Problem 24.35

24.35. a)
We know that the intensity is given by
$I = I_{0} \cos^{2}\theta$
So, the angle $\theta$ is given by
$\theta = \cos^{-1}(\frac{I}{I_{0}})^{1/2}$
So to reduce the transmitted intensity by a factor of 3, we have
$\theta = \cos^{-1}(\frac{1}{3})^{1/2}$
$\theta = 54.7{}^{\circ}$

b)
Similarly, we have
$\theta = \cos^{-1}(\frac{1}{5})^{1/2}$
$\theta = 63.4{}^{\circ}$

c)
Similarly, we have again
$\theta = \cos^{-1}(\frac{1}{10})^{1/2}$
$\theta = 71.6{}^{\circ}$