = fI + 1 2 (1 − f ) I = 1 2 (1 + f ) I . The ratio of I max to I min is I max I min = 1 + f 1 − f . Setting the ratio equal to 5 . 0 and solving for f , we get f = 0 . 67....
View Full Document

This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

Ask a homework question - tutors are online