Unpolarized light of intensity I passes through an ideal polariser A. Another identical polariser B is placed behind A. The intensity of light beyond B is found to be $$\frac{I}{2}$$. Now another identical polariser C is placed between A and B. The intensity beyond B is now found to be $$\frac{I}{8}$$. The angle between polariser A and C is:

We begin with unpolarised light of initial intensity $$I$$ incident on the first ideal polariser A. For unpolarised light, the standard result is that:

$$I_{\text{after A}}=\dfrac{I}{2}$$

This happens because an ideal polariser transmits exactly one‐half of the incident unpolarised intensity.

Now the light emerging from A is plane-polarised. Let the transmission axis of the second polariser B make an angle $$\theta_{AB}$$ with that of A. Malus’s Law states:

$$I_{\text{after B}}=I_{\text{after A}}\cos^{2}\theta_{AB}$$

According to the question, the observed intensity after B (with only A and B present) is still $$\dfrac{I}{2}$$. Substituting $$I_{\text{after A}}=\dfrac{I}{2}$$ gives

$$\dfrac{I}{2}\cos^{2}\theta_{AB}=\dfrac{I}{2}$$

Dividing both sides by $$\dfrac{I}{2}$$ yields

$$\cos^{2}\theta_{AB}=1$$

Hence $$\theta_{AB}=0^{\circ}$$. So the transmission axes of A and B are parallel.

Next, a third identical polariser C is inserted between A and B. Let the angle between the axes of A and C be $$\theta$$. Because B is parallel to A, the angle between C and B is also $$\theta$$.

Step-wise calculation of the intensity now proceeds as follows:

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

2. After C: apply Malus’s Law once more, now with angle $$\theta$$ between A and C:

$$I_{2}=I_{1}\cos^{2}\theta=\dfrac{I}{2}\cos^{2}\theta$$

3. After B: the light from C meets B at the same angle $$\theta$$, so again using Malus’s Law,

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

The experiment tells us that this final intensity is $$\dfrac{I}{8}$$. Therefore,

$$\dfrac{I}{2}\cos^{4}\theta=\dfrac{I}{8}$$

Cancelling the common factor $$\dfrac{I}{2}$$ from both sides gives

$$\cos^{4}\theta=\dfrac{1}{4}$$

Taking the square root of both sides,

$$\cos^{2}\theta=\dfrac{1}{2}$$

Again taking the square root, and noting that the cosine of a small positive angle is positive,

$$\cos\theta=\dfrac{1}{\sqrt{2}}$$

This corresponds to

$$\theta=45^{\circ}$$

Hence, the correct answer is Option D.