Question

Unpolarized light passes through three polarizing filters. The first filter has its transmission axis parallel to the z direction, the second has its transmission axis at an angle of 30.0° from the z direction, and the third has its transmission axis at an angle of 60.0° from the z direction. If the light that emerges from the third filter has an intensity of 978.0 W/m2, what is the original intensity of the light? (Assume both angles are measured in the same direction from the +z axis.)

Answer 1

To solve this problem it is necessary to apply the concepts related to the Law of Malus for which it is understood that

`I=I_(0)\cos ^(2)\theta _(i)`

Where

`I_(0)` indicates the intensity of the light before passing through the polarizer,

I is the resulting intensity, and

`\theta _(i)}` indicates the angle between the axis of the analyzer and the polarization axis of the incident light.

The intensity of light after passing through a series of three polarisers would be given as

`I_3 = (I_0)/(2)cos^2\theta_1*cos^2\theta_2`

Here,

`\theta_1 = 30\°`

`\theta_2 = 90-60 = 30\°`

Therefore
`\theta_1 = \theta_2`

`I_3 = (I_0)/(2)cos^2(30)*cos^2(30)`

`I_0 = (2I_3)/(cos^430)`

`I_0 = (2*380)/(cos^430)`

`I_0 = 1.35*10^3W/m^2`

Therefore the original intensity of light was
`1.35*10^3W/m^2`