1. Initial arrangement of $P_1$ and $P_2$

Since the transmitted intensity through $P_1$ and $P_2$ is zero, the two polarisers must be crossed.

So, the angle between their transmission axes is $90^\circ = \frac{\pi}{2}.$

2. Insert third polariser $P_3$

Let the angle between $P_1$ and $P_3$ be $\theta$. Then, since $P_1$ and $P_2$ are at $\frac{\pi}{2}$ to each other, the angle between $P_3$ and $P_2$ is $\frac{\pi}{2} - \theta.$

If unpolarised light of intensity $I_0$ falls on $P_1$, then after $P_1$: $I_1 = \frac{I_0}{2}.$

After passing through $P_3$, by Malus' law: $I_2 = I_1 \cos^2 \theta = \frac{I_0}{2} \cos^2 \theta.$

After passing through $P_2$: $I = I_2 \cos^2\left(\frac{\pi}{2}-\theta\right).$

Using $\cos\left(\frac{\pi}{2}-\theta\right)=\sin\theta,$ we get $I = \frac{I_0}{2} \cos^2\theta\sin^2\theta.$

3. Maximise transmitted intensity

We need to maximise $I \propto \sin^2\theta\cos^2\theta.$

Now, $\sin^2\theta\cos^2\theta = \frac{1}{4}\sin^2 2\theta.$

This is maximum when $\sin^2 2\theta = 1,$ so $2\theta = \frac{\pi}{2} \quad \Rightarrow \quad \theta = \frac{\pi}{4}.$

Thus,

• angle between $P_1$ and $P_3$ is $\frac{\pi}{4}$,
• angle between $P_2$ and $P_3$ is also $\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}.$

4. Correct option

Therefore, the required angle between $P_2$ and $P_3$ is $\boxed{\frac{\pi}{4}}.$

So the correct option is C.