Unpolarized light of intensity I is incident on a system of two polarizers, A followed by B. The intensity of emergent light is $$\frac{I}{2}$$. If a third polarizer C is placed between A and B, the intensity of emergent light is reduced to $$\frac{I}{3}$$. The angle between the polarizers A and C is $$\theta$$, then:

We start with unpolarized light of intensity $$I$$ incident on polarizer A. The basic result for an unpolarized beam passing through a single polarizer is Malus’ law in its first consequence:

Unpolarized → first polarizer gives $$I_{\text{after A}} \;=\;\dfrac{I}{2}.$$

Now this light meets polarizer B. Let the angle between the transmission axes of A and B be $$\phi$$. Malus’ law in general form states:

For light of intensity $$I_0$$ incident on a polarizer whose axis makes an angle $$\alpha$$ with the light’s existing plane of polarization, the transmitted intensity is $$I = I_0\cos^{2}\alpha.$$

Applying the law to polarizer B we have

$$I_{\text{after B}} = \left(\dfrac{I}{2}\right)\cos^{2}\phi.$$

The problem tells us that the emergent intensity after A and B alone is exactly $$\dfrac{I}{2}$$. Equating, we obtain

$$\left(\dfrac{I}{2}\right)\cos^{2}\phi = \dfrac{I}{2}\;\;\Longrightarrow\;\;\cos^{2}\phi = 1,$$

so $$\phi = 0^{\circ}.$$ Hence polarizers A and B are parallel.

Next, a third polarizer C is inserted between A and B, making an angle $$\theta$$ with A. Because A and B are parallel, the angle between C and B is also $$\theta$$.

Step by step through the three polarizers:

1. After A: $$I_{1}=\dfrac{I}{2}.$$

2. After C: apply Malus’ law with angle $$\theta$$, giving $$I_{2}=I_{1}\cos^{2}\theta =\left(\dfrac{I}{2}\right)\cos^{2}\theta.$$

3. After B: again use Malus’ law with the same angle $$\theta$$ (since C and B differ by $$\theta$$), obtaining $$I_{3}=I_{2}\cos^{2}\theta =\left(\dfrac{I}{2}\right)\cos^{2}\theta\cos^{2}\theta =\left(\dfrac{I}{2}\right)\cos^{4}\theta.$$

According to the question this final intensity equals $$\dfrac{I}{3}$$. Therefore

$$\left(\dfrac{I}{2}\right)\cos^{4}\theta = \dfrac{I}{3} \;\;\Longrightarrow\;\; \cos^{4}\theta = \dfrac{2}{3}.$$

Taking the fourth root,

$$\cos\theta = \left(\dfrac{2}{3}\right)^{\frac14}.$$

Comparing with the given options, this matches option A.

Hence, the correct answer is Option A.