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:
[tex]I = I_0 cos^2\theta[/tex]
Where,
[tex]I_0 =[/tex] Indicates the intensity of the light before passing through the polarizer
I = Resulting intensity
[tex]\theta[/tex]= 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
[tex]I_1= \frac{I_0}{2}[/tex]
In the case of the second polarizer the angle is directly 60 degrees therefore
[tex]I_2 = I_1 cos^2\theta[/tex]
[tex]I_2 = (\frac{I_0}{2} ) cos^2(60)[/tex]
[tex]I_2 = 0.125I_0[/tex]
In the case of the third polarizer, the angle is reflected on the perpendicular, therefore, its angle of index would be
[tex]\theta_3 = 90-60 = 30[/tex]
Then,
[tex]I_3 = I_2 cos^2\theta_3[/tex]
[tex]I_3 = 0.125I_0 cos^2 (30)[/tex]
[tex]I_3 = 0.09375I_0[/tex]
Then the intensity at the end of the polarized lenses will be equivalent to 0.09375 of the initial intensity.
