Three polaroids P_1 , P_2 , and P_3 are placed in the path of a...

When unpolarized light of intensity $I_0$ passes through $P_1$, the intensity becomes $I_1 = \dfrac{I_0}{2}$.

The light then passes through $P_2$, whose pass axis is at an angle $\theta$ with $P_1$. By Malus's law, the intensity after $P_2$ is $I_2 = I_1 \cos^2\theta = \dfrac{I_0}{2} \cos^2\theta$.

The pass axis of $P_3$ is at $90^{\circ}$ to $P_1$, so the angle between $P_2$ and $P_3$ is $(90^{\circ} - \theta)$. The final emergent intensity from $P_3$ is:
$I_3 = I_2 \cos^2(90^{\circ} - \theta) = \dfrac{I_0}{2} \cos^2\theta \sin^2\theta$
Using $\sin(2\theta) = 2\sin\theta\cos\theta$, we get $I_3 = \dfrac{I_0}{8} \sin^2(2\theta)$.

For (A) $\theta = 15^{\circ}$: $I_3 = \dfrac{I_0}{8} \sin^2(30^{\circ}) = \dfrac{I_0}{8} \times \dfrac{1}{4} = \dfrac{I_0}{32}$.
For (B) $\theta = 30^{\circ}$: $I_3 = \dfrac{I_0}{8} \sin^2(60^{\circ}) = \dfrac{I_0}{8} \times \dfrac{3}{4} = \dfrac{3I_0}{32}$.
For (C) $\theta = 45^{\circ}$: $I_3 = \dfrac{I_0}{8} \sin^2(90^{\circ}) = \dfrac{I_0}{8} \times 1 = \dfrac{I_0}{8}$.
For (D) $\theta = 90^{\circ}$: $I_3 = \dfrac{I_0}{8} \sin^2(180^{\circ}) = 0$.

Therefore, the correct matching is (A)-(II), (B)-(III), (C)-(IV), (D)-(I).

Answer: (A)-(II), (B)-(III), (C)-(IV), (D)-(I)