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Step-by-Step Solution
Step 1: Intensity after the first polarizer A
When unpolarized light of intensity $I$ passes through an ideal polarizer, the transmitted intensity becomes
$$
I_1 = \frac{I}{2}.
$$
Step 2: Orientation of polarizer B relative to A (without polarizer C)
The problem states that with only polarizers A and B, the intensity beyond B is still $I/2$.
By Malus's law, the transmitted intensity through the second polarizer would be
$$
I_2 = I_1 \cos^2 \phi,
$$
where $\phi$ is the angle between the axes of A and B. For $I_2$ to remain equal to $I/2$,
we must have
$$
\frac{I}{2} = \frac{I}{2} \cos^2 \phi \quad \Longrightarrow \quad \cos^2 \phi = 1,
$$
implying $\phi = 0^\circ$. Thus, polarizer B is aligned parallel to A.
Step 3: Introduction of the third polarizer C
Now we insert another identical polarizer C between A and B. Let the angle between A and C be $\theta$.
Because B is parallel to A, the angle between C and B is also $\theta$.
Step 4: Intensity after polarizer C
With the intensity after A being $I_1 = \frac{I}{2}$, the intensity after C (using Malus's law) is
$$
I_C = I_1 \cos^2 \theta = \frac{I}{2} \cos^2 \theta.
$$
Step 5: Intensity after polarizer B
The light after C passes through B, which is at the same angle $\theta$ relative to C.
Hence, applying Malus's law once more:
$$
I_{\text{final}} = I_C \cos^2 \theta = \left(\frac{I}{2} \cos^2 \theta\right)\cos^2 \theta
= \frac{I}{2} \cos^4 \theta.
$$
Step 6: Using the given final intensity
The problem states that the final intensity beyond B (when C is present) is $\frac{I}{8}$. So,
$$
\frac{I}{2}\cos^4 \theta = \frac{I}{8}.
$$
Dividing through by $I/2$,
$$
\cos^4 \theta = \frac{\frac{I}{8}}{\frac{I}{2}} = \frac{1}{4}.
$$
Taking the square root,
$$
\cos^2 \theta = \frac{1}{2}
\quad \Longrightarrow \quad
\cos \theta = \frac{1}{\sqrt{2}}
\quad \Longrightarrow \quad
\theta = 45^\circ.
$$
Final Answer
The angle between polarizer A and C is $45^\circ$.
