Answer
$0.67~~$ of the intensity of the original beam is associated with the beam’s polarized light.
Work Step by Step
Let $I_u$ be the intensity of unpolarized light.
Let $I_p$ be the intensity of polarized light.
The intensity will be at the lowest level when all the polarized light is eliminated. We can write an expression for this intensity:
$I_{min} = \frac{1}{2}I_u$
The intensity will be at the highest level when all the polarized light is transmitted. We can write an expression for this intensity:
$I_{max} = \frac{1}{2}I_u+I_p$
We can find an expression for $I_p$:
$I_{max} = 5~I_{min}$
$\frac{1}{2}I_u+I_p = 5~(\frac{1}{2}I_u)$
$I_p = 2~I_u$
We can find the fraction of the intensity of the original beam that is associated with the beam’s polarized light:
$\frac{I_p}{I_u+I_p} = \frac{2~I_u}{I_u+2~I_u} = 0.67$
$0.67~~$ of the intensity of the original beam is associated with the beam’s polarized light.
