Answer:
θ = 66.90°
Explanation:
we know that
[tex]I= \frac{I_0}{2}cos^2\theta[/tex]
I= intensity of polarized light =1
I_o= intensity of unpolarized light = 13
putting vales we get
[tex]1= \frac{13}{2}cos^2\theta[/tex]
⇒[tex]\theta=cos^{-1} \sqrt{\frac{1}{6.5} }[/tex]
therefore θ = 66.90°
