Step 1: Understanding the Concept:
When unpolarized light passes through a polarizer, its intensity is halved.
When polarized light passes through an analyzer (another polaroid), the transmitted intensity is governed by Malus's Law, which depends on the cosine squared of the angle between their transmission axes.
If no light emerges from the second polaroid initially, they must be "crossed" (axes at $90^\circ$ to each other).
Step 2: Key Formula or Approach:
Initial transmission through first polaroid: $I_1 = \frac{I_0}{2}$
Malus's Law: $I_{\text{out}} = I_{\text{in}} \cos^2 \alpha$, where $\alpha$ is the angle between the polarization axis of the incident light and the transmission axis of the polaroid.
Trigonometric identity: $\sin(2\theta) = 2 \sin\theta \cos\theta \implies \sin^2\theta \cos^2\theta = \frac{\sin^2(2\theta)}{4}$.
Step 3: Detailed Explanation:
Let the three polaroids be $P_1$ (first), $P_3$ (middle, introduced later), and $P_2$ (last).
Initially, $P_1$ and $P_2$ are crossed, meaning the angle between their axes is $90^\circ$.
Unpolarized light of intensity $I_0$ passes through $P_1$. The intensity becomes:
\[ I_1 = \frac{I_0}{2} \]
The light is now linearly polarized along the axis of $P_1$.
Now, $P_3$ is placed between $P_1$ and $P_2$. Its axis makes an angle $\theta$ with $P_1$.
The light passing through $P_3$ will have intensity:
\[ I_3 = I_1 \cos^2 \theta = \frac{I_0}{2} \cos^2 \theta \]
The light emerging from $P_3$ is polarized along the axis of $P_3$.
Next, this light hits $P_2$. Since $P_1$ and $P_2$ are at $90^\circ$ to each other, the angle between $P_3$ and $P_2$ is $(90^\circ - \theta)$.
The intensity of light emerging from the last polaroid ($P_2$) is:
\[ I_2 = I_3 \cos^2 (90^\circ - \theta) \]
Since $\cos(90^\circ - \theta) = \sin \theta$:
\[ I_2 = I_3 \sin^2 \theta \]
Substitute the expression for $I_3$:
\[ I_2 = \left( \frac{I_0}{2} \cos^2 \theta \right) \sin^2 \theta \]
\[ I_2 = \frac{I_0}{2} (\sin \theta \cos \theta)^2 \]
Using the double angle formula $\sin(2\theta) = 2 \sin \theta \cos \theta$, we can write $\sin \theta \cos \theta = \frac{\sin(2\theta)}{2}$:
\[ I_2 = \frac{I_0}{2} \left( \frac{\sin(2\theta)}{2} \right)^2 \]
\[ I_2 = \frac{I_0}{2} \left( \frac{\sin^2(2\theta)}{4} \right) \]
\[ I_2 = \frac{I_0}{8} \sin^2(2\theta) \]
This matches the format $\frac{\text{I}_0}{8}(\sin 2\theta)^2$.
Step 4: Final Answer:
The intensity of light emerging from the last polaroid will be $\frac{\text{I}_0}{8}(\sin 2\theta)^2$.