Answer:
The angle is [tex]62.88^{\circ}[/tex]
Solution:
As per the question:
Intensity of the linearly polarized light, I = [tex]900\ W/m^{2}[/tex]
Intensity transmitted, I' = [tex]187\ W/m^{2}[/tex]
Now,
To calculate the angle between the polarization direction of the eye and that of the incident light:
[tex]I' = Icos^{2}\theta [/tex]
Thus Rearranging the eqn for the angle and using the suitable values in the above eqn:
[tex]cos^{2}\theta = \frac{187}{900} = 0.2078[/tex]
[tex]cos\theta = \frac{187}{900} = 0.4558[/tex]
[tex]\theta = cos^{- 1}(0.4558) = 62.88^{\circ}[/tex]