Question

Unpolarizedlight of intensity I_0 is incident on three polarizingfilters. The axis of the first is vertical, that of the secondis 60.0 ^\circ from vertical, andthat of the third is horizontal.

What light intensityemerges from the third filter? (I0)

Answer 1

To solve this problem it is necessary to apply the concepts related to the law of Malus which describe the intensity of light passing through a polarizer. Mathematically this law can be described as:

`I = I_0 cos^2\theta`

Where,

`I_0 =` Indicates the intensity of the light before passing through the polarizer

I = Resulting intensity

`\theta`= Indicates the angle between the axis of the analyzer and the polarization axis of the incident light

From the law of Malus when the light passes at a vertical angle through the first polarizer its intensity is reduced by half therefore

`I_1= (I_0)/(2)`

In the case of the second polarizer the angle is directly 60 degrees therefore

`I_2 = I_1 cos^2\theta`

`I_2 = ((I_0)/(2) ) cos^2(60)`

`I_2 = 0.125I_0`

In the case of the third polarizer, the angle is reflected on the perpendicular, therefore, its angle of index would be

`\theta_3 = 90-60 = 30`

Then,

`I_3 = I_2 cos^2\theta_3`

`I_3 = 0.125I_0 cos^2 (30)`

`I_3 = 0.09375I_0`

Then the intensity at the end of the polarized lenses will be equivalent to 0.09375 of the initial intensity.