Sample calculation

Malus' law: I1 = Io cos2(q)

where q is the angle between the polarization direction of the light and the transmission axis of the polarizer.

Unpolarized light with an intensity of Io = 16 W/m2 is incident on a pair of polarizers. The first polarizer has its transmission axis aligned at 50o from the vertical. The second polarizer has its transmission axis aligned at 20o from the vertical.

What is the intensity of the light when it emerges from the first polarizer?

  1. 4 W/m2
  2. 16 cos250o
  3. 8 W/m2
  4. 12 W/m2
















The light is unpolarized, so I1 = 1/2 Io = 8 W/m2.

What is the intensity of the light when it emerges from the second polarizer?

  1. I2 = 4 W/m2
  2. I2 = 8 cos220o
  3. I2 = 6 W/m2
















When it emerges from the first polarizer the light is linearly polarized at 50o. The angle between this and the transmission axis of the second polarizer is 30o. Therefore:

I2 = I1 cos230o = 8 * 3/4 = 6 W/m2

Now a linearly polarized beam of light, with an intensity of 16 W/m2, is incident on the same pair of polarizers. The polarization direction of the incident light is 20o from the vertical.

What is the intensity of the light emerging from the first polarizer now?

What is the intensity of the light emerging from the second polarizer now?











Going through each polarizer the intensity is reduced by a factor of 3/4. After emerging from the first polarizer the intensity is 16 * 3/4 = 12 W/m2, and is reduced to 12 * 3/4 = 9 W/m2 after going through the second.

One final note: to rotate light from one polarization direction to another you reduce the loss of intensity by using more polarizers.

Let's say we want to rotate the polarization direction by 90o:

N, number of polarizers Angle between successive polarizers I1/Io
1 90o 0
2 45o 1/2 x 1/2 = 1/4
3 30o 3/4 x 3/4 x 3/4 = 27/64
N 90/N [ cos2(90o/N)]N