Light of original intensity

Anonymous

2012-12-07 8:33 am

http://upload.lsforum.net/users/public/e27788YF-33-28f48.jpg

Light of original intensity passes through two ideal polarizing filters
having their polarizing axes oriented as shown in the figure
(Figure 1) . You want to adjust the angle so that the intensity at
point P is equal to ( original intensity / 16.0 ).

1. If the original light is unpolarized, what should the angle be?

2. If the original light is linearly polarized in the same direction
as the polarizing axis of the first polarizer the light reaches,
what should the angle be?

回答 (1)

天同

2012-12-07 6:45 pm

✔ 最佳答案

1. Let Io be the original intensity of unpolarized light.
Hence, intensity of light after passing throught the 1st filter = (Io/2)
After passing through the 2nd filter, light intensity = (Io/2).cos^2(a)
[where a is the angle between the polarizing axes of the two filters]
Thus, Io/16 = (Io/2).cos^2(a)
cos^2(a) = 1/8
a = 69.3 degrees

2. Intensity of light after passing throught the 1st filter = Io
After passing through the 2nd filter, light intensity = (Io).cos^2(a')
[where a' is the angle between the polarizing axes of the two filters]
Thus, Io/16 = (Io).cos^2(a)
cos^2(a) = 1/16
a = 75.5 degrees

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