An unpolarized light beam of initial intensity I_0 travels along an...

When unpolarized light of intensity $I_0$ passes through the first polarizer $P_1$, the intensity of the transmitted light becomes half:

$I_1 = \dfrac{I_0}{2}$

This light is vertically polarized. It then passes through the second polarizer $P_2$, whose transmission axis is at an angle of $60^{\circ}$ with respect to the vertical. According to Malus's law, the intensity of light transmitted through $P_2$ is:

$I_2 = I_1 \cos^2(60^{\circ}) = \dfrac{I_0}{2} \left(\dfrac{1}{2}\right)^2 = \dfrac{I_0}{8}$

The light emerging from $P_2$ is polarized at an angle of $60^{\circ}$ to the vertical. The third polarizer $P_3$ has a horizontal transmission axis, which is at $90^{\circ}$ to the vertical. The angle between the polarization direction of the light incident on $P_3$ and its transmission axis is:

$\theta = 90^{\circ} - 60^{\circ} = 30^{\circ}$

Applying Malus's law again for $P_3$, the final transmitted intensity $I'$ is:

$I' = I_2 \cos^2(30^{\circ}) = \dfrac{I_0}{8} \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{I_0}{8} \times \dfrac{3}{4} = \dfrac{3I_0}{32}$

Answer: $\dfrac{3I_0}{32}$