To solve this problem it is necessary to apply the concepts related to Malus' law. Malus' law indicates that the intensity of a linearly polarized ray of light that passes through a perfect analyzer with a vertical optical axis is equivalent to:
[tex]I = I_0 cos^2(\theta)[/tex]
[tex]I_0 =[/tex] Indicates the intensity of the light before passing through the Polarizer,
I = The resulting intensity, and
[tex]\theta[/tex] = Indicates the angle between the axis of the analyzer and the polarization axis of the incident light.
There is 3 polarizer, then
For the exit of the first polarizer we have that the intensity is,
[tex]I_2 = I_0cos^2(45)[/tex]
For the third polarizer then we have,
[tex]I_3 = I_2 cos^{2}(45)[/tex]
Replacing with the first equation,
[tex]I_3 = I_0cos^2(45)cos^{2}(45)[/tex]
[tex]I_3 = I_0 (\frac{1}{2})(\frac{1}{2})[/tex]
[tex]I_3 = I_0 \frac{1}{4}[/tex]
Therefore the transmitted intensity now is [tex]\frac{1}{4}[/tex] of the initial intensity.
