Show that when unpolarized light passes through a polarizer,the intensity of the emerging light is half that of the intensity of the incident light.

If the intensity of unpolarized light incident on a polarizer is $I_{0}$,the electric field vector of the light oscillates in all possible directions in the plane perpendicular to the direction of propagation.
According to Malus's law,the intensity of light transmitted through a polarizer is $I = I_{0} \cos^{2} \theta$,where $\theta$ is the angle between the electric field vector and the transmission axis of the polarizer.
Since unpolarized light consists of a random distribution of electric field vectors,we must take the average value of $\cos^{2} \theta$ over all possible angles from $0$ to $2\pi$.
$\langle I \rangle = I_{0} \langle \cos^{2} \theta \rangle = I_{0} \frac{1}{2\pi} \int_{0}^{2\pi} \cos^{2} \theta \, d\theta$
Using the identity $\cos^{2} \theta = \frac{1 + \cos 2\theta}{2}$,we get:
$\langle I \rangle = \frac{I_{0}}{2\pi} \int_{0}^{2\pi} \frac{1 + \cos 2\theta}{2} \, d\theta = \frac{I_{0}}{4\pi} \left[ \theta + \frac{\sin 2\theta}{2} \right]_{0}^{2\pi}$
$\langle I \rangle = \frac{I_{0}}{4\pi} \left[ (2\pi + 0) - (0 + 0) \right] = \frac{I_{0}}{4\pi} \times 2\pi = \frac{I_{0}}{2}$
Thus,the intensity of the emerging light is exactly half of the incident intensity.