Three polaroids P_1 , P_2 , and P_3 are placed coaxially. The pass...

The intensity of unpolarized light after passing through the first polaroid $P_1$ becomes half:
$I_1 = \dfrac{I_0}{2}$

Let the angle between the pass axes of $P_1$ and $P_2$ be $\theta$. According to Malus's Law, the intensity after $P_2$ is:
$I_2 = I_1 \cos^2\theta = \dfrac{I_0}{2} \cos^2\theta$

Since $P_1$ and $P_3$ are crossed, the angle between their pass axes is $90^{\circ}$. Thus, the angle between the pass axes of $P_2$ and $P_3$ is $(90^{\circ} - \theta)$. The intensity emerging from $P_3$ is:
$I_3 = I_2 \cos^2(90^{\circ} - \theta) = \left(\dfrac{I_0}{2} \cos^2\theta\right) \sin^2\theta$

Using the trigonometric identity $2\sin\theta \cos\theta = \sin(2\theta)$, we can rewrite this as:
$I_3 = \dfrac{I_0}{8} (2\sin\theta \cos\theta)^2 = \dfrac{I_0}{8} \sin^2(2\theta)$

We are given that the final emergent intensity is $\dfrac{I_0}{32}$. Equating the two expressions:
$\dfrac{I_0}{8} \sin^2(2\theta) = \dfrac{I_0}{32}$

$\sin^2(2\theta) = \dfrac{1}{4}$

$\sin(2\theta) = \dfrac{1}{2}$

$2\theta = 30^{\circ} \Rightarrow \theta = 15^{\circ}$

Answer: $15^{\circ}$