# Light of original intensity

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

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?

### 1 Answer

- 天同Lv 77 years agoFavorite Answer
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