To solve this problem it is necessary to apply the concepts given by Malus regarding the Intensity of light.
From the law of Malus intensity can be defined as
[tex]I' = \frac{I_0}{2} cos^2 \theta[/tex]
Where
[tex]\theta =[/tex]Angle From vertical of the axis of the polarizing filter
[tex]I_0 =[/tex] Intensity of the unpolarized light
The expression for the intensity of the light after passing through the first filter is given by
[tex]I = \frac{I_0}{2}[/tex]
Replacing we have that
[tex]I = \frac{370}{2}[/tex]
[tex]I = 185W/m^2[/tex]
Re-arrange the equation,
[tex]I'= \frac{I_0}{2}cos^2\theta[/tex]
Re-arrange to find \theta
[tex]cos^2\theta = \frac{2I'}{I_0}[/tex]
[tex]cos^2\theta = \frac{2*138}{370}[/tex]
[tex]\theta = cos^{-1}(\sqrt{\frac{2*138}{370}})[/tex]
[tex]\theta = 0.5282rad[/tex]
[tex]\theta = 30.27\°[/tex]
The value of the angle from vertical of the axis of the second polarizing filter is equal to 30.2°
